What is a matrix in Class 12 Mathematics?

A matrix is an ordered rectangular array of numbers or functions arranged in rows and columns. The individual numbers or functions are called elements or entries.

What is the order of a matrix?

The order of a matrix is written as m × n, where m represents the number of rows and n represents the number of columns. An m × n matrix contains mn elements.

What are the different types of matrices in Class 12?

Important types include row matrix, column matrix, rectangular matrix, square matrix, zero matrix, diagonal matrix, scalar matrix and identity matrix. Symmetric and skew-symmetric matrices are also important.

When are two matrices equal?

Two matrices are equal when they have the same order and every corresponding element is equal. Thus, A = B if aij = bij for all valid i and j.

When is matrix multiplication possible?

If A is of order m × n and B is of order n × p, then AB is defined and has order m × p. The number of columns of the first matrix must equal the number of rows of the second matrix.

Is matrix multiplication commutative?

Generally, matrix multiplication is not commutative. In general, AB ? BA. Moreover, one product may be defined while the other may not be defined.

What is the transpose of a matrix?

The transpose of a matrix is obtained by interchanging its rows and columns. If A = [aij]m × n, then A' = [aji]n × m.

What is a symmetric matrix?

A square matrix A is symmetric if A' = A. Equivalently, its corresponding elements satisfy aij = aji for all i and j.

What is a skew-symmetric matrix?

A square matrix A is skew-symmetric if A' = -A. Consequently, every diagonal element of a skew-symmetric matrix is zero.

What is an invertible matrix and its inverse?

A square matrix A is invertible if there exists a matrix A?¹ of the same order such that AA?¹ = A?¹A = I. The inverse, when it exists, is unique.

Class 12 Matrices Notes | NCERT Maths Chapter 3

Chapter 1 · CBSE · Class XII

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Matrices

Matrices NCERT Class 12 Class 12 Mathematics Mathematics Notes CBSE Class 12 NCERT Notes Types of Matrices Matrix Operations Transpose of Matrix Symmetric Matrix Skew Symmetric Matrix Invertible Matrix Inverse Matrix Matrix Theorems JEE Main JEE Advanced CUET Board Exam Competitive Exams

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Table of Contents

17 sections

1 🗺️ Overview ›
2 📘 Definition of a Matrix ›
3 🗂️ Rows and Columns of a Matrix ›
4 📌 Order of a Matrix ›
5 📄 Elements or Entries of a Matrix ›
6 📎 How to Locate an Element of a Matrix ›
7 📜 Number of Elements in a Matrix ›
8 ✏️ Examples: Basic Concepts ›
9 ℹ️ Matrix as a Structured Mathematical Object ›
10 💡 Conceptual Connection with Determinants ›
11 🌟 Significance ›
12 🌟 Why Matrices Are Important for Class 12 Board Examinations ›
13 ⚡ Exam Tip ›
14 ❌ Common Mistakes ›
15 📋 CBSE Case Study Based Question ›
16 🔢 Quick Formula and Concept Sheet ›
17 ⚡ One-Minute Revision ›

🗺️ Overview

Matrices are one of the most important algebraic tools in Class 12 Mathematics. They provide a compact way to arrange numbers, variables, functions and algebraic expressions in rows and columns. The concept of matrices is extensively used in solving simultaneous linear equations, coordinate geometry, transformations, determinants, computer science, statistics, economics, engineering, physics, data science and competitive mathematics.

For CBSE Board examinations and competitive entrance examinations, students should be comfortable with the definition of a matrix, order of a matrix, rows and columns, elements of a matrix, notation, position of an element and different types of matrices. These basic ideas form the foundation for matrix operations and determinants studied later in the chapter.

📘 Definition

Definition of a Matrix

A matrix is an ordered rectangular arrangement of numbers, variables, algebraic expressions or functions, arranged in horizontal rows and vertical columns and enclosed within brackets.

The individual numbers or expressions present in a matrix are called its elements or entries.

Matrices are generally denoted by capital letters such as

\[A,\ B,\ C,\ X,\ Y,\ M\]

while their elements are usually represented by corresponding lowercase letters.

For example,

A= \begin{bmatrix} -2 & 5\\ 0 & \sqrt{5}\\ 3 & 6 \end{bmatrix}, \qquad B= \begin{bmatrix} 2+i & 3 & -\frac{1}{2}\\ 15 & -1 & 2\\ \sqrt{3} & 5 & \frac{5}{7} \end{bmatrix}

C= \begin{bmatrix} 1+x & x^3 & 3\\ \cos x & \sin x+2 & \tan x \end{bmatrix}

In matrix \(A\), the elements are arranged in 3 rows and 2 columns. Matrix \(B\) has 3 rows and 3 columns, whereas matrix \(C\) has 2 rows and 3 columns

🗂️ Types / Category

Rows and Columns of a Matrix

Rows

The horizontal arrangements of elements in a matrix are called its rows.

For example

\[ A= \begin{bmatrix} 2 & 4 & 6\\ 1 & 3 & 5\\ 7 & 8 & 9 \end{bmatrix} \]

has three rows:

\[ R_1=[2\quad4\quad6] \]

\[ R_2=[1\quad3\quad5] \]

\[ R_3=[7\quad8\quad9] \]

Columns

The vertical arrangements of elements in a matrix are called its columns.

For the same matrix,

\[ A= \begin{bmatrix} 2 & 4 & 6\\ 1 & 3 & 5\\ 7 & 8 & 9 \end{bmatrix} \]

the three columns are

\[ C_1= \begin{bmatrix} 2\\ 1\\ 7 \end{bmatrix}, \qquad C_2= \begin{bmatrix} 4\\ 3\\ 8 \end{bmatrix}, \qquad C_3= \begin{bmatrix} 6\\ 5\\ 9 \end{bmatrix} \]

📌 Order of a Matrix

The order of a matrix is the number of rows followed by the number of columns. If a matrix has \(m\) rows and \(n\) columns, then its order is

\[ \boxed{m\times n} \]

Thus, the order of a matrix is written as number of rows \(\times\) number of columns.

For example

\[ A= \begin{bmatrix} 2 & 5 & 7\\ 1 & 4 & 8 \end{bmatrix} \]

has 2 rows and 3 columns. Therefore,

\[ \boxed{\text{Order of }A=2\times3} \]

Similarly,

\[ B= \begin{bmatrix} 1\\ 3\\ 5\\ 7 \end{bmatrix} \]

has 4 rows and 1 column. Hence its order is

\[ \boxed{4\times1} \]

📜 Important Rule for Remembering Matrix Order

Always count rows first and columns second.
\[ \boxed{\text{Order}=\text{Rows}\times\text{Columns}} \]
Therefore, a matrix having 3 rows and 5 columns is a \(3\times5\) matrix, not a \(5\times3\) matrix.
General Form of a Matrix
A matrix having \(m\) rows and \(n\) columns can be represented in the general form

\[ A= \begin{bmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n}\\ a_{21} & a_{22} & a_{23} & \cdots & a_{2n}\\ a_{31} & a_{32} & a_{33} & \cdots & a_{3n}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & a_{m3} & \cdots & a_{mn} \end{bmatrix} \]
Here \(a_{ij}\) represents the element situated in the \(i\)-th row and \(j\)-th column.
Thus, if
\[ A=[a_{ij}]_{m\times n}, \]
then \(\mathrm{A}\) has \(\mathrm{m}\) rows and \(\mathrm{n}\) columns.

🗒️ Elements or Entries of a Matrix

Every individual quantity in a matrix is called an element or entry of the matrix.

In the notation \(a_{ij}\):

\(i\) indicates the row number.
\(j\) indicates the column number.
\(a_{ij}\) is therefore the element at the intersection of the \(i\)-th row and \(j\)-th column.

For example

\[ A= \begin{bmatrix} 4 & 7 & 9\\ 2 & 5 & 8 \end{bmatrix} \]

Here,

\[ a_{11}=4,\qquad a_{12}=7,\quad a_{13}=9 \]

\[ a_{21}=2,\quad a_{22}=5,\quad a_{23}=8 \]

Notice that \(a_{23}=8\) means the element in the 2nd row and 3rd column.

📎 How to Locate an Element of a Matrix

To locate an element \(a_{ij}\), move to the \(i\)-th row and then the \(j\)-th column. The two subscripts should never be interchanged.

Consider

\[ A= \begin{bmatrix} 10 & 20 & 30\\ 40 & 50 & 60\\ 70 & 80 & 90 \end{bmatrix} \]

Then

\[ a_{12}=20,\quad a_{21}=40,\quad a_{31}=70,\quad a_{23}=60 \]

In particular,

\[ a_{12}\neq a_{21} \]

in general. The first subscript identifies the row and the second identifies the column.

📜 Number of Elements in a Matrix

📜 Rules

If a matrix has \(m\) rows and \(n\) columns, then the total number of elements is

\[ \boxed{mn} \]

For example, a \(4\times7\) matrix contains \( 4\times7=28 \) elements.

Therefore:

\[ \boxed{\text{Number of elements}=\text{number of rows}\times\text{number of columns}} \]

✏️ Example

Examples: Basic Concepts

Question

Find the order of the matrix

\[ A= \begin{bmatrix} 2 & 4 & 6 & 8\\ 1 & 3 & 5 & 7\\ 9 & 10 & 11 & 12 \end{bmatrix} \]

The matrix contains 3 horizontal rows and 4 vertical columns.

\[ \boxed{\text{Order of }A=3\times4} \]

Question

\[ A= \begin{bmatrix} 3 & 5 & 7\\ 2 & 4 & 6\\ 1 & 8 & 9 \end{bmatrix}, \]

find \(a_{23}\)

\(a_{23}\) represents the element in the 2nd row and 3rd column.

\[ \boxed{a_{23}=6} \]

Question

How many elements are present in a \(5\times8\) matrix?

\[ \text{Number of elements}=5\times8=40 \]

\[ \boxed{40} \]

Question

If \(A=[a_{ij}]_{4\times6}\), determine the number of rows, columns and elements.

\[ \text{Rows}=4 \]

\[ \text{Columns}=6 \]

\[ \text{Number of elements}=4\times6=24 \]

Question

Construct a \(3\times3\) matrix \(A=[a_{ij}]\) where

\[a_{ij}=i^2+j\]

For \(i=1,2,3\) and \(j=1,2,3\)

\[A=\begin{bmatrix}1^2+1&1^2+2&1^2+3\\ 2^2+1&2^2+2&2^2+3\\ 3^2+1&3^2+2&3^2+3 \end{bmatrix} \]

Therefore,

\[\boxed{A=\begin{bmatrix}2&3&4\\ 5&6&7\\10&11&12 \end{bmatrix}}\]

ℹ️ Matrix as a Structured Mathematical Object

A useful way to understand a matrix is to view it as a structured mathematical object rather than merely an array of numbers. Every element has a definite position determined by its row and column indices.

For a matrix \(A=[a_{ij}]_{m\times n}\), the index restrictions are

\[1\leq i\leq m,\qquad 1\leq j\leq n\]

Consequently, the total number of possible ordered pairs \((i,j)\) is

\[mn,\]

which explains why an \(m\times n\) matrix contains exactly \(mn\) elements.

💡 Conceptual Connection with Determinants

A matrix and a determinant are related but they are not the same mathematical object. A matrix is an arrangement of elements in rows and columns, whereas a determinant is a scalar quantity associated with a square matrix.

For example,

\[A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\]

is a matrix, while its determinant is

\[|A|=ad-bc\]

This distinction becomes important later when studying determinants, inverse matrices and systems of linear equations.

🌟 Significance

Matrices provide a compact method for representing large amounts of related information. A system containing several linear equations can be represented using coefficient matrices, which makes matrix methods particularly useful for solving simultaneous equations.

Matrices are also fundamental in:

solving systems of linear equations;
coordinate transformations;
computer graphics and image transformations;
cryptography and coding theory;
statistics and data analysis;
economics and input-output models;
physics and engineering;
machine learning and artificial intelligence;
network analysis;
linear transformations and vector spaces.

🌟 Why Matrices Are Important for Class 12 Board Examinations

The introductory definitions in this section may appear elementary, but they form the foundation for almost every subsequent concept in the chapter. Board examination questions can test the order of a matrix, identify specific elements, construct matrices from given conditions and distinguish between different types of matrices.

Students should be especially comfortable with the notation \(a_{ij}\), because questions involving equality of matrices, transpose, symmetric matrices, matrix multiplication and inverse matrices depend heavily on correct interpretation of row and column indices.

Importance for JEE and Competitive Entrance Examinations

For competitive examinations, matrix notation is often combined with determinants, algebraic identities, linear equations and functions. A seemingly simple question about the position of an element can become part of a more complex problem involving conditions on \(i\) and \(j\).

Competitive examination preparation should therefore focus not only on memorising definitions but also on interpreting matrix notation quickly and accurately.

⚡ Exam Tip

Always write the order of a matrix as rows \(\times\) columns.
Remember that the first suffix in \(a_{ij}\) represents the row.
Remember that the second suffix represents the column.
Do not reverse \(m\times n\) into \(n\times m\).
Count rows horizontally and columns vertically.
A matrix of order \(m\times n\) contains exactly \(mn\) elements.
A square matrix has equal numbers of rows and columns.
For a square matrix, \(a_{ii}\) represents a principal diagonal element.
Do not confuse a matrix with its determinant.
When constructing \(A=[a_{ij}]\), systematically substitute the allowed values of \(i\) and \(j\).

❌ Common Mistakes

Writing the order incorrectly: A matrix with 4 rows and 3 columns has order \(4\times3\), not \(3\times4\).
Reversing subscripts: \(a_{23}\) means second row and third column, not third row and second column.
Confusing number of elements with order: A \(3\times4\) matrix has 12 elements, but its order is \(3\times4\).
Assuming every matrix is square: Matrices can be rectangular, row matrices or column matrices.
Ignoring position: Two matrices containing the same numbers in different positions are generally different matrices.
Confusing matrix and determinant: A matrix is an array, whereas a determinant is a scalar associated with a square matrix.

📋 CBSE Case Study Based Question

A school laboratory records the number of students participating in three mathematical activities over two academic terms. The data is represented by

\[ A= \begin{bmatrix} 24&18&30\ 28&22&35 \end{bmatrix}, \]

where the rows represent two terms and the columns represent three activities.

Questions

What is the order of matrix \(A\)?
How many total entries are present in \(A\)?
What does \(a_{12}\) represent?
Find \(a_{23}\).
Is \(A\) a square matrix?

Answers

1. Matrix \(A\) has 2 rows and 3 columns. Therefore,

\[ \boxed{\text{Order of }A=2\times3}. \]

2. The number of elements is

\[ 2\times3=6. \]

Hence,

\[ \boxed{6}. \]

3. \(a_{12}\) represents the element in the first row and second column:

\[ \boxed{a_{12}=18}. \]

4. \(a_{23}\) is the element in the second row and third column:

\[ \boxed{a_{23}=35}. \]

5. No. Since the number of rows and columns are different, \(A\) is a rectangular matrix, not a square matrix.

Higher-Order Thinking Question

Question: A matrix \(A=[a_{ij}]\) has order \(4\times5\). Is it possible for \(a_{54}\) to exist?

Solution

Since \(A\) has 4 rows, the row index can only take values

\[ i=1,2,3,4. \]

The notation \(a_{54}\) requires the fifth row, but the matrix has only four rows.

Therefore,

\[ \boxed{a_{54}\text{ does not exist}}. \]

This type of index-based reasoning is useful in higher-level competitive examination problems.

🔢 Quick Formula and Concept Sheet

General matrix: \(\displaystyle A=[a_{ij}]_{m\times n}\)
Order: \(\displaystyle m\times n\)
Number of elements: \(\displaystyle mn\)
Element \(a_{ij}\): \(i\)-th row and \(j\)-th column
Row matrix: order \(1\times n\)
Column matrix: order \(m\times1\)
Square matrix: order \(n\times n\)
Principal diagonal: \(a_{11},a_{22},\ldots,a_{nn}\)
Diagonal element: \(a_{ij}\) where \(i=j\)
Off-diagonal element: \(a_{ij}\) where \(i\ne j\)
Trace: \(\displaystyle \operatorname{tr}(A)=a_{11}+a_{22}+\cdots+a_{nn}\)

⚡ One-Minute Revision

A matrix is an ordered rectangular arrangement of elements in rows and columns. If a matrix has \(m\) rows and \(n\) columns, its order is \(m\times n\), and it contains \(mn\) elements. The element \(a_{ij}\) is located at the intersection of the \(i\)-th row and \(j\)-th column. A matrix with one row is a row matrix, a matrix with one column is a column matrix, and a matrix with equal numbers of rows and columns is a square matrix. In a square matrix, \(a_{11},a_{22},\ldots,a_{nn}\) form the principal diagonal. These basic ideas are essential for understanding every subsequent operation and property of matrices.

Key Concepts

NCERT Class 12 Mathematics Chapter 3 Matrices introduces the fundamental language of matrix algebra, including matrix definition, order of a matrix, rows and columns, elements of a matrix, matrix notation, general form of a matrix, row matrix, column matrix, square matrix, diagonal elements and principal diagonal. Understanding these concepts thoroughly helps students prepare for CBSE Class 12 Mathematics board examinations, JEE Main, JEE Advanced and other competitive entrance examinations.

These concepts also provide the foundation for advanced topics such as types of matrices, equality of matrices, addition and multiplication of matrices, transpose of a matrix, symmetric and skew-symmetric matrices, elementary transformations, inverse of a matrix and applications of matrices.

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Order of a Matrix

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Table of Contents

16 sections

1 📘 Definition ›
2 💡 Number of Elements in a Matrix ›
3 🔢 General Form of an Matrix ›
4 💡 Important Concept: Row Index and Column Index ›
5 🔗 Key Relationship Between Order and Number of Elements ›
6 💡 Representation of Points Using Matrices ›
7 ✏️ Example: Matrix Representation of a Point ›
8 📌 Representing Geometrical Figures Using Matrices ›
9 👁️ Important Observation ›
10 ✏️ Example ›
11 🌟 Competitive-Exam Insight ›
12 📋 CBSE Case Study / HOTS ›
13 ❌ Common Mistakes ›
14 ⚡ Exam Tip ›
15 ⚡ Quick Revision ›
16 🔑 Key Takeaway ›

📘 Definition

A matrix having \(m\) rows and \(n\) columns is called a matrix of order \(m\times n\), or simply an \(m\times n\) matrix, read as “\(m\) by \(n\) matrix”.

The first number in the order represents the number of rows, while the second number represents the number of columns.

\[ \boxed{\text{Order of a matrix}=\text{Number of rows}\times\text{Number of columns}} \]

For example, consider

\[ A= \begin{bmatrix} -2 & 5\\ 0 & \sqrt{5}\\ 3 & 6 \end{bmatrix}\]

\[B= \begin{bmatrix} 2+i & 3 & -\frac12\\ 15 & -1 & 2\\ \sqrt3 & 5 & \frac57 \end{bmatrix} \]

\[ C= \begin{bmatrix} 1+x & x^3 & 3\\ \cos x & \sin x+2 & \tan x \end{bmatrix} \]

We have:

\(A\) has 3 rows and 2 columns, so \(A\) is a \(3\times2\) matrix.
\(B\) has 3 rows and 3 columns, so \(B\) is a \(3\times3\) matrix.
\(C\) has 2 rows and 3 columns, so \(C\) is a \(2\times3\) matrix.

💡 Number of Elements in a Matrix

If a matrix has \(m\) rows and \(n\) columns, then every row contains \(n\) elements and there are \(m\) such rows. Therefore, the total number of elements is

\[ \boxed{mn} \]

Thus:

\[ \begin{aligned} A&: 3\times2=6\text{ elements},\\ B&: 3\times3=9\text{ elements},\\ C&: 2\times3=6\text{ elements}. \end{aligned} \]

This gives the fundamental relation:

\[ \boxed{\text{Number of elements}=(\text{number of rows})(\text{number of columns})} \]

🔢 General Form of an \(m\times n\) Matrix

An arbitrary matrix \(A\) of order \(m\times n\) can be written as

\[ A= \begin{bmatrix} a_{11} & a_{12} & a_{13} & \cdots & a_{1n}\\ a_{21} & a_{22} & a_{23} & \cdots & a_{2n}\\ a_{31} & a_{32} & a_{33} & \cdots & a_{3n}\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & a_{m3} & \cdots & a_{mn} \end{bmatrix} \]

In compact notation,

\[ \boxed{A=[a_{ij}]_{m\times n}} \]

where
\( \boxed{1\leq i\leq m,\qquad 1\leq j\leq n} \) and
\(i,j\in\mathbb N\) when the row and column indices are taken from the natural numbers.

Here, \(a_{ij}\) denotes the element located at the intersection of the \(i\)-th row and the \(j\)-th column.

Meaning of the Subscripts in \(a_{ij}\)

The notation \(a_{ij}\) is an indexing system used to identify every element uniquely.

\[ \boxed{a_{ij}=\text{element in the }i\text{-th row and }j\text{-th column}} \]

The first subscript always indicates the row, and the second subscript always indicates the column.

For example, in

\[ A= \begin{bmatrix} 2 & 4 & 6 & 8\\ 1 & 3 & 5 & 7\\ 9 & 10 & 11 & 12 \end{bmatrix}, \]

we have

\[ a_{12}=4,\quad a_{23}=5,\quad a_{31}=9,\quad a_{34}=12. \]

For instance, \(a_{23}=5\) means that \(5\) is located in the 2nd row and 3rd column.

The \(i\)-th Row

The \(i\)-th row of an \(m\times n\) matrix consists of the elements

\[ a_{i1},a_{i2},a_{i3},\ldots,a_{in}. \]

Thus, the \(i\)-th row can be written as

\[ \boxed{[a_{i1}\quad a_{i2}\quad a_{i3}\quad\cdots\quad a_{in}]} \]

For example, if

\[ A= \begin{bmatrix} 2 & 5 & 7\\ 4 & 6 & 8\\ 1 & 3 & 9 \end{bmatrix}, \]

then the second row is

\[ \boxed{[4\quad6\quad8]} \]

The \(j\)-th Column

The \(j\)-th column of an \(m\times n\) matrix consists of the elements

\[ a_{1j},a_{2j},a_{3j},\ldots,a_{mj}. \]

It can be written as

\[ \boxed{ \begin{bmatrix} a_{1j}\\ a_{2j}\\ a_{3j}\\ \vdots\\ a_{mj} \end{bmatrix}} \]

In the matrix above, the third column is

\[ \boxed{ \begin{bmatrix} 7\\ 8\\ 9 \end{bmatrix}} \]

💡 Important Concept: Row Index and Column Index

🔗 Key Relationship Between Order and Number of Elements

If \(A\) is an \(m\times n\) matrix, then:

\[ \boxed{ \begin{aligned} \text{Number of rows}&=m\\ \text{Number of columns}&=n\\ \text{Order}&=m\times n\\ \text{Number of elements}&=mn \end{aligned}} \]

💡 Representation of Points Using Matrices

Matrices are not restricted to arranging numbers for algebraic calculations. They can also be used to represent points and geometric figures in a coordinate plane.

A point \(P(x,y)\) can be represented in matrix form either as a column matrix or as a row matrix:

\[ \boxed{ P= \begin{bmatrix} x\\ y \end{bmatrix}} \]

or,

\[\boxed{P=[x\quad y]}\]

Both matrices contain the same coordinate information, although they have different orders:

\[ \begin{bmatrix} x\\ y \end{bmatrix} \quad\text{has order }2\times1 \]

\[ [x\quad y] \quad\text{has order }1\times2 \]

✏️ Example: Matrix Representation of a Point

Consider the point

\[P(0,1)\]

Its column-matrix representation is

\[ \boxed{ P= \begin{bmatrix} 0\\ 1 \end{bmatrix}} \]

and its row-matrix representation is

\[ \boxed{P=[0\quad1]} \]

The column representation is particularly useful in linear algebra, coordinate transformations and matrix multiplication because points can be transformed systematically using matrices.

📌 Representing Geometrical Figures Using Matrices

The coordinates of several points can be collected into a single matrix. This provides a convenient algebraic representation of a geometrical figure.

Consider a quadrilateral \(ABCD\) whose vertices are

\[ A(1,0),\qquad B(3,2),\qquad C(1,3),\qquad D(-1,2)\]

Column-Wise Coordinate Representation If each vertex is represented as a column vector, the four vertices can be placed as columns of a single matrix:

\[ \boxed{ X= \begin{bmatrix} 1 & 3 & 1 & -1\\ 0 & 2 & 3 & 2 \end{bmatrix}} \]

Here:

\[ \begin{bmatrix} 1\\ 0 \end{bmatrix} \]

represents \(A(1,0)\), while

\[ \begin{bmatrix} 3\ 2 \end{bmatrix} \]

represents \(B(3,2)\), and so on.
Therefore, the matrix \(X\) has order

\[ \boxed{2\times4} \]

because there are 2 coordinate components, \(x\) and \(y\), and 4 vertices.

Row-Wise Coordinate Representation

Alternatively, each vertex can be written as a row matrix. Placing the four row vectors one below another gives

\[ \boxed{ Y= \begin{bmatrix} 1 & 0\\ 3 & 2\\ 1 & 3\\ -1 & 2 \end{bmatrix}} \]

This matrix has order

\[ \boxed{4\times2} \]

Here each row corresponds to one vertex:

\[ \begin{aligned} A&\longrightarrow[1\quad0]\\ B&\longrightarrow[3\quad2]\\ C&\longrightarrow[1\quad3]\\ D&\longrightarrow[-1\quad2] \end{aligned} \]

👁️ Important Observation

✏️ Example

Question

Find the order and number of elements of

\[ A= \begin{bmatrix} 1&2&3&4&5\\ 6&7&8&9&10\\ 11&12&13&14&15 \end{bmatrix}. \]

The matrix has 3 rows and 5 columns.

\[ \boxed{\text{Order}=3\times5} \]

\[ \boxed{\text{Number of elements}=3\times5=15} \]

Question

Let

\[ A= \begin{bmatrix} 4&8&12\\ 3&6&9\\ 5&10&15 \end{bmatrix}. \]

Find \(a_{32}\)

\(a_{32}\) means the element in the 3rd row and 2nd column.

\[ \boxed{a_{32}=10} \]

Question

Construct the \(2\times3\) matrix \(A=[a_{ij}]\), where

\[ a_{ij}=2i+j\]

Since \(A\) is \(2\times3\),

\[ i=1,2,\qquad j=1,2,3. \]

Calculate the entries:

\[ a_{11}=2(1)+1=3 \]

\[ a_{12}=2(1)+2=4 \]

\[ a_{13}=2(1)+3=5 \]

\[ a_{21}=2(2)+1=5 \]

\[ a_{22}=2(2)+2=6 \]

\[ a_{23}=2(2)+3=7 \]

Hence,

\[ \boxed{ A= \begin{bmatrix} 3&4&5\\ 5&6&7 \end{bmatrix}} \]

Question

A matrix contains 30 elements. Write some possible orders of the matrix.

Since

\[ mn=30, \]

possible ordered dimensions include

\[ \boxed{ 1\times30,\quad 2\times15,\quad 3\times10,\quad 5\times6,\quad 6\times5,\quad 10\times3,\quad 15\times2,\quad 30\times1} \]

🌟 Competitive-Exam Insight

A common objective-type question may provide the number of elements and ask for the possible order of a matrix. If a matrix has \(N\) elements, every possible order corresponds to an ordered factor pair of \(N\).

For example, if a matrix has \(12\) elements, possible orders include

\[ 1\times12,\quad2\times6,\quad3\times4,\quad4\times3,\quad6\times2,\quad12\times1. \]

If the question additionally states that the matrix is square, then only

\[ \boxed{\sqrt{12}\times\sqrt{12}} \]

would be required, which is not possible because \(\sqrt{12}\notin\mathbb N\). Therefore, a square matrix cannot have 12 elements.

More generally, a square matrix with \(n\) rows and \(n\) columns contains

\[ \boxed{n^2} \]

elements. Hence, the number of elements of a square matrix must be a perfect square.

📋 CBSE Case Study / HOTS

The vertices of a quadrilateral are

\[ A(1,0),\quad B(3,2),\quad C(1,3),\quad D(-1,2). \]

They are represented by

\[ X= \begin{bmatrix} 1&3&1&-1\\ 0&2&3&2 \end{bmatrix}. \]

Answer the following:

Find the order of \(X\).
How many elements does \(X\) contain?
Which column represents vertex \(C\)?
Write the matrix representation of vertex \(D\) as a column matrix.
Write the corresponding row-wise matrix \(Y\).

Solution

\(X\) has 2 rows and 4 columns.

\[ \boxed{\text{Order of }X=2\times4} \]

\[ \boxed{\text{Number of elements}=2\times4=8} \]

Vertex \(C(1,3)\) is represented by the third column:

\[ \boxed{ \begin{bmatrix} 1\\ 3 \end{bmatrix}} \]

Vertex \(D(-1,2)\) is represented by

\[ \boxed{ \begin{bmatrix} -1\\ 2 \end{bmatrix}} \]

The row-wise representation is

\[ \boxed{ Y= \begin{bmatrix} 1&0\\ 3&2\\ 1&3\\ -1&2 \end{bmatrix}} \]

❌ Common Mistakes

Reversing the order: 4 rows and 3 columns means \(4\times3\), not \(3\times4\).
Misreading \(a_{ij}\): \(i\) is the row index and \(j\) is the column index.
Confusing order with number of elements: A \(4\times5\) matrix has order \(4\times5\), but it contains 20 elements.
Assuming \(m\times n\) means multiplication: In “order \(m\times n\)”, the notation describes dimensions; the actual number of elements is obtained by multiplying \(m\) and \(n\).
Ignoring the order when representing coordinates:
\[ \begin{bmatrix}x\\y\end{bmatrix} \]
is \(2\times1\), whereas
\[ [x\quad y] \]
is \(1\times2\).

⚡ Exam Tip

Memorise the rule:
\[ \boxed{\text{Rows}\times\text{Columns}} \]
Read \(a_{ij}\) aloud as “row \(i\), column \(j\)”.
For an \(m\times n\) matrix, immediately note:
\[ 1\leq i\leq m,\qquad1\leq j\leq n. \]
If the number of elements is given, use factor pairs to determine possible orders.
If the matrix is square and contains \(N\) elements, check whether \(N\) is a perfect square.
In coordinate problems, decide whether the points are being arranged as rows or columns before interpreting the matrix order.

⚡ Quick Revision

Concept	Key Result
Order of matrix	\(m\times n\)
Number of rows	\(m\)
Number of columns	\(n\)
Total elements	\(mn\)
General notation	\(A=[a_{ij}]_{m\times n}\)
\(a_{ij}\)	Element in row \(i\), column \(j\)
Column representation of \(P(x,y)\)	\(\begin{bmatrix}x\\y\end{bmatrix}\)
Row representation of \(P(x,y)\)	\([x\quad y]\)
Square matrix	\(n\times n\)
Elements in square matrix	\(n^2\)

🔑 Key Takeaway

A matrix with \(m\) rows and \(n\) columns has order \(m\times n\).
An \(m\times n\) matrix contains exactly \(mn\) elements.
The element \(a_{ij}\) lies in row \(i\) and column \(j\).
The \(i\)-th row contains \(a_{i1},a_{i2},\ldots,a_{in}\).
The \(j\)-th column contains \(a_{1j},a_{2j},\ldots,a_{mj}\).
A point \((x,y)\) can be represented by either a \(2\times1\) column matrix or a \(1\times2\) row matrix.
The coordinates of several vertices can be collected into a single matrix.
The row-wise and column-wise coordinate representations are related through transposition.
These concepts provide the foundation for matrix operations, transpose, symmetric matrices, inverse matrices and determinants studied later.

▦

Example 1

📋

Table of Contents

9 sections

1 ❓ Question ›
2 💡 Concept ›
3 🗺️ Roadmap ›
4 🧩 Solution ›
5 🔍 Interpreting the Required Entry ›
6 🤔 Why This Example Is Important ›
7 ⚡ Exam Tip ›
8 ❌ Common Mistakes ›
9 🌟 Competitive Exam Insight ›

❓ Question

Consider the following information regarding the number of men and women workers in three factories I, II and III:

Factory	Men Workers	Women Workers
I	30	25
II	25	31
III	27	26

Represent the above information in the form of a \(3\times2\) matrix. Also, state what the entry in the third row and second column represents.

💡 Concept

🗺️ Roadmap

Identify the objects being represented: three factories.
Identify the categories of data: men workers and women workers.
Assign each factory to one row.
Use the first column for men workers and the second column for women workers.
Place the numerical data in the corresponding positions.
Locate the required entry using its row and column indices.

🧩 Solution

Since there are three factories and two categories of workers, the required matrix is of order \(3\times2\).

\[ \boxed{ A= \begin{bmatrix} 30&25\\ 25&31\\ 27&26 \end{bmatrix}} \]

The rows correspond respectively to factories I, II and III:

\[ \begin{aligned} \text{Factory I}&\longrightarrow[30\quad25]\\ \text{Factory II}&\longrightarrow[25\quad31]\\ \text{Factory III}&\longrightarrow[27\quad26] \end{aligned} \]

The columns represent:

\[ \begin{aligned} \text{Column 1}&\longrightarrow\text{Men workers}\\\text{Column 2}&\longrightarrow\text{Women workers} \end{aligned} \]

🔍 Interpreting the Required Entry

The entry in the third row and second column is denoted by \(a_{32}\).

\[ a_{32}=26 \]

The first subscript \(3\) indicates Factory III, while the second subscript \(2\) indicates the women workers column.

\[ \boxed{a_{32}=26} \]

Therefore, the entry in the third row and second column represents the number of women workers in Factory III, which is 26 women workers.

🤔 Why This Example Is Important

This example illustrates one of the most fundamental applications of matrices: converting structured real-world information into a mathematical form. Instead of describing the worker data repeatedly in words, the matrix stores the same information systematically using rows and columns.
This representation becomes especially useful when dealing with large datasets, because mathematical operations can subsequently be performed on the matrix.

⚡ Exam Tip

❌ Common Mistakes

A frequent error is to interchange the two categories or reverse the row and column interpretation. For example, \(a_{32}\) does not mean Factory II and the third category. It means third row and second column.

🌟 Competitive Exam Insight

In objective-type examinations, the numerical entries may remain unchanged while the interpretation of rows and columns is altered. Always identify the convention used in the question before reading a matrix element.

▦

Example 2

📋

Table of Contents

9 sections

1 ❓ Question ›
2 💡 Concept ›
3 🗺️ Roadmap ›
4 🧩 Solution ›
5 🌟 Important Concept: Order Matters ›
6 🌟 Competitive Exam Insight ›
7 ⚡ Exam Tip ›
8 ❌ Common Mistakes ›
9 🔑 Key Takeaway ›

❓ Question

If a matrix has 8 elements, what are the possible orders it can have?

💡 Concept

🗺️ Roadmap

Let the order of the matrix be \(m\times n\).
Use the formula \(\text{Number of elements}=mn\).
Since the matrix contains 8 elements, set \(mn=8\).
Find all ordered factor pairs of 8.
Convert each ordered pair into a possible matrix order.

🧩 Solution

Let the matrix be of order \(m\times n\). Then

\[ \text{Number of elements}=mn. \]

Since the matrix has 8 elements,

\[ mn=8\]

The positive factor pairs of 8 are

\[ 1\times8=8 \]

\[ 2\times4=8 \]

\[ 4\times2=8 \]

\[ 8\times1=8 \]

Therefore, the corresponding ordered pairs \((m,n)\) are

\[ \boxed{(1,8),\ (2,4),\ (4,2),\ (8,1)} \]

Hence, the possible orders of the matrix are

\[ \boxed{ 1\times8,\quad 8\times1,\quad 2\times4,\quad 4\times2 } \]

🌟 Important Concept: Order Matters

Notice that \(2\times4\) and \(4\times2\) are considered different orders. Although both matrices contain the same number of elements, their dimensions are different.

\[ \boxed{2\times4\neq4\times2} \]

For example, a \(2\times4\) matrix has 2 rows and 4 columns:

\[ A= \begin{bmatrix} a_{11}&a_{12}&a_{13}&a_{14}\\ a_{21}&a_{22}&a_{23}&a_{24} \end{bmatrix} \]

whereas a \(4\times2\) matrix has 4 rows and 2 columns:

\[ B= \begin{bmatrix} b_{11}&b_{12}\\ b_{21}&b_{22}\\ b_{31}&b_{32}\\ b_{41}&b_{42} \end{bmatrix} \]

Both contain 8 elements, but their arrangements are different.

🌟 Competitive Exam Insight

For a matrix containing \(N\) elements, the number of possible orders is determined by the positive divisors of \(N\). If

\[ N=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k} \]

is the prime factorisation of \(N\), then the number of positive divisors of \(N\) is

\[ \boxed{(\alpha_1+1)(\alpha_2+1)\cdots(\alpha_k+1)}. \]

Each positive divisor \(m\) determines exactly one ordered matrix order

\[ m\times\frac{N}{m}. \]

Therefore, the number of possible ordered matrix dimensions equals the number of positive divisors of \(N\).

For \(8=2^3\), the number of positive divisors is

\[ 3+1=4. \]

Hence there are exactly four possible orders:

\[ \boxed{1\times8,\ 2\times4,\ 4\times2,\ 8\times1}. \]

⚡ Exam Tip

❌ Common Mistakes

Students sometimes write only the unordered factor pairs

\[ (1,8),\ (2,4) \]

and omit

\[ (8,1),\ (4,2). \]

This is incorrect when the question asks for all possible orders, because \(m\times n\) and \(n\times m\) represent different matrix dimensions whenever \(m\neq n\).

Answer

\[ \boxed{ 1\times8,\quad 8\times1,\quad 2\times4,\quad 4\times2 } \]

🔑 Key Takeaway

▦

Example 3

📋

Table of Contents

9 sections

1 ❓ Question ›
2 💡 Concept ›
3 🗺️ Roadmap ›
4 🧩 Solution ›
5 ⚡ Exam Tip ›
6 ❌ Common Mistakes ›
7 🌟 CBSE Board Exam Significance ›
8 🌟 Competitive Exam Insight ›
9 ⚡ Quick Revision ›

❓ Question

Construct a \(3\times2\) matrix whose elements are given by

\[ \boxed{a_{ij}=\frac{1}{2}\left|i-3j\right|} \]

💡 Concept

When the general element \(a_{ij}\) of a matrix is given by a formula, each element is obtained by substituting the corresponding values of \(i\) and \(j\).
Since the required matrix is of order \(3\times2\),

\[ \boxed{i=1,2,3} \]

\[ \boxed{j=1,2} \]

Therefore, the six required entries are

\[ a_{11},\quad a_{12},\quad a_{21},\quad a_{22},\quad a_{31},\quad a_{32}. \]

🗺️ Roadmap

Write the general form of a \(3\times2\) matrix.
Identify all possible values of \(i\) and \(j\).
Substitute each pair \((i,j)\) into \(a_{ij}=\frac12|i-3j|\).
Place each calculated value in its correct position.
Write the resulting matrix.

🧩 Solution

Step 1: Write the General Matrix

A general \(3\times2\) matrix is

\[ A= \begin{bmatrix} a_{11}&a_{12}\\ a_{21}&a_{22}\\ a_{31}&a_{32} \end{bmatrix}. \]

Step 2: Calculate the Entries

We are given

\[ a_{ij}=\frac12|i-3j|\]
First Row → For \(a_{11}\), put \(i=1\) and \(j=1\):

\[ \begin{aligned} a_{11} &=\frac12|1-3(1)|\\ &=\frac12|-2|\\ &=1. \end{aligned} \]
For \(a_{12}\), put \(i=1\) and \(j=2\):

\[ \begin{aligned} a_{12} &=\frac12|1-3(2)|\\ &=\frac12|-5|\\ &=\frac52. \end{aligned} \]
Second Row → For \(a_{21}\), put \(i=2\) and \(j=1\):

\[ \begin{aligned} a_{21} &=\frac12|2-3(1)|\\ &=\frac12|-1|\\ &=\frac12. \end{aligned} \]
For \(a_{22}\), put \(i=2\) and \(j=2\):

\[ \begin{aligned} a_{22} &=\frac12|2-3(2)|\ &=\frac12|-4|\\ &=2. \end{aligned} \]
Third Row → For \(a_{31}\), put \(i=3\) and \(j=1\):

\[ \begin{aligned} a_{31} &=\frac12|3-3(1)|\\ &=\frac12|0|\\ &=0. \end{aligned} \]
For \(a_{32}\), put \(i=3\) and \(j=2\):

\[ \begin{aligned} a_{32} &=\frac12|3-3(2)|\\ &=\frac12|-3|\\ &=\frac32. \end{aligned} \]

Step 3: Organise the Entries

The calculated values are:

Element	\(i\)	\(j\)	Calculation	Value
\(a_{11}\)	1	1	\(\frac12\|1-3\|\)	\(1\)
\(a_{12}\)	1	2	\(\frac12\|1-6\|\)	\(\frac52\)
\(a_{21}\)	2	1	\(\frac12\|2-3\|\)	\(\frac12\)
\(a_{22}\)	2	2	\(\frac12\|2-6\|\)	\(2\)
\(a_{31}\)	3	1	\(\frac12\|3-3\|\)	\(0\)
\(a_{32}\)	3	2	\(\frac12\|3-6\|\)	\(\frac32\)

Final Matrix
Placing the calculated entries in their corresponding positions gives
\[A= \begin{bmatrix} 1&\frac52\\ \frac12&2\\ 0&\frac32 \end{bmatrix}\]
Therefore, the required \(3\times2\) matrix is

\[\begin{bmatrix} 1&\frac52\\ \frac12&2\\ 0&\frac32 \end{bmatrix}\]

⚡ Exam Tip

❌ Common Mistakes

Using \(j=3\): A \(3\times2\) matrix has only two columns, so \(j\) can only be 1 or 2.
Interchanging \(i\) and \(j\): \(i\) represents the row and \(j\) represents the column.
Forgetting the modulus: For example,
\[ \frac12(-5)\neq\frac52. \]
The correct calculation is
\[ \frac12|-5|=\frac52. \]
Putting entries in the wrong positions: After calculating each \(a_{ij}\), place it at exactly the \(i\)-th row and \(j\)-th column.
Stopping after calculating only some entries: A \(3\times2\) matrix has \(3\times2=6\) entries, so all six must be calculated.

🌟 CBSE Board Exam Significance

This type of question tests whether a student can translate a mathematical rule defining \(a_{ij}\) into an explicit matrix. It combines the concepts of matrix order, element notation, substitution, absolute value and systematic arrangement. The same skill is useful in questions involving matrices defined by algebraic, trigonometric or piecewise expressions.

🌟 Competitive Exam Insight

⚡ Quick Revision

Required order: \(3\times2\).
Therefore, \(i=1,2,3\) and \(j=1,2\).
Given formula:
\[ a_{ij}=\frac12|i-3j|. \]
The six entries are:
\[ a_{11}=1,\quad a_{12}=\frac52,\quad a_{21}=\frac12,\quad a_{22}=2,\quad a_{31}=0,\quad a_{32}=\frac32. \]
Required matrix:
\[ \boxed{ A= \begin{bmatrix} 1&\frac52\ \frac12&2\ 0&\frac32 \end{bmatrix}} \]

▦

Types of Matrices

📋

Table of Contents

12 sections

1 🗺️ Overview ›
2 🗂️ Square Matrix ›
3 ⚖️ Comparison of Matrix Types ›
4 🔑 How to Identify the Type of a Matrix Quickly ›
5 ✏️ Example ›
6 🔢 Important Formulae and Conditions ›
7 🌟 Significance for CBSE Board Examinations ›
8 ❌ Common Mistakes ›
9 📋 Case Study: Classification of Matrices ›
10 ⚡ Exam Tip ›
11 ⚡ Quick Revision ›
12 🔑 Key Takeaway ›

🗺️ Overview

Matrices can be classified according to their order, number of rows and columns, diagonal elements and the pattern of their entries. Understanding these classifications is essential for NCERT Class 12 Mathematics Chapter 3 because several later concepts, including transpose, symmetric matrices, skew-symmetric matrices, matrix multiplication and determinants, depend on recognising the structure of a matrix.

The most important types of matrices in this section are:

Column Matrix
Row Matrix
Square Matrix
Diagonal Matrix
Scalar Matrix
Identity Matrix
Zero or Null Matrix

A useful way to remember their relationship is:

\[ \boxed{ \text{Identity Matrix} \subset \text{Scalar Matrix} \subset \text{Diagonal Matrix} \subset \text{Square Matrix} } \]

This hierarchy is important: every identity matrix is scalar, every scalar matrix is diagonal, and every diagonal matrix is square. The converse statements are generally not true.

🗂️ Types of Matrices

Column Matrix Row Matrix Square Matrix Diagonal Matrix Scalar Matrix

Identity Matrix & Zero Matrix

⚖️ Comparison of Matrix Types

Type	Defining Condition	Typical Order	Example
Column Matrix	Exactly one column	\(m\times1\)	\(\begin{bmatrix}1\\2\\3\end{bmatrix}\)
Row Matrix	Exactly one row	\(1\times n\)	\(\begin{bmatrix}1&2&3\end{bmatrix}\)
Square Matrix	Rows = columns	\(n\times n\)	\(\begin{bmatrix}1&2\\3&4\end{bmatrix}\)
Diagonal Matrix	\(a_{ij}=0\) for \(i\neq j\)	\(n\times n\)	\(\begin{bmatrix}2&0\\0&5\end{bmatrix}\)
Scalar Matrix	All diagonal entries equal	\(n\times n\)	\(\begin{bmatrix}3&0\\0&3\end{bmatrix}\)
Identity Matrix	Diagonal entries 1; others 0	\(n\times n\)	\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)
Zero Matrix	Every element is zero	\(m\times n\)	\(\begin{bmatrix}0&0\\0&0\end{bmatrix}\)

🔑 How to Identify the Type of a Matrix Quickly

For examination questions, use the following decision process:

Count the rows and columns. If there is one column, it is a column matrix. If there is one row, it is a row matrix.
Check whether rows equal columns. If yes, it is a square matrix.
Check the non-diagonal entries. If all are zero, it is diagonal.
Check the diagonal entries. If all diagonal entries are equal, it is scalar.
Check whether the common diagonal value is 1. If yes, it is an identity matrix.
Check whether every entry is zero. If yes, it is a zero matrix.

✏️ Example

Question

Identify the type of Matrix

\[ A= \begin{bmatrix} 5&0&0\\ 0&5&0\\ 0&0&5 \end{bmatrix}. \]

It has three rows and three columns, so it is square. All non-diagonal elements are zero, so it is diagonal. Its three diagonal elements are equal to 5, so it is scalar.

\[A\text{ is a square, diagonal and scalar matrix.}\]

It is not an identity matrix because the diagonal entries are \(5\), not \(1\).

Question

Identify the Matrix Type

\[ B= \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix}. \]

\(B\) is:

a square matrix;
a diagonal matrix;
a scalar matrix with \(k=1\);
an identity matrix of order 3.

\[ \boxed{B=I_3} \]

Question

A Diagonal Matrix That Is Not Scalar

\[ C= \begin{bmatrix} 2&0&0\\ 0&5&0\\ 0&0&7 \end{bmatrix}. \]

\(C\) is diagonal because all non-diagonal entries are zero. However, \(2,5,7\) are not equal. Therefore, it is not scalar.

\[ \boxed{C\text{ is diagonal but not scalar.}} \]

Question

Zero Matrix

\[ O= \begin{bmatrix} 0&0\\ 0&0 \end{bmatrix}. \]

Every entry is zero. Hence \(O\) is a zero matrix. Since it is square, it is also a diagonal matrix and a scalar matrix with \(k=0\).

\[ \boxed{O=0I_2} \]

It is not an identity matrix.

🔢 Important Formulae and Conditions

Column matrix:
\[ \boxed{A=[a_{ij}]_{m\times1}} \]
Row matrix:
\[ \boxed{A=[a_{ij}]_{1\times n}} \]
Square matrix:
\[ \boxed{m=n} \]
Diagonal matrix:
\[ \boxed{a_{ij}=0,\quad i\neq j} \]
Scalar matrix:
\[ \boxed{A=kI_n} \]
Identity matrix:
\[ \boxed{I_n=[\delta_{ij}]} \]
where
\[ \delta_{ij}= \begin{cases} 1,&i=j,\\ 0,&i\neq j. \end{cases} \]
Zero matrix:
\[ \boxed{a_{ij}=0\quad\text{for all }i,j} \]

🌟 Significance for CBSE Board Examinations

Questions based on the types of matrices are often used to test conceptual understanding rather than lengthy calculations. A question may ask students to identify a matrix, determine an unknown parameter so that a matrix becomes scalar or identity, or distinguish between diagonal, scalar and identity matrices.

Students should be able to explain why a matrix belongs to a particular category instead of merely memorising its name.

Significance for Competitive Entrance Examinations

Competitive examinations frequently exploit the hierarchy between matrix types. A matrix may satisfy several classifications simultaneously.

For example,

\[ I_3= \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} \]

is simultaneously:

\[ \boxed{ \text{Identity} \Rightarrow \text{Scalar} \Rightarrow \text{Diagonal} \Rightarrow \text{Square}} \]

Recognising this hierarchy can eliminate several answer choices rapidly in multiple-choice questions.

❌ Common Mistakes

Calling every square matrix diagonal: A square matrix can have non-zero non-diagonal elements.
Calling every diagonal matrix scalar: A diagonal matrix is scalar only when all its diagonal elements are equal.
Calling every scalar matrix identity: It is identity only when the common diagonal value is \(1\).
Assuming a zero matrix must be square: A zero matrix can have any order \(m\times n\).
Confusing row and column matrices:
\[ 1\times n \]
is a row matrix, whereas
\[ n\times1 \]
is a column matrix.

📋 Case Study: Classification of Matrices

Consider the matrices

\[ A= \begin{bmatrix} 3&0&0\\ 0&3&0\\ 0&0&3 \end{bmatrix}, \qquad B= \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix}, \]

\[ C= \begin{bmatrix} 2&0&0\\ 0&4&0\\ 0&0&2 \end{bmatrix}, \qquad D= \begin{bmatrix} 0&0\\ 0&0 \end{bmatrix}. \]

Answer the following:

Which matrix is an identity matrix?
Which matrix is scalar but not identity?
Which matrix is diagonal but not scalar?
Which matrix is a zero matrix?

Solution

\(B\) has all diagonal elements equal to \(1\) and all other elements zero. Hence,

\[ \boxed{B\text{ is an identity matrix}.} \]

\(A\) has equal diagonal elements \(3\), but they are not \(1\). Hence,

\[ \boxed{A\text{ is scalar but not identity}.} \]

\(C\) is diagonal, but its diagonal elements \(2,4,2\) are not all equal. Hence,

\[ \boxed{C\text{ is diagonal but not scalar}.} \]

Every element of \(D\) is zero. Hence,

\[ \boxed{D\text{ is a zero matrix}.} \]

⚡ Exam Tip

⚡ Quick Revision

Matrix Type	Condition to Remember
Column Matrix	Exactly one column
Row Matrix	Exactly one row
Square Matrix	Rows = columns
Diagonal Matrix	\(a_{ij}=0\) for \(i\neq j\)
Scalar Matrix	Diagonal entries are equal
Identity Matrix	Diagonal entries are all \(1\), others \(0\)
Zero Matrix	Every entry is \(0\)

🔑 Key Takeaway

A column matrix has exactly one column and order \(m\times1\).
A row matrix has exactly one row and order \(1\times n\).
A square matrix has equal numbers of rows and columns.
A diagonal matrix has zero entries outside its principal diagonal.
A scalar matrix is a diagonal matrix whose diagonal elements are all equal.
An identity matrix is a scalar matrix with common diagonal element \(1\).
A zero matrix has every element equal to zero and may be rectangular or square.
The important hierarchy is
\[ \boxed{ \text{Identity}\subset\text{Scalar}\subset\text{Diagonal}\subset\text{Square}} \]
Classification of matrices is based on structure, so a single matrix can simultaneously satisfy several definitions.

▦

Equality of Matrices

📋

Table of Contents

17 sections

1 📘 Definition ›
2 📌 Understanding Corresponding Elements ›
3 🧭 Conditions for Equality of Matrices ›
4 📌 Finding Unknowns Using Equality of Matrices ›
5 ✏️ Finding Unknowns ›
6 ✏️ Matrix Equality with Multiple Variables ›
7 📌 Equality and Number of Elements ›
8 📎 Equality and Transpose: An Important Distinction ›
9 🏷️ Properties of Equality of Matrices ›
10 🧰 Exam Method for Matrix Equality Questions ›
11 🌟 Significance for CBSE Board Examinations ›
12 📋 CBSE Case Study / HOTS ›
13 ❌ Common Mistakes ›
14 ⚡ Exam Tip ›
15 🔢 Key Formula ›
16 ⚡ Quick Revision ›
17 🔑 Key Takeaway ›

📘 Definition

Two matrices \(A=[a_{ij}]\) and \(B=[b_{ij}]\) are said to be equal matrices if and only if they satisfy both of the following conditions:

Same order: \(A\) and \(B\) must have the same number of rows and the same number of columns.
Equal corresponding elements: Every element of \(A\) must be equal to the corresponding element of \(B\).

In mathematical notation,

\[A=B \iff \begin{cases} \text{Order of }A=\text{Order of }B,\\ a_{ij}=b_{ij}\quad\text{for every valid }i,j. \end{cases}\]

📌 Note

Understanding Corresponding Elements

Two elements are called corresponding elements when they occupy the same position in their respective matrices.

Thus, \(a_{11}\) corresponds to \(b_{11}\), \(a_{12}\) corresponds to \(b_{12}\), \(a_{21}\) corresponds to \(b_{21}\), and so on.

\[ \boxed{a_{ij}=b_{ij}\quad\text{for all }i,j} \]

Equality of matrices is therefore a position-by-position comparison. It is not enough for two matrices to contain the same numbers; those numbers must occur in the same corresponding positions.

🧭 Condition

Conditions for Equality of Matrices

🧭

Conditions for Equality of Matrices

Condition 1: Same Order

\[ A\text{ is of order }m\times n \]

and

\[ B\text{ is of order }p\times q, \]

then \(A=B\) is possible only if

\[ \boxed{m=p,\quad n=q}. \]

Matrices having different orders can never be equal.

Condition 2: Equal Corresponding Elements

Once the orders are the same, every corresponding entry must be equal:

\[ \boxed{ a_{11}=b_{11},\quad a_{12}=b_{12},\quad a_{21}=b_{21},\quad\ldots } \]

If even one corresponding pair is unequal, the matrices are not equal.

Example 1: Equal Matrices

Consider

\[ A= \begin{bmatrix} 2&5\\ 1&2 \end{bmatrix}, \qquad B= \begin{bmatrix} 2&5\\ 1&2 \end{bmatrix}. \]

Both matrices have order \(2\times2\), and

\[ a_{11}=b_{11}=2, \]

\[ a_{12}=b_{12}=5, \]

\[ a_{21}=b_{21}=1, \]

\[ a_{22}=b_{22}=2. \]

Hence all corresponding elements are equal.

\[ \boxed{A=B} \]

Example 2: Different Orders

Consider

\[A= \begin{bmatrix} 1&2\\ 3&4 \end{bmatrix}, \quad B= \begin{bmatrix} 1&2&3\\ 4&5&6 \end{bmatrix}.\]

Here,

\[\text{Order of }A=2\times2\]

\[\text{Order of }B=2\times3.\]

Since their orders are different,

\[ \boxed{A\neq B}\]

There is no need to compare the individual entries once different orders have been established.

Example 3: Same Elements but Different Positions

Consider

\[ A= \begin{bmatrix} 4&5\\ 2&1 \end{bmatrix}, \qquad B= \begin{bmatrix} 1&2\\ 4&5 \end{bmatrix}\]

Both matrices have order \(2\times2\), but their corresponding entries are not equal:

\[ a_{11}=4\neq1=b_{11}\]

Therefore,

\[ \boxed{A\neq B}\]

Although both matrices contain the same four numbers \(1,2,4,5\), they are not equal because the numbers occur in different positions.

📌 Note

Finding Unknowns Using Equality of Matrices

✏️ Example

Finding Unknowns

\[\begin{bmatrix} x+2&5\\ 7&y-3 \end{bmatrix} = \begin{bmatrix} 8&5\\ 7&4 \end{bmatrix}\]

find \(x\) and \(y\)

Equating corresponding elements:

\[ x+2=8 \]

\[ x=6 \]

and

\[ y-3=4 \]

\[ y=7. \]

Therefore,

\[ \boxed{x=6,\qquad y=7}\]

Why This Method Works

Matrix equality reduces a matrix equation to a system of scalar equations. If

\[ A=B, \]

then every corresponding entry must satisfy the same equality.

\[A=B\Rightarrow a_{ij}=b_{ij}\text{ for every }i,j\]

This principle is frequently used in CBSE and competitive examination questions.

✏️ Example

Matrix Equality with Multiple Variables

\[ \begin{bmatrix} x+y&2x-y\\ 3&x-y \end{bmatrix} = \begin{bmatrix} 7&5\\ 3&1 \end{bmatrix}, \]

find \(x\) and \(y\).

From the corresponding entries,

\[ x+y=7 \]

\[ 2x-y=5\]

Adding the equations:

\[3x=12\]

\[x=4\]

Substituting \(x=4\) into \(x+y=7\):

\[4+y=7\]

\[y=3\]

The remaining entry verifies the result:

\[x-y=4-3=1\]

Hence,

\[\boxed{x=4,\quad y=3}\]

📌 Equality and Number of Elements

If two matrices are equal, they must have the same order and consequently the same number of elements.

For example, if \(A=B\) and \(A\) is \(3\times4\), then \(B\) must also be \(3\times4\). Therefore, both contain \(3\times4=12\) elements.

\[\boxed{A=B\Rightarrow\text{same order}\Rightarrow\text{same number of elements}}\]

Same Elements Does Not Mean Equal Matrices

A common conceptual trap is to think that two matrices are equal when they contain the same numbers. This is false.

For example,

\[A= \begin{bmatrix} 1&2\\ 3&4 \end{bmatrix}\]

\[B= \begin{bmatrix} 4&3\\ 2&1 \end{bmatrix}\]

Both contain the numbers \(1,2,3,4\), but

\[a_{11}=1\neq4=b_{11}\]

Therefore,

\[\boxed{A\neq B}\]

📎 Equality and Transpose: An Important Distinction

A matrix and its transpose generally have different arrangements of elements. Even when they have the same order, they need not be equal.

For example,

\[A= \begin{bmatrix} 1&2\\ 3&4 \end{bmatrix}, \qquad A^T= \begin{bmatrix} 1&3\\ 2&4 \end{bmatrix}\]

Since

\[2\neq3\]

we have

\[\boxed{A\neq A^T}\]

Later, matrices satisfying \(A=A^T\) will be called symmetric matrices.

🏷️ Properties of Equality of Matrices

Properties

Reflexive Property

Every matrix is equal to itself:

\[\boxed{A=A}\]

Symmetric Property

\[A=B,\]

then

\[\boxed{B=A}\]

Transitive Property

\[A=B\]

and

\[B=C,\]

then

\[\boxed{A=C}\]

🧰 Exam Method for Matrix Equality Questions

When asked whether two matrices are equal, use this sequence:

Compare the orders.
If the orders differ, immediately conclude \(A\neq B\).
If the orders are the same, compare corresponding entries.
If every corresponding pair is equal, conclude \(A=B\).
If even one corresponding pair differs, conclude \(A\neq B\).

🌟 Significance for CBSE Board Examinations

Equality of matrices is a high-value foundational concept because it is frequently used to find unknown variables. A typical CBSE question may provide two matrices containing algebraic expressions and ask for unknown constants. The solution is obtained by equating corresponding entries and solving the resulting equations.

Students should remember that matrix equality is an element-wise condition, not merely a comparison of the overall appearance of two matrices.

Significance for Competitive Entrance Examinations

Competitive questions may combine matrix equality with algebraic equations, functions, parameters or special matrix types. A rapid comparison of dimensions and corresponding elements can substantially reduce calculation time.

A particularly useful observation is:

\[ \boxed{ A=B\Rightarrow \text{same order and equal corresponding entries}} \]

📋 Case Study

CBSE Case Study / HOTS

Let

\[ A= \begin{bmatrix} x+1&2y\\ 3x-y&5 \end{bmatrix} \]

and

\[ B= \begin{bmatrix} 5&8\\ 9&5 \end{bmatrix}. \]

If \(A=B\), determine \(x\) and \(y\).

Solution

From the first corresponding entries,

\[ x+1=5 \]

\[ x=4. \]

From the second corresponding entries,

\[ 2y=8 \]

\[ y=4. \]

Verify the third corresponding entries:

\[ 3x-y=3(4)-4=8. \]

But the corresponding entry of \(B\) is \(9\). Therefore, the given matrices cannot be equal for any \(x,y\).

\[ \boxed{\text{No values of }x,y\text{ satisfy }A=B.} \]

This is a useful HOTS observation: not every matrix equation involving unknowns has a solution. All corresponding-entry equations must be simultaneously consistent.

❌ Common Mistakes

Comparing only the number of elements: Equal matrices must have the same order, not merely the same number of entries.
Ignoring positions: The same set of numbers in a different arrangement does not make two matrices equal.
Comparing non-corresponding elements: Compare \(a_{ij}\) only with \(b_{ij}\).
Solving only some equations: When variables are involved, all corresponding entries must satisfy the equality.
Forgetting the order condition: Matrices of different orders can never be equal.

⚡ Exam Tip

🔢 Key Formula

⚡ Quick Revision

Requirement	Condition
Order	Same number of rows and columns
Corresponding entries	\(a_{ij}=b_{ij}\)
Different order	Always \(A\neq B\)
One unequal corresponding entry	Always \(A\neq B\)
All corresponding entries equal	\(A=B\)

🔑 Key Takeaway

Two matrices are equal if and only if they have the same order and equal corresponding elements.
Matrix equality is checked position by position.
Two matrices containing the same numbers can still be unequal if the numbers occupy different positions.
Matrix equations can be converted into ordinary algebraic equations by equating corresponding entries.
All resulting equations must be simultaneously satisfied.
The fundamental criterion is:
\[ \boxed{A=B\iff a_{ij}=b_{ij}\text{ for every }i,j} \]
together with equality of orders.

▦

Example 4

📋

Table of Contents

8 sections

1 ❓ Question ›
2 💡 Concept ›
3 🗺️ Roadmap ›
4 🧩 Solution ›
5 🌟 Important Exam Insight ›
6 ⚡ Exam Tip ›
7 ❌ Common Mistakes ›
8 🌟 CBSE Board Exam Significance ›

❓ Question

\[ \begin{bmatrix} x+3 & z+4 & 2y-7\\ -6 & a-1 & 0\\ b-3 & -21 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 6 & 3y-2\\ -6 & -3 & 2c+2\\ 2b+4 & -21 & 0 \end{bmatrix}, \]

find the values of \(a,b,c,x,y\) and \(z\).

💡 Concept

🗺️ Roadmap

Compare the corresponding entries of the two matrices.
Form one algebraic equation from each pair containing an unknown.
Solve the resulting equations individually.
Substitute the obtained values into the original matrices to verify the result.

🧩 Solution

Step 1: Compare the First Row

Comparing the first entries:

\[\begin{aligned} x+3&=0\\&=-3 \end{aligned}\]
Comparing the second entries:

\[\begin{aligned} z+4=6\\&=2 \end{aligned}\]
Comparing the third entries:

\[ 2y-7=3y-2 \]
Rearranging,
\[\begin{aligned}2y-3y&=-2+7\\-y&=5\\y&=-5 \end{aligned}\]

Step 2: Compare the Second Row

The first entries are already equal:

\[ -6=-6 \]
Comparing the second entries:

\[ a-1=-3 \]

\[ a=-2 \]

\[ \boxed{a=-2} \]
Comparing the third entries:

\[ 0=2c+2 \]

\[ 2c=-2 \]

\[ \boxed{c=-1} \]

Step 3: Compare the Third Row

Comparing the first entries:

\[ b-3=2b+4 \]
Rearranging,
\[\begin{aligned} b-2b&=4+3\\-b&=7\\b&=-7\end{aligned}\]
The second and third corresponding entries are already identical:

\[ -21=-21 \]

\[ 0=0 \]

Step 4: Collecting the Values

Variable	Corresponding-Entry Equation	Value
\(x\)	\(x+3=0\)	\(\boxed{-3}\)
\(z\)	\(z+4=6\)	\(\boxed{2}\)
\(y\)	\(2y-7=3y-2\)	\(\boxed{-5}\)
\(a\)	\(a-1=-3\)	\(\boxed{-2}\)
\(c\)	\(2c+2=0\)	\(\boxed{-1}\)
\(b\)	\(b-3=2b+4\)	\(\boxed{-7}\)

Verification

We obtained

\[ \boxed{ a=-2,\quad b=-7,\quad c=-1,\quad x=-3,\quad y=-5,\quad z=2 } \]
Substitute these values into the left-hand matrix:

\[ \begin{aligned} x+3&=-3+3=0,\\ z+4&=2+4=6,\\ 2y-7&=2(-5)-7=-17,\\ a-1&=-2-1=-3,\\ 2c+2&=2(-1)+2=0,\\ b-3&=-7-3=-10. \end{aligned} \]
Therefore, the left-hand matrix becomes

\[ \begin{bmatrix} 0&6&-17\\ -6&-3&0\\ -10&-21&0 \end{bmatrix}. \]
Now substitute the values into the right-hand matrix:

\[ 3y-2=3(-5)-2=-17\]

\[ 2b+4=2(-7)+4=-10\]

\[ 2c+2=2(-1)+2=0\]
Hence the right-hand matrix also becomes

\[ \begin{bmatrix} 0&6&-17\\ -6&-3&0\\ -10&-21&0 \end{bmatrix}\]
Thus the equality is verified.

\[\begin{bmatrix} 0&6&-17\\ -6&-3&0\\ -10&-21&0 \end{bmatrix} = \begin{bmatrix} 0&6&-17\\ -6&-3&0\\ -10&-21&0 \end{bmatrix}\]

\[ \boxed{ a=-2,\qquad b=-7,\qquad c=-1,\qquad x=-3,\qquad y=-5,\qquad z=2 } \]

🌟 Important Exam Insight

⚡ Exam Tip

❌ Common Mistakes

Comparing non-corresponding entries: Compare entries at exactly the same row and column positions.
Sign error in \(b-3=2b+4\):
\[ b-2b=4+3 \]
gives
\[ -b=7 \]
and therefore \(b=-7\).
Sign error in the \(y\)-equation:
\[ 2y-7=3y-2 \]
gives
\[ -y=5, \]
hence
\[ y=-5. \]
Not verifying the answer: Substitution into both matrices provides a quick check and can reveal an algebraic sign error.

🌟 CBSE Board Exam Significance

This is a standard application of equality of matrices. Questions of this form test whether students understand that corresponding entries must be equal. It is also an efficient way of converting a matrix equation into a system of algebraic equations.

Competitive Exam Significance

In entrance examinations, several unknowns may be embedded in a matrix, with only selected entries requiring comparison. The fastest approach is to identify the positions containing variables and immediately equate the corresponding entries.

▦

Addition of Matrices

📋

Table of Contents

17 sections

1 🗺️ Overview ›
2 📘 Definition ›
3 🧭 Necessary Condition for Matrix Addition ›
4 💡 Concept: Corresponding Elements ›
5 ✏️ Example ›
6 🏷️ Properties of Matrix Addition ›
7 📄 Zero Matrix And Matrix Addition ›
8 📌 Negative of a Matrix ›
9 📄 Connection With Matrix Subtraction ›
10 ✏️ Worked Concept Example: Verify Commutativity ›
11 🌟 Significance for CBSE Board Examinations ›
12 🌟 Significance for Competitive Entrance Examinations ›
13 ❌ Common Mistakes ›
14 ⚡ Exam Tip ›
15 ⚡ Quick Revision ›
16 📋 CBSE Case Study / HOTS ›
17 🔑 Key Takeaway ›

🗺️ Overview

Matrix addition is one of the fundamental operations on matrices. It is performed by adding the corresponding elements of two matrices. The operation is possible only when the two matrices have the same order.

📘 Definition

Let

\[ A=[a_{ij}]_{m\times n} \]

and

\[ B=[b_{ij}]_{m\times n} \]

be two matrices of the same order \(m\times n\). Their sum is defined as the matrix

\[ C=[c_{ij}]_{m\times n}, \]

where each element of \(C\) is obtained by adding the corresponding elements of \(A\) and \(B\).

\[\boxed{c_{ij}=a_{ij}+b_{ij}}\]

Therefore,

\[\boxed{A+B=[a_{ij}+b_{ij}]_{m\times n}}\]

🧭 Necessary Condition for Matrix Addition

🧭

Condition Two matrices can be added if and only if they have the same order.

\[\boxed{ A_{m\times n}+B_{p\times q}\text{ is defined only when }m=p\text{ and }n=q }\]

For example, a \(2\times3\) matrix can be added to another \(2\times3\) matrix, but it cannot be directly added to a \(3\times2\), \(2\times2\) or \(3\times3\) matrix.

General Formula

Let

\[ A=[a_{ij}]_{m\times n}, \quad B=[b_{ij}]_{m\times n}. \]

Then

\[\boxed{A+B=[a_{ij}+b_{ij}]_{m\times n}}\]

In expanded form, for a \(3\times3\) matrix:

\[ A= \begin{bmatrix} a_{11}&a_{12}&a_{13}\\ a_{21}&a_{22}&a_{23}\\ a_{31}&a_{32}&a_{33} \end{bmatrix} \]

\[ B= \begin{bmatrix} b_{11}&b_{12}&b_{13}\\ b_{21}&b_{22}&b_{23}\\ b_{31}&b_{32}&b_{33} \end{bmatrix} \]

Therefore,

\[ \boxed{ A+B= \begin{bmatrix} a_{11}+b_{11}&a_{12}+b_{12}&a_{13}+b_{13}\\ a_{21}+b_{21}&a_{22}+b_{22}&a_{23}+b_{23}\\ a_{31}+b_{31}&a_{32}+b_{32}&a_{33}+b_{33} \end{bmatrix}} \]

Notice that the \((i,j)\)-th element of \(A+B\) is obtained from only the \((i,j)\)-th elements of \(A\) and \(B\).

💡 Concept: Corresponding Elements

✏️ Example

Question

Find \(A+B\) where

\[ A= \begin{bmatrix} 2&-1\\ 3&4 \end{bmatrix}, \qquad B= \begin{bmatrix} 5&6\\ -2&1 \end{bmatrix}. \]

Both matrices are of order \(2\times2\), so addition is defined.

\[ \begin{aligned} A+B &= \begin{bmatrix} 2+5&-1+6\\ 3+(-2)&4+1 \end{bmatrix}\\ &= \boxed{ \begin{bmatrix} 7&5\\ 1&5 \end{bmatrix}} \end{aligned} \]

Question

Find Sum of

\[ A= \begin{bmatrix} 1&2&3\\ 4&5&6 \end{bmatrix} \]

and

\[ B= \begin{bmatrix} 1&2\\ 3&4\\ 5&6 \end{bmatrix}. \]

Here,

\[\text{Order of }A=2\times3\]

\[\text{Order of }B=3\times2\]

Since

\[ 2\times3\neq3\times2 \]

as matrix dimensions, \(A+B\) is not defined.

\[\boxed{A+B\text{ is not defined}}\]

Question

Addition with Algebraic Entries

\[ A= \begin{bmatrix} x&2y\\ 3&z \end{bmatrix}, \qquad B= \begin{bmatrix} 2&5\\ 4&7 \end{bmatrix}, \]

then

\[ \begin{aligned} A+B &= \begin{bmatrix} x+2&2y+5\\ 3+4&z+7 \end{bmatrix}\\ &= \boxed{ \begin{bmatrix} x+2&2y+5\\ 7&z+7 \end{bmatrix}}. \end{aligned} \]

🏷️ Properties of Matrix Addition

📌

Note Matrix addition inherits the familiar algebraic properties of ordinary addition. If \(A\), \(B\) and \(C\) are matrices of the same order, then the following properties hold.

Properties

1. Closure Property

The sum of two matrices of the same order is again a matrix of that same order.

\[ \boxed{ A,B\in M_{m\times n} \Rightarrow A+B\in M_{m\times n}} \]

2. Commutative Property

Matrix addition is commutative:

\[ \boxed{A+B=B+A} \]

This follows from the commutative property of scalar addition:

\[ a_{ij}+b_{ij}=b_{ij}+a_{ij}. \]

3. Associative Property

Matrix addition is associative:

\[\boxed{(A+B)+C=A+(B+C)}\]

Element-wise,

\[ (a_{ij}+b_{ij})+c_{ij} = a_{ij}+(b_{ij}+c_{ij}) \]

4. Additive Identity

The zero matrix acts as the additive identity:

\[\boxed{A+O=O+A=A}\]

where \(\mathrm{O}\) is the zero matrix of the same order as \(\mathrm{A}\).

5. Additive Inverse

For every matrix \(A=[a_{ij}]\), its additive inverse is

\[-A=[-a_{ij}]\]

Therefore,

\[\boxed{A+(-A)=O}\]

6. Cancellation Property

If \(A,B,C\) are matrices of the same order and

\[A+C=B+C\]

then

\[\boxed{A=B}\]

🗒️ Zero Matrix And Matrix Addition

The zero matrix plays the same role in matrix addition as the number \(0\) does in ordinary addition.

\[A+O=A\]

For example,

\[ \begin{bmatrix} 2&-1\\ 4&5 \end{bmatrix} + \begin{bmatrix} 0&0\\ 0&0 \end{bmatrix} = \begin{bmatrix} 2&-1\\ 4&5 \end{bmatrix}. \]

📌 Negative of a Matrix

The negative of a matrix is obtained by changing the sign of every element.

\[ A= \begin{bmatrix} 2&-3\\ 5&1 \end{bmatrix} \]

then

\[ \boxed{ -A= \begin{bmatrix} -2&3\\ -5&-1 \end{bmatrix}} \]

Consequently,

\[ A+(-A)= \begin{bmatrix} 0&0\\ 0&0 \end{bmatrix} \]

🗒️ Connection With Matrix Subtraction

Matrix subtraction is defined using matrix addition and the additive inverse:

\[\boxed{A-B=A+(-B)}\]

Therefore, corresponding elements are subtracted:

\[ \boxed{ A-B=[a_{ij}-b_{ij}]_{m\times n}} \]

This operation is also possible only when \(A\) and \(B\) have the same order.

✏️ Worked Concept Example: Verify Commutativity

Let

\[ A= \begin{bmatrix} 1&2\\ 3&4 \end{bmatrix}, \qquad B= \begin{bmatrix} 5&6\\ 7&8 \end{bmatrix} \]

Then

\[ A+B= \begin{bmatrix} 6&8\\ 10&12 \end{bmatrix} \]

Also,

\[ B+A= \begin{bmatrix} 6&8\\ 10&12 \end{bmatrix}. \]

Hence,

\[\boxed{A+B=B+A}\]

🌟 Significance for CBSE Board Examinations

Matrix addition is a basic operation that frequently appears as a direct computational question as well as within questions involving matrix equations. Students should be able to identify whether addition is defined, calculate the sum correctly and use equality of matrices to determine unknown quantities.

A typical board-exam strategy is:

\[ \boxed{ \text{Check order} \rightarrow \text{add corresponding entries} \rightarrow \text{simplify} \rightarrow \text{verify if required} } \]

🌟 Significance for Competitive Entrance Examinations

Competitive entrance questions often test whether a student can distinguish between operations that are defined and those that are not. The fastest check for matrix addition is the order.

For example:

\[ (3\times4)+(3\times4) \]

is defined, whereas

\[ (3\times4)+(4\times3) \]

is not defined.

The numerical values of the entries are irrelevant until the dimension check has been completed.

❌ Common Mistakes

Adding matrices of different orders: Matrix addition requires identical dimensions.
Adding non-corresponding elements: Always add \(a_{ij}\) to \(b_{ij}\).
Changing the order: The sum of two \(m\times n\) matrices is also \(m\times n\).
Confusing addition with multiplication: Matrix addition is element-wise; matrix multiplication follows a completely different rule.
Sign errors: Carefully preserve negative signs while adding algebraic or numerical entries.
Using the wrong matrix in an equality question: If the equation is \(A=B\), compare corresponding entries. Do not add the two sides.

⚡ Exam Tip

Check the order before performing any calculation.
Remember:
\[ \boxed{(m\times n)+(m\times n)=m\times n}. \]
Use a position-by-position approach to avoid misplaced entries.
For algebraic entries, simplify each corresponding sum carefully.
In multiple-choice questions, checking matrix dimensions can eliminate incorrect options immediately.
Do not confuse matrix addition with matrix multiplication. The condition for multiplication is different from the condition for addition.

⚡ Quick Revision

Concept	Key Result
Condition for addition	Both matrices must have the same order
Element of sum	\(c_{ij}=a_{ij}+b_{ij}\)
General form	\(A+B=[a_{ij}+b_{ij}]\)
Order of sum	Same as the order of \(A\) and \(B\)
Commutative property	\(A+B=B+A\)
Associative property	\((A+B)+C=A+(B+C)\)
Additive identity	\(A+O=A\)
Additive inverse	\(A+(-A)=O\)
Subtraction	\(A-B=A+(-B)\)

📋 CBSE Case Study / HOTS

Two departments record the number of items processed during two shifts. Their data is represented by

\[ A= \begin{bmatrix} 12&15\\ 18&20 \end{bmatrix}, \qquad B= \begin{bmatrix} 8&10\\ 7&12 \end{bmatrix}. \]

Here the rows represent the two departments and the columns represent the two shifts. The combined data is represented by \(A+B\).

Calculate the combined matrix.

Solution

\[ \begin{aligned} A+B &= \begin{bmatrix} 12+8&15+10\\ 18+7&20+12 \end{bmatrix}\\ &= \boxed{ \begin{bmatrix} 20&25\\ 25&32 \end{bmatrix}}. \end{aligned} \]

Thus, matrix addition provides a compact mathematical method for combining corresponding categories of data.

🔑 Key Takeaway

Matrix addition is defined only for matrices having the same order.
Corresponding elements are added:
\[ \boxed{(A+B)_{ij}=a_{ij}+b_{ij}}. \]
The order of \(A+B\) is the same as the order of \(A\) and \(B\).
Matrix addition is commutative and associative.
The zero matrix is the additive identity:
\[ \boxed{A+O=A}. \]
The additive inverse of \(A\) is \(-A\):
\[ \boxed{A+(-A)=O}. \]
Matrix subtraction is defined through addition:
\[ \boxed{A-B=A+(-B)}. \]
In matrix-equation questions, equality of corresponding entries is used to determine unknown variables.

▦

Example 5

📋

Table of Contents

5 sections

1 ❓ Question ›
2 💡 Concept ›
3 🗺️ Roadmap ›
4 🧩 Solution ›
5 📌 Important Note About This Example ›

❓ Question

Find the values of \(a,b,c\) and \(d\) from

\[ \boxed{ \begin{bmatrix} 2a+b&a-2b\\ 5c-d&4c+3d \end{bmatrix} = \begin{bmatrix} 4&-3\\ 11&24 \end{bmatrix}} \]

💡 Concept

🗺️ Roadmap

Equate the corresponding entries of the two matrices.
Solve the first two equations for \(a\) and \(b\).
Solve the remaining two equations for \(c\) and \(d\).
Substitute the answers back into the original matrix equation for verification.

🧩 Solution

Step 1: Find \(a\) and \(b\)

From the first corresponding entries,

\[2a+b=4\]
Hence,

\[b=4-2a\]
From the second corresponding entries,

\[a-2b=-3\]
Substitute \(b=4-2a\):

\[\begin{aligned}a-2(4-2a)&=-3\\a-8+4a&=-3\\5a&=5\\a&=1\end{aligned}\]
Therefore,

\[\begin{aligned}b&=4-2(1)\\&=2\end{aligned}\]

Step 2: Find \(c\) and \(d\)

From the third corresponding entries,

\[5c-d=11\]
Hence,

\[d=5c-11\]
From the fourth corresponding entries,

\[4c+3d=24\]
Substitute \(d=5c-11\):

\[\begin{aligned}4c+3(5c-11)&=24\\4c+15c-33&=24\\19c&=57\\c&=3\end{aligned}\]
Therefore,

\[\begin{aligned}d&=5(3)-11\\&=15-11\\&=4\end{aligned}\]

Verification

Substituting

\[ a=1,\quad b=2,\quad c=3,\quad d=4 \]
into the left-hand matrix:
\[ \begin{aligned} 2a+b&=2(1)+2=4,\\ a-2b&=1-2(2)=-3,\\ 5c-d&=5(3)-4=11,\\ 4c+3d&=4(3)+3(4)=24. \end{aligned} \]
Thus,

\[ \begin{bmatrix} 2a+b&a-2b\\ 5c-d&4c+3d \end{bmatrix} = \begin{bmatrix} 4&-3\\ 11&24 \end{bmatrix}. \]
Hence the values are

\[\boxed{a=1,\qquad b=2,\qquad c=3,\qquad d=4}.\]

📌 Important Note About This Example

▦

Exercise 6

📋

Table of Contents

9 sections

1 ❓ Question ›
2 💡 Concept ›
3 🗺️ Roadmap ›
4 🧩 Solution ›
5 👁️ Important Observation ›
6 ⚡ Exam Tip ›
7 ❌ Common Mistakes ›
8 🌟 CBSE Board Exam Significance ›
9 🔑 Key Takeaway ›

❓ Question

Given

\[ A= \begin{bmatrix} \sqrt{3}&1&-1\\ 2&3&0 \end{bmatrix}, \]

\[ B= \begin{bmatrix} 2&\sqrt{5}&1\\ -2&3&\frac12 \end{bmatrix}, \]

find \(A+B\).

💡 Concept

🗺️ Roadmap

Check that both matrices have the same order.

🧩 Solution

Step 1: Check the Orders

The matrix \(A\) is

\[ A= \begin{bmatrix} \sqrt{3}&1&-1\\ 2&3&0 \end{bmatrix} \]
and therefore has order \(2\times3\).
Similarly,

\[ B= \begin{bmatrix} 2&\sqrt{5}&1\\ -2&3&\frac12 \end{bmatrix} \]
also has order \(2\times3\).
Hence,

\[\boxed{A+B\text{ is defined}}\]

Step 2: Add Corresponding Elements

Using

\[(A+B)_{ij}=a_{ij}+b_{ij}\]
we obtain

\[ \begin{aligned} A+B &= \begin{bmatrix} \sqrt{3}&1&-1\\ 2&3&0 \end{bmatrix} + \begin{bmatrix} 2&\sqrt{5}&1\\ -2&3&\frac12 \end{bmatrix}\\[6pt] &= \begin{bmatrix} \sqrt{3}+2&1+\sqrt{5}&-1+1\\ 2+(-2)&3+3&0+\frac12 \end{bmatrix} \end{aligned} \]

Step 3: Simplify Each Entry

First row:

\[\sqrt{3}+2=\sqrt{3}+2\]

\[1+\sqrt{5}=1+\sqrt{5}\]

\[-1+1=0\]
Second row:

\[2+(-2)=0\]

\[3+3=6\]

\[0+\frac12=\frac12\]

\[ A+B= \begin{bmatrix} \sqrt{3}+2&1+\sqrt{5}&0\\ 0&6&\frac12 \end{bmatrix} \]

👁️ Important Observation

⚡ Exam Tip

In a matrix-addition question, always perform a quick order check before calculating. If the orders are different, addition is not defined. If the orders are the same, add entries directly at the same row-column positions.

\[ \boxed{ \text{Same order} \rightarrow \text{Add corresponding elements} \rightarrow \text{Simplify} } \]

❌ Common Mistakes

Adding non-corresponding entries: The first element of \(A\) must be added to the first element of \(B\), and so on.
Ignoring negative signs:
\[ -1+1=0,\qquad 2+(-2)=0. \]
Changing the order: A \(2\times3\) matrix added to another \(2\times3\) matrix produces a \(2\times3\) matrix.
Incorrectly combining unlike terms:
\[ \sqrt{3}+2 \]
cannot be simplified further, and neither can
\[ 1+\sqrt{5}. \]
Confusing matrix addition with matrix multiplication: Addition is element-wise; multiplication follows a different rule.

🌟 CBSE Board Exam Significance

This example directly tests the definition of matrix addition and the concept of corresponding elements. Such questions are useful for building accuracy before solving matrix equations and more advanced operations.

Competitive Entrance Exam Significance

In competitive examinations, a direct matrix-addition question can often be solved quickly once the order condition is recognised. Questions may also combine irrational numbers, fractions and negative values to test computational accuracy.

🔑 Key Takeaway

The essential rule illustrated by this example is

\[ \boxed{(A+B)_{ij}=a_{ij}+b_{ij}} \]

provided \(A\) and \(B\) have the same order. Every corresponding pair of entries is added independently, while the row-column structure of the matrices remains unchanged.

▦

Exercise 7

📋

Table of Contents

9 sections

1 ❓ Question ›
2 💡 Concept ›
3 🗺️ Roadmap ›
4 🧩 Solution ›
5 🤔 Why Is Also of Order ? ›
6 🌟 Board Examination Significance ›
7 ❌ Common Mistakes ›
8 ⚡ Exam Tip ›
9 📋 CBSE Case Study / HOTS ›

❓ Question

\[ A= \begin{bmatrix} 8&0\\ 4&-2\\ 3&6 \end{bmatrix}, \qquad B= \begin{bmatrix} 2&-2\\ 4&2\\ -5&1 \end{bmatrix}, \]

find the matrix \(X\), such that

\[\boxed{2A+3X=5B}\]

💡 Concept

🗺️ Roadmap

Write the given matrix equation.

🧩 Solution

Step 1: Isolate \(X\)

Subtract \(2A\) from both sides:

\[3X=5B-2A\]
Dividing both sides by \(3\):

\[X=\frac13(5B-2A)\]

Step 2: Calculate \(5B\)

\[ B= \begin{bmatrix} 2&-2\\ 4&2\\ -5&1 \end{bmatrix} \]
Multiplying every element of \(B\) by \(5\):

\[ \begin{aligned} 5B &= 5 \begin{bmatrix} 2&-2\\ 4&2\\ -5&1 \end{bmatrix}\\[4pt] &= \begin{bmatrix} 10&-10\\ 20&10\\ -25&5 \end{bmatrix}. \end{aligned} \]

Step 3: Calculate \(2A\)

\[ A= \begin{bmatrix} 8&0\\ 4&-2\\ 3&6 \end{bmatrix} \]
Therefore,

\[ \begin{aligned} 2A &= 2 \begin{bmatrix} 8&0\\ 4&-2\\ 3&6 \end{bmatrix}\\[4pt] &= \begin{bmatrix} 16&0\\ 8&-4\\ 6&12 \end{bmatrix}. \end{aligned} \]

Step 4: Calculate \(5B-2A\)

\[ \begin{aligned} 5B-2A &= \begin{bmatrix} 10&-10\\ 20&10\\ -25&5 \end{bmatrix} - \begin{bmatrix} 16&0\\ 8&-4\\ 6&12 \end{bmatrix}\\[6pt] &= \begin{bmatrix} 10-16&-10-0\\ 20-8&10-(-4)\\ -25-6&5-12 \end{bmatrix}\\[6pt] &= \begin{bmatrix} -6&-10\\ 12&14\\ -31&-7 \end{bmatrix} \end{aligned} \]

Step 5: Divide by \(3\)

\[ \begin{aligned} X &= \frac13 \begin{bmatrix} -6&-10\\ 12&14\\ -31&-7 \end{bmatrix}\\[6pt] &= \begin{bmatrix} -2&-\dfrac{10}{3}\\ 4&\dfrac{14}{3}\\ -\dfrac{31}{3}&-\dfrac{7}{3} \end{bmatrix} \end{aligned} \]

\[ X= \begin{bmatrix} -2&-\dfrac{10}{3}\\ 4&\dfrac{14}{3}\\ -\dfrac{31}{3}&-\dfrac{7}{3} \end{bmatrix} \]

🤔 Why Is \(X\) Also of Order \(3\times2\)?

Both \(A\) and \(B\) are of order \(3\times2\). Therefore, \(2A\) and \(5B\) are also \(3\times2\) matrices. Hence their difference is a \(3\times2\) matrix.

\[5B-2A\text{ is of order }3\times2\]

Multiplication by the scalar \(\frac13\) does not change the order. Therefore,

\[\boxed{X\text{ is of order }3\times2}\]

🌟 Board Examination Significance

This type of question tests several foundational skills simultaneously:

scalar multiplication of matrices,
matrix addition and subtraction,
algebraic manipulation of a matrix equation,
understanding of matrix order, and
accurate handling of negative numbers and fractions.

It is therefore an important model for questions involving unknown matrices in Class 12 Mathematics examinations.

Competitive Entrance Exam Significance

For competitive examinations, the key skill is recognising that the equation can be treated algebraically:

\[2A+3X=5B\]

\[\boxed{X=\frac{5B-2A}{3}}\]

Once this form is recognised, the problem reduces to scalar multiplication, matrix subtraction and division by a scalar. This avoids solving six separate equations unless required.

❌ Common Mistakes

Incorrect sign in subtraction:
\[ 10-16=-6, \]
while
\[ -25-6=-31. \]
Incorrectly subtracting a negative entry:
\[ 10-(-4)=14. \]
Dividing only some entries by \(3\): Every entry of the matrix must be divided by \(3\).
Confusing scalar multiplication with matrix multiplication: \(5B\) means that every entry of \(B\) is multiplied by \(5\).
Forgetting the order: \(X\) must have the same order as \(A\) and \(B\) in this equation.

⚡ Exam Tip

Isolate the unknown matrix before doing numerical calculations.
Calculate \(5B\) and \(2A\) separately to reduce arithmetic errors.
Keep the subtraction signs visible when calculating \(5B-2A\).
Divide the complete resulting matrix by \(3\).
For a high-confidence answer, verify by calculating \(2A+3X\).

📋 CBSE Case Study / HOTS

Suppose two data matrices \(A\) and \(B\) represent the output of two production stages. A combined production relationship is modelled by

\[ 2A+3X=5B. \]

Here \(X\) represents an unknown adjustment matrix. Rearranging gives

\[ \boxed{ X=\frac13(5B-2A) } \]

This illustrates how matrices can represent several related quantities simultaneously and how a matrix equation can be solved without determining each entry independently.

▦

Exercise 8

📋

Table of Contents

9 sections

1 ❓ Question ›
2 💡 Concept ›
3 🗺️ Roadmap ›
4 🧩 Solution ›
5 🔍 Conceptual Interpretation ›
6 🌟 Board Examination Significance ›
7 ❌ Common Mistakes ›
8 ⚡ Exam Tip ›
9 🔑 Key Takeaway ›

❓ Question

Find the matrices \(X\) and \(Y\), if

\[ X+Y= \begin{bmatrix} 5&2\\ 0&9 \end{bmatrix} \]

and

\[ X-Y= \begin{bmatrix} 3&6\\ 0&-1 \end{bmatrix}. \]

💡 Concept

This problem is the matrix analogue of solving two simultaneous equations:

\[X+Y=A\]

\[X-Y=B\]

Adding the two equations eliminates \(Y\):

\[(X+Y)+(X-Y)=A+B\]

\[2X=A+B\]

Hence,

\[\boxed{X=\frac12(A+B)}\]

Similarly, subtracting the second equation from the first eliminates \(X\):

\[(X+Y)-(X-Y)=A-B\]

\[2Y=A-B\]

Therefore,

\[\boxed{Y=\frac12(A-B)}\]

General Formula

\[X+Y=A\]

\[X-Y=B\]

then the matrices \(X\) and \(Y\) are given directly by

\[X=\frac{1}{2}(A+B)\]

\[Y=\frac{1}{2}(A-B)\]

These formulas are extremely useful in board examinations and competitive entrance examinations because they eliminate the need to solve for every matrix element separately.

🗺️ Roadmap

Represent the two given matrices by \(A\) and \(B\).

🧩 Solution

Step 1: Find \(X\)

We are given

\[X+Y=\begin{bmatrix}5&2\\ 0&9\end{bmatrix}\]

\[X-Y=\begin{bmatrix}3&6\\ 0&-1\end{bmatrix}\]
Add the two equations:

\[(X+Y)+(X-Y)=\begin{bmatrix}5&2\\ 0&9\end{bmatrix}+\begin{bmatrix}3&6\\ 0&-1\end{bmatrix}\]
On the left-hand side,

\[X+Y+X-Y=2X\]
Therefore,

\[2X=\begin{bmatrix}5+3&2+6\\ 0+0&9+(-1)\end{bmatrix}\]

\[2X=\begin{bmatrix}8&8\\ 0&8\end{bmatrix}\]
Dividing by \(2\):

\[X=\frac12\begin{bmatrix}8&8\\ 0&8\end{bmatrix}\]

\[\boxed{X=\begin{bmatrix}4&4\\ 0&4\end{bmatrix}}\]

Step 2: Find \(Y\)

We can use the first equation:

\[X+Y=\begin{bmatrix}5&2\\ 0&9\end{bmatrix}\]
Hence,

\[Y=\begin{bmatrix}5&2\\ 0&9\end{bmatrix}-X\]
Substituting the value of \(X\):

\[Y=\begin{bmatrix}5&2\\ 0&9\end{bmatrix}-\begin{bmatrix}4&4\\ 0&4\end{bmatrix}\]
Subtract corresponding elements:

\[Y=\begin{bmatrix}5-4&2-4\\ 0-0&9-4\end{bmatrix}\]

\[Y=\begin{bmatrix}1&-2\\ 0&5\end{bmatrix}\]

Alternative Method: Direct Formula for \(Y\)

A faster method is to subtract the second equation from the first:

\[(X+Y)-(X-Y)=\begin{bmatrix}5&2\\ 0&9\end{bmatrix}-\begin{bmatrix}3&6\\ 0&-1\end{bmatrix}\]
Since,
X+Y-X+Y=2Y
we obtain

\[2Y=\begin{bmatrix}5-3&2-6\\ 0-0&9-(-1)\end{bmatrix}\]

\[2Y=\begin{bmatrix}2&-4\\ 0&10\end{bmatrix}\]
Therefore,

\[Y=\frac12\begin{bmatrix}2&-4\\ 0&10\end{bmatrix}\]

\[Y=\begin{bmatrix}1&-2\\ 0&5\end{bmatrix}\]

\[X=\begin{bmatrix}4&4\\ 0&4\end{bmatrix}\]

\[Y=\begin{bmatrix}1&-2\\ 0&5\end{bmatrix}\]

🔍 Conceptual Interpretation

The equations \(X+Y=A\) and \(X-Y=B\) contain two unknown matrices. Adding them eliminates \(Y\), while subtracting them eliminates \(X\).

\[\boxed{\begin{array}{rcl}X+Y&=&A\\X-Y&=&B\\\hline2X&=&A+B\end{array}}\]

Similarly,

\[\boxed{\begin{array}{rcl}X+Y&=&A\\-(X-Y)&=&-B\\\hline2Y&=&A-B\end{array}}\]

Thus, this question is essentially a system of simultaneous equations in matrix form.

🌟 Board Examination Significance

This is an important Class 12 Matrices pattern because it combines matrix addition, subtraction and scalar multiplication with algebraic manipulation. A student who knows the formulas can solve the question quickly and present a logically complete solution.

The most important result to remember is:

\[ \boxed{ X+Y=A,\quad X-Y=B \Rightarrow X=\frac{A+B}{2},\quad Y=\frac{A-B}{2} }. \]

Competitive Entrance Exam Significance

This pattern is particularly useful in time-bound examinations. Instead of calculating the entries of \(X\) and \(Y\) independently, use the sum-difference identities immediately.

The recognition pattern is:

\[ \boxed{ \text{Sum}+\text{Difference}\Rightarrow2X } \]

\[ \boxed{ \text{Sum}-\text{Difference}\Rightarrow2Y } \]

❌ Common Mistakes

Forgetting to divide by \(2\): After adding the equations, the result is \(2X\), not \(X\).
Using the wrong sign for \(Y\):
\[ (X+Y)-(X-Y)=2Y. \]
Incorrect subtraction of negative entries:
\[ 9-(-1)=10. \]
Subtracting the matrices in the wrong order: To obtain \(Y\), use \(A-B\), not \(B-A\).
Ignoring matrix order: Matrix addition and subtraction require matrices of the same order.

⚡ Exam Tip

If you see both \(X+Y\) and \(X-Y\), immediately think of the sum-difference formulas.
Add the equations to find \(X\).
Subtract the equations to find \(Y\).
Divide the resulting matrix by \(2\) only after performing the addition or subtraction.
Verify both \(X+Y\) and \(X-Y\) if time permits.

🔑 Key Takeaway

Whenever two matrices are given in the form

\[ X+Y=A \]

and

\[ X-Y=B, \]

treat them as simultaneous matrix equations. Adding eliminates \(Y\), while subtracting eliminates \(X\):

\[ \boxed{ \begin{aligned} X&=\frac{A+B}{2},\\ Y&=\frac{A-B}{2}. \end{aligned}} \]

▦

Example 9

📋

Table of Contents

9 sections

1 ❓ Question ›
2 💡 Concept ›
3 🗺️ Roadmap ›
4 🧩 Solution ›
5 🌟 Board Examination Significance ›
6 ❌ Common Mistakes ›
7 ⚡ Exam Tip ›
8 📋 CBSE Case Study / HOTS ›
9 🔑 Key Takeaway ›

❓ Question

Find the values of \(x\) and \(y\) from the following matrix equation:

\[2\begin{bmatrix}x&5\\ 7&y-3\end{bmatrix}+\begin{bmatrix}3&-4\\ 1&2\end{bmatrix}=\begin{bmatrix}7&6\\ 15&14\end{bmatrix} \]

💡 Concept

This question combines two fundamental operations:

Scalar multiplication of a matrix
Addition of matrices

A scalar multiplies every element of the matrix. Therefore,

\[k\begin{bmatrix}a&b\\ c&d\end{bmatrix}=\begin{bmatrix}ka&kb\\ kc&kd\end{bmatrix}\]

After performing the scalar multiplication, corresponding elements are added and then compared with the corresponding entries of the matrix on the right-hand side.

🗺️ Roadmap

Multiply the first matrix by \(2\).

🧩 Solution

Step 1: Perform Scalar Multiplication

\[2\begin{bmatrix}x&5\\ 7&y-3\end{bmatrix}=\begin{bmatrix}2x&10\\ 14&2y-6\end{bmatrix}\]
Hence the original equation becomes

\[\begin{bmatrix}2x&10\\ 14&2y-6\end{bmatrix}+\begin{bmatrix}3&-4\\ 1&2\end{bmatrix}=\begin{bmatrix}7&6\\ 15&14\end{bmatrix}\]

Step 2: Add Corresponding Elements

\[\begin{aligned}&\begin{bmatrix}2x&10\\ 14&2y-6\end{bmatrix}+\begin{bmatrix}3&-4\\ 1&2\end{bmatrix}\\[6pt] &=\begin{bmatrix}2x+3&10+(-4)\\ 14+1&(2y-6)+2\end{bmatrix}.\end{aligned} \]
Simplifying:

\[10+(-4)=6\]

\[14+1=15\]

\[(2y-6)+2=2y-4\]
Therefore,

\[\begin{bmatrix}2x+3&6\\ 15&2y-4\end{bmatrix}=\begin{bmatrix}7&6\\ 15&14\end{bmatrix}\]

Step 3: Compare Corresponding Entries

Since the two matrices are equal, corresponding elements must be equal.
From the \((1,1)\)-th entries:
\[2x+3=7\]
Therefore,

\[2x=7-3\]

\[2x=4\]

\[\boxed{x=2}\]
From the \((2,2)\)-th entries:

\[2y-4=14\]
Therefore,

\[2y=14+4\]

\[2y=18\]

\[\boxed{y=9}\]

\[\boxed{x=2,\qquad y=9}\]

🌟 Significance

Board Examination Significance

This example is important because it combines three core ideas from the Matrices chapter: scalar multiplication, matrix addition and equality of matrices. It is representative of questions in which algebraic expressions are embedded inside matrix entries.

A reliable board-exam sequence is:

\[\boxed{\text{Scalar multiplication}\rightarrow\text{Matrix addition}\rightarrow\text{Equate corresponding entries}\rightarrow\text{Solve}}\]

Competitive Entrance Exam Significance

In competitive examinations, the question can often be solved by immediately identifying the entries containing the unknowns. There is no need to solve equations arising from entries that are already identical.

In this problem:

\[\boxed{(1,1)\text{ entry}\Rightarrow x}\]

\[\boxed{(2,2)\text{ entry}\Rightarrow y}\]

❌ Common Mistakes

Multiplying only selected entries by \(2\): The scalar must multiply every entry.
Sign error:
\[ 10+(-4)=6, \]
not \(14\).
Incorrect simplification of the \(y\)-entry:
\[ 2y-6+2=2y-4. \]
Comparing non-corresponding entries: Equality must be checked at identical row-column positions.
Forgetting to verify: Substituting \(x=2\) and \(y=9\) into the original equation is a quick way to confirm the result.

⚡ Exam Tip

Apply scalar multiplication before performing matrix addition.
Write every intermediate matrix clearly.
After obtaining the final matrix, compare corresponding entries.
Focus on entries containing unknowns; entries such as \(6=6\) and \(15=15\) are identities.
Check signs carefully when negative entries are involved.

📋 CBSE Case Study / HOTS

Suppose a mathematical model is represented by

\[2M+N=\begin{bmatrix}7&6\\15&14\end{bmatrix}\]

where

\[M=\begin{bmatrix}x&5\\7&y-3\end{bmatrix},\quad N=\begin{bmatrix}3&-4\\1&2\end{bmatrix}\]

The matrix equation becomes

\[\begin{bmatrix}2x+3&6\\15&2y-4\end{bmatrix}=\begin{bmatrix}7&6\\15&14\end{bmatrix}\]

Hence,

\[2x+3=7\]

\[2y-4=14\]

\[\boxed{x=2,\qquad y=9}\]

🔑 Key Takeaway

▦

Example 10

📋

Table of Contents

9 sections

1 ❓ Question ›
2 💡 Concept ›
3 🗺️ Roadmap ›
4 🧩 Solution ›
5 ⚡ Exam Tip ›
6 ❌ Common Mistakes ›
7 🌟 CBSE Board Examination Significance ›
8 📋 CBSE Case Study / HOTS Extension ›
9 🔑 Key Takeaway ›

❓ Question

Two farmers, Ramkishan and Gurcharan Singh, cultivate only three varieties of rice, namely Basmati, Permal and Naura. The sales (in rupees) of these varieties by the two farmers in the months of September and October are represented by the following matrices \(A\) and \(B\):

\[A=\begin{bmatrix}10000&20000&30000\\ 50000&30000&10000\end{bmatrix}\]

\[B=\begin{bmatrix}5000&10000&6000\\ 20000&10000&10000\end{bmatrix}\]

Here, matrix \(A\) represents the sales in September, while matrix \(B\) represents the sales in October. The rows represent the two farmers and the columns represent the three varieties of rice. Therefore:

Row	Farmer
1st row	Ramkishan
2nd row	Gurcharan Singh

Column	Rice Variety
1st column	Basmati
2nd column	Permal
3rd column	Naura

💡 Concept

This problem demonstrates how matrices can be used to represent and process real-world tabular data. Since \(A\) and \(B\) have the same order, \(2\times3\), their corresponding entries can be added or subtracted.

\[\boxed{\text{September sales}=A,\quad\text{October sales}=B}\]

Therefore:

\[\boxed{A+B=\text{combined sales of September and October}}\]

\[\boxed{A-B=\text{decrease in sales from September to October}}\]

For profit, a percentage is represented by scalar multiplication:

\[\boxed{\text{Profit}=2\%\times B=\frac{2}{100}B}\]

🗺️ Roadmap

🧩 Solution

Part (i): Combined Sales in September and October

The combined sales are obtained by adding the corresponding entries of \(A\) and \(B\):

\[\text{Combined Sales}=A+B\]
Therefore,

\[\begin{aligned}A+B&=\begin{bmatrix}10000&20000&30000\\ 50000&30000&10000\end{bmatrix}+\begin{bmatrix}5000&10000&6000\\ 20000&10000&10000\end{bmatrix}\\[6pt] &=\begin{bmatrix}10000+5000&20000+10000&30000+6000\\ 50000+20000&30000+10000&10000+10000\end{bmatrix}\\[6pt] &=\begin{bmatrix}15000&30000&36000\\70000&40000&20000\end{bmatrix}\end{aligned}\]
Interpretation of \(A+B\)
The first row corresponds to Ramkishan:
\[\begin{bmatrix}15000&30000&36000\end{bmatrix}\]
Therefore, Ramkishan's combined September and October sales were:
- Basmati: ₹15,000
- Permal: ₹30,000
- Naura: ₹36,000
The second row corresponds to Gurcharan Singh:

\[\begin{bmatrix}70000&40000&20000\end{bmatrix}\]
Therefore, Gurcharan Singh's combined September and October sales were:
- Basmati: ₹70,000
- Permal: ₹40,000
- Naura: ₹20,000

Part (ii): Decrease in Sales from September to October

Since \(A\) represents September and \(B\) represents October, the decrease from September to October is calculated as:

\[\boxed{\text{Decrease}=A-B}\]
Hence,

\[\begin{aligned}A-B&=\begin{bmatrix}10000&20000&30000\\ 50000&30000&10000\end{bmatrix}-\begin{bmatrix}5000&10000&6000\\ 20000&10000&10000\end{bmatrix}\\[6pt] &=\begin{bmatrix}10000-5000&20000-10000&30000-6000\\ 50000-20000&30000-10000&10000-10000\end{bmatrix}\\[6pt] &=\begin{bmatrix}5000&10000&24000\\ 30000&20000&0\end{bmatrix}\end{aligned}\]
Interpretation of the Decrease Matrix
The first row gives the decrease in Ramkishan's sales:
\[\begin{bmatrix}5000&10000&24000\end{bmatrix}\]
Thus:
- Decrease in Basmati sales = ₹5,000
- Decrease in Permal sales = ₹10,000
- Decrease in Naura sales = ₹24,000
The second row gives the decrease in Gurcharan Singh's sales:

\[\begin{bmatrix}30000&20000&0\end{bmatrix}\]
Thus:
- Decrease in Basmati sales = ₹30,000
- Decrease in Permal sales = ₹20,000
- Decrease in Naura sales = ₹0
The zero in the third column of the second row means that Gurcharan Singh's Naura sales remained unchanged:
10000-10000=0

Part (iii): Profit at 2% on October Sales

The October sales are represented by \(B\). Therefore, the profit at \(2\%\) is

\[\text{Profit}=2\%\text{ of }B\]

\[\text{Profit}=\frac{2}{100}B\]
Thus,

\[\begin{aligned}\text{Profit}&=\frac{2}{100}\begin{bmatrix}5000&10000&6000\\ 20000&10000&10000\end{bmatrix}\\[6pt] &=\begin{bmatrix}100&200&120\\ 400&200&200\end{bmatrix}\end{aligned}\]

\[\text{Profit Matrix}=\begin{bmatrix}100&200&120\\ 400&200&200\end{bmatrix}\]
Interpretation of the Profit Matrix
The first row represents Ramkishan:
\[\begin{bmatrix}100&200&120\end{bmatrix}\]
Therefore, Ramkishan receives:
- Basmati: ₹100
- Permal: ₹200
- Naura: ₹120
The second row represents Gurcharan Singh:

\[\begin{bmatrix}400&200&200\end{bmatrix}\]
Therefore, Gurcharan Singh receives:
- Basmati: ₹400
- Permal: ₹200
- Naura: ₹200

Quantity	Matrix
Combined September and October sales	\[ \begin{bmatrix} 15000&30000&36000\\ 70000&40000&20000 \end{bmatrix} \]
Decrease from September to October	\[ \begin{bmatrix} 5000&10000&24000\\ 30000&20000&0 \end{bmatrix} \]
2% profit on October sales	\[ \begin{bmatrix} 100&200&120\\ 400&200&200 \end{bmatrix} \]

⚡ Exam Tip

Before performing any operation, identify what each matrix represents. In this question, \(A\) represents September and \(B\) represents October. Therefore, the direction of change matters:

\[ \boxed{ A-B=\text{September sales}-\text{October sales} } \]

Reversing the order would produce the negative of the decrease matrix.

❌ Common Mistakes

Interchanging September and October: Always identify which matrix belongs to which month before subtracting.
Adding non-corresponding entries: Each farmer-variety combination must retain its row and column position.
Using \(A\) instead of \(B\) for October profit: Since \(B\) represents October, the profit is
\[ \frac{2}{100}B. \]
Forgetting percentage conversion:
\[ 2\%=\frac{2}{100}=0.02. \]
Ignoring the meaning of a zero: A zero in the decrease matrix means that sales did not decrease for that particular farmer and rice variety.

🌟 CBSE Board Examination Significance

This example is especially important because it demonstrates a common application-based question in which a matrix represents real-world data. It tests whether a student can translate a verbal situation into matrix operations and correctly interpret the resulting entries.

The three operations to remember are:

\[ \boxed{ \begin{aligned} \text{Combined sales}&=A+B,\\ \text{Decrease}&=A-B,\\ \text{Profit at 2%}&=\frac{2}{100}B. \end{aligned}} \]

Competitive Entrance Exam Significance

Application-based matrix questions in entrance examinations frequently encode information about sales, production, expenditure, population, inventory or other datasets. The important skill is to map the rows and columns correctly before performing the matrix operation.

Once the representation is understood, the calculations are straightforward:

\[ \boxed{ \text{Represent data} \rightarrow \text{identify operation} \rightarrow \text{perform element-wise calculation} \rightarrow \text{interpret result} } \]

📋 CBSE Case Study / HOTS Extension

Suppose the two farmers want to know which rice variety experienced the largest total decrease in sales from September to October.

From the decrease matrix:

\[ A-B= \begin{bmatrix} 5000&10000&24000\\ 30000&20000&0 \end{bmatrix}. \]

The total decrease for each variety is obtained by adding the corresponding column entries.

For Basmati:

\[ 5000+30000=35000. \]

For Permal:

\[ 10000+20000=30000. \]

For Naura:

\[ 24000+0=24000. \]

Therefore, Basmati experienced the greatest combined decrease:

\[ \boxed{\text{Greatest decrease =₹}35,000\text{ in Basmati sales}} \]

Further HOTS: Farmer-Wise Decrease

The total decrease for Ramkishan is

\[ 5000+10000+24000=39000. \]

\[ \boxed{\text{Ramkishan's total decrease =₹}39,000} \]

For Gurcharan Singh:

\[ 30000+20000+0=50000. \]

\[ \boxed{\text{Gurcharan Singh's total decrease =₹}50,000} \]

Thus, Gurcharan Singh experienced the greater reduction in sales between September and October.

🔑 Key Takeaway

Matrices provide an efficient way to represent structured real-world data.
\(A+B\) combines corresponding sales figures.
\(A-B\) measures the decrease when \(A\) represents the earlier month and \(B\) the later month.
A percentage of a matrix is calculated using scalar multiplication.
The meaning of every matrix entry depends on its row and column position.
Matrix operations are useful in analysing sales, production, finance, inventory and other structured datasets.

▦

Multiplication of Matrices

📋

Table of Contents

19 sections

1 📘 Definition ›
2 📜 Most Important Rule: Compatibility of Orders ›
3 ✏️ Example of Compatible Orders ›
4 ✏️ Example of Incompatible Orders ›
5 📐 Derivation of the Formula for Matrix Multiplication ›
6 🤔 Why Is the Number of Columns of Equal to the Number of Rows of ? ›
7 ✏️ Example: Multiplying Two Matrices ›
8 🏷️ Important Properties of Matrix Multiplication ›
9 📌 Matrix Multiplication and Dimensions ›
10 ⭐ Special Case: Square Matrices ›
11 ✏️ Example: and Need Not Be Equal ›
12 🤔 Why Matrix Multiplication Is Important ›
13 📎 Matrix Multiplication as Composition of Transformations ›
14 🌟 Board Examination Significance ›
15 ❌ Common Mistakes ›
16 ⚡ Exam Tip ›
17 📋 CBSE Case Study / HOTS ›
18 ⚡ Quick Revision ›
19 🔑 Key Takeaway ›

📘 Definition

The product of two matrices is defined only when the number of columns of the first matrix is equal to the number of rows of the second matrix. Let

\[A=[a_{ij}]_{m\times n}\]

and

\[B=[b_{jk}]_{n\times p}\]

Since \(A\) has \(n\) columns and \(B\) has \(n\) rows, the product \(AB\) is defined. The resulting matrix \(C=AB\) has order \(m\times p\).

\[(m\times n)(n\times p)=m\times p\]

The \((i,k)\)-th element of \(C=AB\) is obtained by multiplying the elements of the \(i\)-th row of \(A\) with the corresponding elements of the \(k\)-th column of \(B\) and then adding those products.

\[c_{ik}=\sum_{j=1}^{n}a_{ij}b_{jk}\]

Therefore,

\[AB=C=[c_{ik}]_{m\times p}\]

Matrix Multiplication in Simple Words

Matrix multiplication is commonly described as the row-by-column rule.

To find any entry of \(AB\):

Choose one row from the first matrix \(A\).
Choose one column from the second matrix \(B\).
Multiply corresponding elements.
Add all the resulting products.

\[\text{Entry of }AB=\text{corresponding row of }A\cdot\text{corresponding column of }B\]

📜 Most Important Rule: Compatibility of Orders

📜 Rules

Suppose

\[A_{m\times n}\quad\text{and}\quad B_{r\times p}\]

The product \(AB\) exists if and only if

\[\boxed{n=r}\]

If the product exists, its order is determined by the outer dimensions:

\[A_{m\times n}B_{n\times p}=AB_{m\times p}\]

A useful memory rule is:

\[\boxed{\underbrace{(m\times n)}_{\text{first matrix}}\underbrace{(n\times p)}_{\text{second matrix}}=\underbrace{(m\times p)}_{\text{result}}}\]

The two inner numbers must be equal, and the two outer numbers give the order of the product.

✏️ Example of Compatible Orders

\[A_{2\times3}B_{3\times4}\]

Since the number of columns of \(A\) is \(3\), which equals the number of rows of \(B\), the product is defined.

\[\boxed{A_{2\times3}B_{3\times4}=C_{2\times4}}\]

✏️ Example of Incompatible Orders

\[A_{2\times3}B_{2\times4}\]

Here, \(A\) has \(3\) columns whereas \(B\) has \(2\) rows. Since \(3\neq2\), the product \(AB\) is not defined.

\[\boxed{(2\times3)(2\times4)\text{ is not defined}}\]

📐 Derivation of the Formula for Matrix Multiplication

Let

\[A=[a_{ij}]_{m\times n}\quad\text{and}\]

\[B=[b_{jk}]_{n\times p}\]

The \(i\)-th row of \(A\) is

\[[a_{i1}\quad a_{i2}\quad\cdots\quad a_{in}]\]

and the \(k\)-th column of \(B\) is

\[\begin{bmatrix}b_{1k}\\b_{2k}\\\vdots\\b_{nk}\end{bmatrix}\]

Their row-by-column product is

\[\begin{aligned}c_{ik}&=a_{i1}b_{1k}+a_{i2}b_{2k}+a_{i3}b_{3k}+\cdots+a_{in}b_{nk}\\ &=\boxed{\sum_{j=1}^{n}a_{ij}b_{jk}}\end{aligned}\]

Hence,

\[AB=[c_{ik}]_{m\times p}\]

🤔 Why Is the Number of Columns of \(A\) Equal to the Number of Rows of \(B\)?

Each row of \(A\) contains \(n\) elements, while each column of \(B\) must also contain \(n\) elements so that the corresponding elements can be multiplied.

\[[a_{i1},a_{i2},\ldots,a_{in}]\]

can be multiplied with

\[\begin{bmatrix}b_{1k}\\b_{2k}\\\vdots\\b_{nk}\end{bmatrix}\]

because both contain exactly \(n\) elements.

This is the fundamental reason behind the compatibility condition for matrix multiplication.

✏️ Example: Multiplying Two Matrices

Consider

\[C=\begin{bmatrix}1&-1&2\\0&3&4\end{bmatrix}\]

and

\[D=\begin{bmatrix}2&7\\-1&1\\5&-9\end{bmatrix}\]

Step 1: Check Whether \(CD\) Is Defined

The order of \(C\) is

\[2\times3\]
and the order of \(D\) is

\[3\times2\]
Therefore,

\[(2\times3)(3\times2)\]
is defined because the inner dimensions are equal:

\[3=3\]
The resulting matrix will have order

\[\boxed{2\times2}\]

Step 2: Find the \((1,1)\)-th Element

Take the first row of \(C\):

\[[1\quad -1\quad2]\]
and the first column of \(D\):

\[\begin{bmatrix}2\\-1\\5\end{bmatrix}\]
Therefore,

\[\begin{aligned}(CD)_{11}&=(1)(2)+(-1)(-1)+(2)(5)\\&=2+1+10\\&=13\end{aligned}\]

\[\boxed{(CD)_{11}=13}\]

Step 3: Find the \((1,2)\)-th Element

Take the first row of \(C\) and the second column of \(D\):

\[[1\quad-1\quad2]\quad \text{and}\]

\[\begin{bmatrix}7\\1\\-9\end{bmatrix}\]
Hence,

\[\begin{aligned}(CD)_{12}&=(1)(7)+(-1)(1)+(2)(-9)\\&=7-1-18\\&=-12\end{aligned}\]

\[\boxed{(CD)_{12}=-12}\]

Step 4: Find the \((2,1)\)-th Element

Take the second row of \(C\):

\[[0\quad3\quad4]\]
and the first column of \(D\):
\[\begin{bmatrix}2\\-1\\5\end{bmatrix}\]
Therefore,

\[\begin{aligned}(CD)_{21}&=(0)(2)+(3)(-1)+(4)(5)\\&=0-3+20\\&=17\end{aligned}\]

\[\boxed{(CD)_{21}=17}\]

Step 5: Find the \((2,2)\)-th Element

Take the second row of \(C\) and the second column of \(D\):

\[[0\quad3\quad4]\quad\text{and}\]

\[\begin{bmatrix}7\\1\\-9\end{bmatrix}\]
Thus,

\[\begin{aligned}(CD)_{22}&=(0)(7)+(3)(1)+(4)(-9)\\&=0+3-36\\&=-33\end{aligned}\]

\[\boxed{(CD)_{22}=-33}\]

Final Product:

\[CD=\begin{bmatrix}13&-12\\17&-33\end{bmatrix}\]

📌

Note Notice that the result has order \(2\times2\), exactly as predicted by the order rule:

\[\boxed{(2\times3)(3\times2)=2\times2}\]

Complete Row-by-Column Calculation

\[\begin{aligned}CD&=\begin{bmatrix}1&-1&2\\ 0&3&4\end{bmatrix}\begin{bmatrix}2&7\\ -1&1\\ 5&-9\end{bmatrix}\\[8pt] &=\begin{bmatrix}(1)(2)+(-1)(-1)+(2)(5)&(1)(7)+(-1)(1)+(2)(-9)\\[5pt] (0)(2)+(3)(-1)+(4)(5)&(0)(7)+(3)(1)+(4)(-9)\end{bmatrix}\\[8pt] &=\begin{bmatrix}2+1+10&7-1-18\\ 0-3+20&0+3-36\end{bmatrix}\\[8pt] &=\begin{bmatrix}13&-12\\ 17&-33\end{bmatrix}\end{aligned}\]

🏷️ Important Properties of Matrix Multiplication

Properties

1. Closure Property

If \(A\) is \(m\times n\) and \(B\) is \(n\times p\), then \(AB\) is an \(m\times p\) matrix.

Therefore, whenever the product is defined, the result is again a matrix.

\[\boxed{A_{m\times n}B_{n\times p}=AB_{m\times p}}\]

2. Matrix Multiplication Is Generally Not Commutative

In ordinary arithmetic,

\[ab=ba\]

However, for matrices, it is generally true that

\[\boxed{AB\neq BA}\]

In fact, one of the products may exist while the other does not.

3. Associative Property

Whenever the relevant products are defined,

\[\boxed{(AB)C=A(BC)}\]

4. Distributive Property

Matrix multiplication distributes over matrix addition:

\[\boxed{A(B+C)=AB+AC}\]

and

\[\boxed{(A+B)C=AC+BC}\]

5. Multiplication by the Zero Matrix

Whenever the products are defined,

\[\boxed{A0=0,\qquad0A=0}\]

6. Multiplication by the Identity Matrix

If \(I\) is the identity matrix of the appropriate order, then

\[\boxed{AI=A,\qquad IA=A}\]

The identity matrix therefore plays the role of \(1\) in matrix multiplication.

📌 Matrix Multiplication and Dimensions

The following table is useful for quickly checking whether a product exists:

First Matrix	Second Matrix	Product	Defined?
\(2\times3\)	\(3\times4\)	\(2\times4\)	Yes
\(3\times2\)	\(2\times5\)	\(3\times5\)	Yes
\(4\times4\)	\(4\times4\)	\(4\times4\)	Yes
\(2\times3\)	\(2\times4\)	Not defined	No
\(3\times4\)	\(2\times3\)	Not defined	No

⭐ Special Case: Square Matrices

✏️ Example: \(AB\) and \(BA\) Need Not Be Equal

Let

\[A=\begin{bmatrix}1&1\\0&1\end{bmatrix},\]

\[B=\begin{bmatrix}1&0\\1&1\end{bmatrix}\]

Then

\[\begin{aligned}AB&=\begin{bmatrix}1&1\\ 0&1\end{bmatrix}\begin{bmatrix}1&0\\ 1&1\end{bmatrix}\\ &=\begin{bmatrix}2&1\\ 1&1\end{bmatrix}\end{aligned}\]

Whereas

\[\begin{aligned}BA&=\begin{bmatrix}1&0\\ 1&1\end{bmatrix}\begin{bmatrix}1&1\\ 0&1\end{bmatrix}\\ &=\begin{bmatrix}1&1\\ 1&2\end{bmatrix}\end{aligned}\]

Therefore,

\[\boxed{AB\neq BA}\]

🤔 Why Matrix Multiplication Is Important

Matrix multiplication is one of the most important operations in linear algebra. It provides a compact mathematical mechanism for combining transformations, systems of equations and structured datasets.

It is extensively used in:

solving systems of linear equations,
coordinate transformations in geometry,
computer graphics and image transformations,
statistics and data analysis,
economics and input-output models,
physics and engineering,
computer science and algorithms,
machine learning and neural networks,
cryptography, and
linear transformations and vector spaces.

📎 Matrix Multiplication as Composition of Transformations

A deeper interpretation of matrix multiplication comes from linear transformations. If a matrix \(B\) transforms a vector first and matrix \(A\) transforms the resulting vector, then the combined transformation is represented by \(AB\).

\[\boxed{\mathbf{x}\xrightarrow{B}B\mathbf{x}\xrightarrow{A}A(B\mathbf{x})=(AB)\mathbf{x}}\]

This also explains why the order of matrix multiplication matters. In general, performing transformation \(B\) followed by \(A\) is different from performing \(A\) followed by \(B\).

🌟 Board Examination Significance

Matrix multiplication is a high-value concept in Class 12 Mathematics because questions commonly test:

checking whether a product is defined,
finding the order of a product,
calculating individual entries,
finding the complete matrix product,
using matrix multiplication in matrix equations,
verifying properties such as associativity and distributivity, and
distinguishing matrix multiplication from scalar or element-wise multiplication.

A particularly important examination skill is checking dimensions before beginning the multiplication.

\[\text{Check dimensions first; calculate second}\]

Competitive Entrance Exam Significance

In JEE and other entrance examinations, matrix multiplication questions frequently test speed, dimensional reasoning and conceptual understanding rather than lengthy arithmetic.

The fastest approach is to inspect the dimensions:

\[(m\times n)(n\times p)\Rightarrow(m\times p)\]

If the inner dimensions do not match, stop immediately:

\[\boxed{(m\times n)(r\times p),\quad n\neq r\Rightarrow AB\text{ does not exist}}\]

❌ Common Mistakes

Checking the wrong dimensions: The columns of the first matrix must equal the rows of the second matrix.
Using column-by-column multiplication: Standard matrix multiplication follows the row-by-column rule.
Multiplying corresponding entries: This is not standard matrix multiplication.
Writing the wrong order of the result: The result has the outer dimensions.
Assuming \(AB=BA\): Matrix multiplication is generally non-commutative.
Changing the order of matrices: \(AB\) and \(BA\) are different products and may even have different dimensions.
Sign errors: Negative entries must be handled carefully during the row-by-column calculation.
Arithmetic errors: Calculate one entry at a time rather than attempting the entire product mentally.

⚡ Exam Tip

Write the orders of both matrices before multiplying.
Check that the inner dimensions are equal.
Write the order of the result before calculating:
\[ (2\times3)(3\times2)\Rightarrow2\times2. \]
Calculate each entry using one row and one column.
Keep negative signs inside parentheses while multiplying.
For competitive examinations, eliminate impossible products immediately using dimensions.

📋 CBSE Case Study / HOTS

A company stores production data in a \(2\times3\) matrix \(A\), while a pricing or transformation matrix \(B\) has order \(3\times2\). The combined calculation is represented by \(AB\).

Since

\[ (2\times3)(3\times2) \]

is compatible, the result is a \(2\times2\) matrix.

\[ \boxed{ AB\text{ exists and has order }2\times2 } \]

If the order of the second matrix were changed to \(2\times3\), then

\[ (2\times3)(2\times3) \]

would not be defined because \(3\neq2\).

HOTS Question

If \(A\) is of order \(3\times5\) and \(B\) is of order \(5\times4\), determine the order of \(AB\).

Since

\[ (3\times5)(5\times4), \]

the inner dimensions match. Therefore,

\[ \boxed{AB\text{ is of order }3\times4}. \]

Advanced Insight: Entry Formula

The compact formula

\[ \boxed{ (AB)_{ik}=\sum_{j=1}^{n}a_{ij}b_{jk} } \]

is fundamental. It tells us exactly which entries participate in calculating a particular entry of the product.

For example, for \(A_{2\times3}\) and \(B_{3\times2}\):

\[ (AB)_{12} = a_{11}b_{12} + a_{12}b_{22} + a_{13}b_{32}. \]

Thus, the first row of \(A\) is paired with the second column of \(B\).

⚡ Quick Revision

\[ A_{m\times n} \quad\text{and}\quad B_{n\times p}, \]

then

\[ \boxed{ AB=C_{m\times p} } \]

\[ \boxed{ c_{ik}=\sum_{j=1}^{n}a_{ij}b_{jk} } \]

Compatibility condition:

\[ \boxed{ \text{Columns of }A=\text{Rows of }B } \]

Order of product:

\[ \boxed{ (m\times n)(n\times p)=m\times p } \]

Important properties:

\[ \boxed{A(B+C)=AB+AC} \]

\[ \boxed{(A+B)C=AC+BC} \]

\[ \boxed{(AB)C=A(BC)} \]

\[ \boxed{AB\neq BA\text{ in general}} \]

\[ \boxed{AI=IA=A} \]

\[ \boxed{A0=0A=0} \]

🔑 Key Takeaway

Matrix multiplication is defined only when the columns of the first matrix equal the rows of the second matrix.
The product follows the row-by-column rule.
If \(A\) is \(m\times n\) and \(B\) is \(n\times p\), then \(AB\) is \(m\times p\).
Every entry of \(AB\) is a dot product of a row of \(A\) with a column of \(B\).
Matrix multiplication is generally not commutative:
\[ AB\neq BA. \]
Matrix multiplication is associative and distributive over matrix addition.
Always check dimensions before carrying out numerical multiplication.

▦

Example 11

📋

Table of Contents

10 sections

1 ❓ Question ›
2 💡 Concept ›
3 🗺️ Roadmap ›
4 🧩 Solution ›
5 🔍 Understanding the Result ›
6 📄 Row By Column Map ›
7 🌟 Board Examination Significance ›
8 ❌ Common Mistakes ›
9 ⚡ Exam Tip ›
10 🔑 Key Takeaway ›

❓ Question

Find \(AB\), if

\[A=\begin{bmatrix}6&9\\2&3\end{bmatrix},\]

\[B=\begin{bmatrix}2&6&0\\7&9&8\end{bmatrix}\]

💡 Concept

Matrix multiplication follows the row-by-column rule. To find an entry of \(AB\), multiply the elements of a row of \(A\) by the corresponding elements of a column of \(B\), and then add the products.

Before performing the multiplication, always check the orders of the matrices.

\[A:\;2\times2\]

\[B:\;2\times3\]

Therefore,

\[(2\times2)(2\times3)\]

is defined because the number of columns of \(A\) is equal to the number of rows of \(B\):

\[2=2\]

The resulting matrix has the order given by the outer dimensions:

\[\boxed{AB\text{ is of order }2\times3}\]

🗺️ Roadmap

Identify the orders of \(A\) and \(B\).
Verify that \(AB\) is defined.
Determine the order of the product.
Multiply each row of \(A\) by each column of \(B\).
Place the four resulting row-column products in their corresponding positions.

🧩 Solution

Step 1: Check Compatibility

\[A_{2\times2}B_{2\times3}\]
Since the inner dimensions are equal, the product exists:

\[(2\times2)(2\times3)=2\times3\]

Step 2: Write the Product

\[\begin{aligned}AB&=\begin{bmatrix}6&9\\ 2&3\end{bmatrix}\begin{bmatrix}2&6&0\\ 7&9&8\end{bmatrix}\end{aligned}\]
The first row of \(A\) is

\[[6\quad9]\]
and the three columns of \(B\) are

\[\begin{bmatrix}2\\ 7\end{bmatrix},\qquad\begin{bmatrix}6\\ 9\end{bmatrix},\qquad\begin{bmatrix}0\\ 8\end{bmatrix}\]

Step 3: Find the First Row of \(AB\)

First entry:

\[\begin{aligned}(AB)_{11}&=(6)(2)+(9)(7)\\ &=12+63\\ &=75\end{aligned}\]

\[\boxed{(AB)_{11}=75}\]
Second entry:

\[\begin{aligned}(AB)_{12}&=(6)(6)+(9)(9)\\ &=36+81\\ &=117\end{aligned}\]

\[\boxed{(AB)_{12}=117}\]
Third entry:

\[\begin{aligned}(AB)_{13}&=(6)(0)+(9)(8)\\ &=0+72\\ &=72\end{aligned}\]

\[\boxed{(AB)_{13}=72}\]
Therefore, the first row of \(AB\) is

\[\boxed{\begin{bmatrix}75&117&72\end{bmatrix}}\]

Step 4: Find the Second Row of \(AB\)

The second row of \(A\) is

\[[2\quad3]\]
First entry:

\[\begin{aligned}(AB)_{21}&=(2)(2)+(3)(7)\\ &=4+21\\ &=25\end{aligned}\]

\[\boxed{(AB)_{21}=25}\]
Second entry:

\[\begin{aligned}(AB)_{22}&=(2)(6)+(3)(9)\\ &=12+27\\&=39\end{aligned}\]

\[\boxed{(AB)_{22}=39}\]
Third entry:

\[\begin{aligned}(AB)_{23}&=(2)(0)+(3)(8)\\&=0+24\\&=24\end{aligned}\]

\[\boxed{(AB)_{23}=24}\]
Therefore, the second row of \(AB\) is

\[\boxed{\begin{bmatrix}25&39&24\end{bmatrix}}\]

Hence, Final Answer:

\[ \begin{aligned} AB &= \begin{bmatrix} 6&9\\ 2&3 \end{bmatrix} \begin{bmatrix} 2&6&0\\ 7&9&8 \end{bmatrix}\\[6pt] &= \begin{bmatrix} 6(2)+9(7)&6(6)+9(9)&6(0)+9(8)\\ 2(2)+3(7)&2(6)+3(9)&2(0)+3(8) \end{bmatrix}\\[6pt] &= \begin{bmatrix} 75&117&72\\ 25&39&24 \end{bmatrix}. \end{aligned} \]

\[ \boxed{ AB= \begin{bmatrix} 75&117&72\\ 25&39&24 \end{bmatrix}} \]

🔍 Understanding the Result

The result is a \(2\times3\) matrix because

\[ \boxed{ (2\times2)(2\times3)=2\times3. } \]

Notice that each of the three columns of \(B\) produces one column of the resulting matrix. Similarly, each of the two rows of \(A\) produces one row of the resulting matrix.

\[ \boxed{ \text{Rows of result}=\text{rows of }A } \]

\[ \boxed{ \text{Columns of result}=\text{columns of }B } \]

🗒️ Row By Column Map

The six entries of the product can be visualised as follows:

Entry	Row from \(A\)	Column from \(B\)	Calculation	Result
\((1,1)\)	\([6,9]\)	\([2,7]^T\)	\(6(2)+9(7)\)	\(75\)
\((1,2)\)	\([6,9]\)	\([6,9]^T\)	\(6(6)+9(9)\)	\(117\)
\((1,3)\)	\([6,9]\)	\([0,8]^T\)	\(6(0)+9(8)\)	\(72\)
\((2,1)\)	\([2,3]\)	\([2,7]^T\)	\(2(2)+3(7)\)	\(25\)
\((2,2)\)	\([2,3]\)	\([6,9]^T\)	\(2(6)+3(9)\)	\(39\)
\((2,3)\)	\([2,3]\)	\([0,8]^T\)	\(2(0)+3(8)\)	\(24\)

🌟 Board Examination Significance

This type of question directly tests the fundamental procedure of matrix multiplication. A complete answer should ideally show the matrix orders, the compatibility condition and the row-by-column calculations. This prevents errors and makes the solution mathematically transparent.

Competitive Entrance Exam Significance
Questions involving matrix products often test whether a student can quickly determine compatibility and calculate selected entries without unnecessarily expanding the entire product.

For example, if only \((AB)_{23}\) is required, there is no need to calculate all six entries:

\[ (AB)_{23} = \begin{bmatrix} 2&3 \end{bmatrix} \begin{bmatrix} 0\\ 8 \end{bmatrix} = 2(0)+3(8) = \boxed{24}. \]

❌ Common Mistakes

Multiplying corresponding entries: Matrix multiplication uses rows of \(A\) and columns of \(B\), not corresponding positions.
Ignoring dimensions: Always verify \(2=2\) before multiplying.
Writing a \(2\times2\) answer: The product is \(2\times3\), not \(2\times2\).
Mixing rows and columns: For \((AB)_{12}\), use row 1 of \(A\) and column 2 of \(B\).
Arithmetic errors:
\[ 6(6)+9(9)=36+81=117, \]
not \(126\).
Forgetting zero terms:
\[ 6(0)+9(8)=72. \]

⚡ Exam Tip

Write the order of each matrix first.
Check the inner dimensions.
Write the order of the result before calculating.
Calculate one row of the answer at a time.
Use parentheses around negative numbers.
For a question asking for only one entry, calculate only that entry.
Check that the final matrix contains exactly the expected number of entries.

🔑 Key Takeaway

\(AB\) is defined because the number of columns of \(A\) equals the number of rows of \(B\).
The product has order \(2\times3\).
Each entry is obtained using the row-by-column rule.
The first row of \(A\) determines the first row of \(AB\), and the second row of \(A\) determines the second row of \(AB\).
The three columns of \(B\) determine the three columns of \(AB\).
Individual entries can be calculated independently, which is useful in competitive examinations.

▦

Example 12

📋

Table of Contents

7 sections

1 ❓ Question ›
2 💡 Concept ›
3 🧩 Solution ›
4 🌟 Board Examination Significance ›
5 ❌ Common Mistakes ›
6 ⚡ Exam Tip ›
7 🔑 Key Takeaway ›

❓ Question

\[A=\begin{bmatrix}1&-2&3\\-4&2&5\end{bmatrix}\]

and

\[B=\begin{bmatrix}2&3\\4&5\\2&1\end{bmatrix},\]

find \(AB\) and \(BA\). Hence, show that

\[AB\neq BA\]

💡 Concept

This example illustrates one of the most important properties of matrix multiplication: matrix multiplication is not commutative in general.

In ordinary arithmetic,

\[ ab=ba, \]

but for matrices, we generally have

\[ \boxed{AB\neq BA}. \]

To demonstrate this, we must calculate both products separately. It is not sufficient to calculate only \(AB\).

🧩 Solution

Step 1: Identify the Orders

Matrix \(A\) has 2 rows and 3 columns:

\[A_{2\times3}\]
Matrix \(B\) has 3 rows and 2 columns:

\[B_{3\times2}\]
Therefore,

\[AB=(2\times3)(3\times2)\]
Since the inner dimensions are equal, \(AB\) is defined:

\[\boxed{AB\text{ is of order }2\times2}\]
Similarly,

\[BA=(3\times2)(2\times3)\]
Therefore, \(BA\) is also defined and has order

\[\boxed{BA\text{ is of order }3\times3}\]
Important Observation Before Calculation
We already know something significant:
\[AB\text{ is }2\times2,\quad BA\text{ is }3\times3\]
Hence, \(AB\) and \(BA\) do not even have the same order. Therefore, they cannot be equal.

\[\boxed{AB\neq BA}\]
However, since the question asks us to find both products, we calculate them explicitly. This also provides a complete verification.

Part A: Find \(AB\)

\[\begin{aligned}AB&=\begin{bmatrix}1&-2&3\\-4&2&5\end{bmatrix} \begin{bmatrix}2&3\\4&5\\2&1\end{bmatrix}\end{aligned}\]
Finding the \((1,1)\)-th Entry
Take the first row of \(A\) and the first column of \(B\):
\[[1\quad-2\quad3]\]

\[\begin{bmatrix}2\\4\\2\end{bmatrix}\]

\[\begin{aligned}(AB)_{11}&=(1)(2)+(-2)(4)+(3)(2)\\&=2-8+6\\&=0\end{aligned}\]

\[\boxed{(AB)_{11}=0}\]
Finding the \((1,2)\)-th Entry
Take the first row of \(A\) and the second column of \(B\):
\[[1\quad-2\quad3]\]

\[\begin{bmatrix}3\\5\\1\end{bmatrix}\]

\[\begin{aligned}(AB)_{12}&=(1)(3)+(-2)(5)+(3)(1)\\&=3-10+3\\&=-4\end{aligned}\]

\[\boxed{(AB)_{12}=-4}\]
Finding the \((2,1)\)-th Entry

\[\begin{aligned}(AB)_{21}&=(-4)(2)+(2)(4)+(5)(2)\\&=-8+8+10\\&=10\end{aligned}\]

\[\boxed{(AB)_{21}=10}\]
Finding the \((2,2)\)-th Entry

\[\begin{aligned}(AB)_{22}&=(-4)(3)+(2)(5)+(5)(1)\\&=-12+10+5\\&=3\end{aligned}\]

\[\boxed{(AB)_{22}=3}\]
Therefore,

\[\begin{aligned}AB&=\begin{bmatrix}1&-2&3\\ -4&2&5\end{bmatrix}\begin{bmatrix}2&3\\4&5\\2&1\end{bmatrix}\\[6pt] &=\begin{bmatrix}2-8+6&3-10+3\\ -8+8+10&-12+10+5\end{bmatrix}\\[6pt] &=\begin{bmatrix}0&-4\\10&3\end{bmatrix}\end{aligned}\]

Part B: Find \(BA\)

Now reverse the order of multiplication:

\[BA=\begin{bmatrix}2&3\\4&5\\2&1\end{bmatrix} \begin{bmatrix}1&-2&3\\-4&2&5\end{bmatrix}\]
The result will be a \(3\times3\) matrix because

\[(3\times2)(2\times3)=3\times3\]
First Row of \(BA\)
The first row of \(B\) is
\[[2\quad3]\]
Multiplying it with the three columns of \(A\):

\[\begin{aligned}(BA)_{11}&=(2)(1)+(3)(-4)\\&=2-12\\&=-10\end{aligned}\]

\[\begin{aligned}(BA)_{12}&=(2)(-2)+(3)(2)\\&=-4+6\\&=2\end{aligned}\]

\[\begin{aligned}(BA)_{13}&=(2)(3)+(3)(5)\\&=6+15\\&=21\end{aligned}\]
Hence, the first row is

\[[-10\quad2\quad21]\]
Second Row of \(BA\)
The second row of \(B\) is
\[[4\quad5]\]

\[\begin{aligned}(BA)_{21}&=(4)(1)+(5)(-4)\\&=4-20\\&=-16\end{aligned}\]

\[\begin{aligned}(BA)_{22}&=(4)(-2)+(5)(2)\\&=-8+10\\&=2\end{aligned}\]

\[\begin{aligned}(BA)_{23}&=(4)(3)+(5)(5)\\&=12+25\\&=37\end{aligned}\]
Hence, the second row is

\[[-16\quad2\quad37]\]
Third Row of \(BA\)
The third row of \(B\) is
\[[2\quad1]\]

\[\begin{aligned}(BA)_{31}&=(2)(1)+(1)(-4)\\&=2-4\\&=-2\end{aligned}\]

\[\begin{aligned}(BA)_{32}&=(2)(-2)+(1)(2)\\&=-4+2\\&=-2\end{aligned}\]

\[\begin{aligned}(BA)_{33}&=(2)(3)+(1)(5)\\&=6+5\\&=11\end{aligned}\]
Hence, the third row is

\[\boxed{[-2\quad-2\quad11]}\]

Therefore,

\[ \begin{aligned} BA &= \begin{bmatrix} 2&3\\ 4&5\\ 2&1 \end{bmatrix} \begin{bmatrix} 1&-2&3\\ -4&2&5 \end{bmatrix}\\[6pt] &= \begin{bmatrix} -10&2&21\\ -16&2&37\\ -2&-2&11 \end{bmatrix} \end{aligned} \]

Comparison of \(AB\) and \(BA\)

We obtained:

\[AB=\begin{bmatrix}0&-4\\10&3\end{bmatrix}\]
and

\[BA=\begin{bmatrix}-10&2&21\\-16&2&37\\-2&-2&11\end{bmatrix}\]
Their orders are different:

\[ \operatorname{ord}(AB)=2\times2 \]

\[ \operatorname{ord}(BA)=3\times3. \]
Two matrices can be equal only if they have the same order. Since

\[2\times2\neq3\times3\]
it immediately follows that

\[\boxed{AB\neq BA}\]

🌟 Board Examination Significance

This is a fundamental example for questions based on the non-commutative property of matrix multiplication. In board examinations, students may be asked to calculate \(AB\) and \(BA\), or simply determine whether \(AB=BA\).

Remember that the equality of matrices requires both:

the same order, and
equal corresponding entries.

Therefore, if the orders are different, comparison can stop immediately.

\[ \boxed{ \text{Different orders}\Rightarrow\text{matrices cannot be equal} } \]

Competitive Entrance Exam Significance

This example contains a useful time-saving technique. Before performing lengthy calculations, check the dimensions of \(AB\) and \(BA\).

\[ A_{2\times3},\quad B_{3\times2} \]

\[ AB_{2\times2},\quad BA_{3\times3}. \]

Hence, \(AB\neq BA\) can be concluded instantly.

\[ \boxed{ AB\neq BA\quad\text{because}\quad \operatorname{ord}(AB)\neq\operatorname{ord}(BA) } \]

❌ Common Mistakes

Assuming \(AB=BA\): This is true for ordinary numbers but not generally true for matrices.
Calculating \(BA\) using the row-column structure of \(AB\): The order of multiplication must be reversed completely.
Ignoring dimensions: The order of \(AB\) and \(BA\) can be different.
Comparing entries when the orders are different: Different-order matrices cannot be equal.
Sign errors: For example,
\[ (-2)(4)=-8, \]
not \(8\).

⚡ Exam Tip

🔑 Key Takeaway

Matrix multiplication is generally not commutative.
For this example, both \(AB\) and \(BA\) are defined.
\(AB\) has order \(2\times2\), whereas \(BA\) has order \(3\times3\).
Since matrices of different orders cannot be equal, \(AB\neq BA\).
Checking dimensions can sometimes prove non-commutativity without performing any multiplication.
The statement is not that \(AB\) and \(BA\) are always unequal; rather, \(AB=BA\) is not a general property of matrices.

▦

Example 13

📋

Table of Contents

8 sections

1 📄 Question ›
2 💡 Concept Used: Associative Law of Matrix Multiplication ›
3 🧩 Solution ›
4 📄 Dimension Analysis ›
5 🌟 Board Examination Significance ›
6 ❌ Common Mistakes ›
7 ⚡ Exam Tip ›
8 🔑 Key Takeaway ›

🗒️ Question

\[ A= \begin{bmatrix} 1&1&-1\\ 2&0&3\\ 3&-1&2 \end{bmatrix} \]

\[ B= \begin{bmatrix} 1&3\\ 0&2\\ -1&4 \end{bmatrix} \]

\[ C= \begin{bmatrix} 1&2&3&-4\\ 2&0&-2&1 \end{bmatrix}, \]

find \(A(BC)\) and \((AB)C\), and hence show that

\[{A(BC)=(AB)C\]

💡 Concept Used: Associative Law of Matrix Multiplication

Matrix multiplication satisfies the associative law. If the required products are defined, then

\[\boxed{(AB)C=A(BC)}\]

The associative law allows us to change the grouping of matrices without changing their order.

Thus,

\[\boxed{(AB)C=A(BC)}\]

but this should not be confused with the commutative law. Associativity does not permit changing the order of the matrices:

\[\boxed{(AB)C\neq(BA)C\text{ in general}}\]

🧩 Solution

Step 1: Check the Orders of the Matrices

\[A_{3\times3},\qquad B_{3\times2},\qquad C_{2\times4}\]
Therefore,

\[BC=(3\times2)(2\times4)=3\times4\]
and hence

\[A(BC)=(3\times3)(3\times4)=3\times4\]
On the other hand,

\[AB=(3\times3)(3\times2)=3\times2\]
and therefore

\[(AB)C=(3\times2)(2\times4)=3\times4\]
Thus, both sides of the associative identity are defined and both must have order \(3\times4\):

\[A(BC),\;(AB)C\text{ are both of order }3\times4\]

Part I: Find \(A(BC)\)

Step 1: Calculate \(BC\)

\[\begin{aligned}BC&=\begin{bmatrix}1&3\\0&2\\-1&4\end{bmatrix}\begin{bmatrix} 1&2&3&-4\\2&0&-2&1\end{bmatrix}\end{aligned}\]
The product has order \(3\times4\)

\[ \begin{aligned} BC &= \begin{bmatrix} (1)(1)+(3)(2) & (1)(2)+(3)(0) & (1)(3)+(3)(-2) & (1)(-4)+(3)(1) \\[4pt] (0)(1)+(2)(2) & (0)(2)+(2)(0) & (0)(3)+(2)(-2) & (0)(-4)+(2)(1) \\[4pt] (-1)(1)+(4)(2) & (-1)(2)+(4)(0) & (-1)(3)+(4)(-2) & (-1)(-4)+(4)(1) \end{bmatrix}\\[15pt] &= \begin{bmatrix} 1+6&2+0&3-6&-4+3\\ 0+4&0+0&0-4&0+2\\ -1+8&-2+0&-3-8&4+4 \end{bmatrix}\\[15pt] &= \begin{bmatrix} 7&2&-3&-1\\ 4&0&-4&2\\ 7&-2&-11&8 \end{bmatrix} \end{aligned} \]

Step 2: Calculate \(A(BC)\)

\[\begin{aligned}A(BC)&=\begin{bmatrix}1&1&-1\\2&0&3\\3&-1&2\end{bmatrix} \begin{bmatrix}7&2&-3&-1\\4&0&-4&2\\7&-2&-11&8\end{bmatrix}\end{aligned}\]
Calculate each entry using the row-by-column rule.
First Row
\[\begin{aligned}(A(BC))_{11}&=(1)(7)+(1)(4)+(-1)(7)\\&=7+4-7\\&=4\end{aligned}\]

\[\begin{aligned}(A(BC))_{12}&=(1)(2)+(1)(0)+(-1)(-2)\\&=2+0+2\\&=4\end{aligned}\]

\[\begin{aligned}(A(BC))_{13}&=(1)(-3)+(1)(-4)+(-1)(-11)\\&=-3-4+11\\&=4\end{aligned}\]

\[\begin{aligned}(A(BC))_{14}&=(1)(-1)+(1)(2)+(-1)(8)\\&=-1+2-8\\&=-7\end{aligned}\]
Hence, the first row is

\[\begin{bmatrix}4&4&4&-7\end{bmatrix}\]
Second Row

\[\begin{aligned}(A(BC))_{21}&=(2)(7)+(0)(4)+(3)(7)\\&=14+0+21\\&=35,\end{aligned}\]

\[\begin{aligned}(A(BC))_{22}&=(2)(2)+(0)(0)+(3)(-2)\\&=4+0-6\\&=-2\end{aligned}\]

\[\begin{aligned}(A(BC))_{23}&=(2)(-3)+(0)(-4)+(3)(-11)\\&=-6+0-33\\&=-39\end{aligned}\]

\[\begin{aligned}(A(BC))_{24}&=(2)(-1)+(0)(2)+(3)(8)\\&=-2+0+24\\&=22\end{aligned}\]
Hence, the second row is

\[\begin{bmatrix}35&-2&-39&22\end{bmatrix}\]
Third Row

\[\begin{aligned}(A(BC))_{31}&=(3)(7)+(-1)(4)+(2)(7)\\&=21-4+14\\&=31\end{aligned}\]

\[\begin{aligned}(A(BC))_{32}&=(3)(2)+(-1)(0)+(2)(-2)\\&=6+0-4\\&=2\end{aligned}\]

\[\begin{aligned}(A(BC))_{33}&=(3)(-3)+(-1)(-4)+(2)(-11)\\&=-9+4-22\\&=-27\end{aligned}\]

\[\begin{aligned}(A(BC))_{34}&=(3)(-1)+(-1)(2)+(2)(8)\\&=-3-2+16\\&=11\end{aligned}\]
Hence, the third row is

\[\begin{bmatrix}31&2&-27&11\end{bmatrix}\]
Therefore,

\[A(BC)=\begin{bmatrix}4&4&4&-7\\35&-2&-39&22\\31&2&-27&11\end{bmatrix}\]

Part II: Find \((AB)C\)

Step 1: Calculate \(AB\)

\[ \begin{aligned} AB &= \begin{bmatrix} 1&1&-1\\ 2&0&3\\ 3&-1&2 \end{bmatrix} \begin{bmatrix} 1&3\\ 0&2\\ -1&4 \end{bmatrix}\\[6pt] &= \begin{bmatrix} (1)(1)+(1)(0)+(-1)(-1) & (1)(3)+(1)(2)+(-1)(4) \\[4pt] (2)(1)+(0)(0)+(3)(-1) & (2)(3)+(0)(2)+(3)(4) \\[4pt] (3)(1)+(-1)(0)+(2)(-1) & (3)(3)+(-1)(2)+(2)(4) \end{bmatrix}\\[8pt] &= \begin{bmatrix} 1+0+1&3+2-4\\ 2+0-3&6+0+12\\ 3+0-2&9-2+8 \end{bmatrix}\\[8pt] &= \boxed{ \begin{bmatrix} 2&1\\ -1&18\\ 1&15 \end{bmatrix}} \end{aligned} \]

Step 2: Calculate \((AB)C\)

\[\begin{aligned}(AB)C&=\begin{bmatrix}2&1\\-1&18\\1&15\end{bmatrix} \begin{bmatrix}1&2&3&-4\\2&0&-2&1\end{bmatrix}\end{aligned}\]
Now calculate each entry.
First Row
\[\begin{aligned}((AB)C)_{11}&=(2)(1)+(1)(2)=4,\\ ((AB)C)_{12}&=(2)(2)+(1)(0)=4,\\ ((AB)C)_{13}&=(2)(3)+(1)(-2)=4,\\ ((AB)C)_{14}&=(2)(-4)+(1)(1)=-7\end{aligned}\]
Thus, the first row is

\[\begin{bmatrix}4&4&4&-7\end{bmatrix}\]
Second Row

\[\begin{aligned}((AB)C)_{21}&=(-1)(1)+(18)(2)\\&=-1+36\\&=35\end{aligned}\]

\[\begin{aligned}((AB)C)_{22}&=(-1)(2)+(18)(0)\\&=-2\end{aligned}\]

\[\begin{aligned}((AB)C)_{23}&=(-1)(3)+(18)(-2)\\&=-3-36\\&=-39\end{aligned}\]

\[\begin{aligned}((AB)C)_{24}&=(-1)(-4)+(18)(1)\\&=4+18\\&=22\end{aligned}\]
Thus, the second row is

\[\begin{bmatrix}35&-2&-39&22\end{bmatrix}\]
Third Row

\[\begin{aligned}((AB)C)_{31}&=(1)(1)+(15)(2)\\&=1+30\\&=31\end{aligned}\]

\[\begin{aligned}((AB)C)_{32}&=(1)(2)+(15)(0)\\&=2\end{aligned}\]

\[\begin{aligned}((AB)C)_{33}&=(1)(3)+(15)(-2)\\&=3-30\\&=-27\end{aligned}\]

\[\begin{aligned}((AB)C)_{34}&=(1)(-4)+(15)(1)\\&=-4+15\\&=11\end{aligned}\]
Thus, the third row is

\[\begin{bmatrix}31&2&-27&11\end{bmatrix}\]
Therefore,

\[\boxed{(AB)C=\begin{bmatrix}4&4&4&-7\\35&-2&-39&22\\31&2&-27&11\end{bmatrix}}\]

Verification of the Associative Law

We have obtained

\[ A(BC)= \begin{bmatrix} 4&4&4&-7\\ 35&-2&-39&22\\ 31&2&-27&11 \end{bmatrix} \]
and
\[ (AB)C= \begin{bmatrix} 4&4&4&-7\\ 35&-2&-39&22\\ 31&2&-27&11 \end{bmatrix} \]
Since all corresponding entries are equal,

\[\boxed{A(BC)=(AB)C}\]

🗒️ Dimension Analysis

The dimensions provide an important check:

\[A_{3\times3},\quad B_{3\times2},\quad C_{2\times4}\]

For \(A(BC)\):

\[BC=(3\times2)(2\times4)=3\times4\]

\[A(BC)=(3\times3)(3\times4)=3\times4\]

For \((AB)C\):

\[AB=(3\times3)(3\times2)=3\times2\]

\[(AB)C=(3\times2)(2\times4)=3\times4\]

Therefore, both sides have the same order:

\[A(BC),(AB)C\in M_{3\times4}\]

🌟 Board Examination Significance

This is a standard verification-type question for the associative property of matrix multiplication. In a board examination, students should clearly show both products:

\[A(BC)\]

and

\[(AB)C\]

The final matrices must then be compared element by element.

Competitive Entrance Exam Significance

For competitive examinations, dimension analysis can quickly determine whether both expressions are meaningful. In this example:

\[A_{3\times3}B_{3\times2}C_{2\times4}\]

gives

\[\boxed{3\times3\times2\times4\Rightarrow3\times4}\]

for either grouping. This allows students to check the structure before performing numerical calculations.

❌ Common Mistakes

Confusing associativity with commutativity:
\[ (AB)C=A(BC) \]
does not mean
\[ AB=BA. \]
Changing the order of matrices: Associativity only changes brackets, not the sequence \(A,B,C\).
Skipping the dimension check: Both products must be defined before the associative property can be applied.
Arithmetic errors: The third-row fourth-column entry is
\[ -3-2+16=11, \]
not \(17\).
Writing different orders for the two sides: Both \(A(BC)\) and \((AB)C\) have order \(3\times4\) in this example.

⚡ Exam Tip

Write the order of \(A\), \(B\) and \(C\) first.
Check that \(BC\) and \(AB\) are defined.
Calculate \(BC\) and then \(A(BC)\).
Separately calculate \(AB\) and then \((AB)C\).
Compare the final matrices entry by entry.
Use the expected order \(3\times4\) as a quick error check.

🔑 Key Takeaway

Matrix multiplication is associative whenever all the required products are defined.
The grouping of factors may change:
\[ (AB)C=A(BC). \]
The order of factors cannot generally be changed.
In this example,
\[ A_{3\times3},\quad B_{3\times2},\quad C_{2\times4}. \]
Both \(A(BC)\) and \((AB)C\) are \(3\times4\) matrices.
Both products give exactly the same matrix, verifying the associative law.
The corrected third-row fourth-column entry is \(11\), not \(17\).

▦

Example 14

📋

Table of Contents

9 sections

1 ❓ Question ›
2 💡 Concept ›
3 🗺️ Roadmap ›
4 🧩 Solution ›
5 🌟 Board Examination Significance ›
6 ❌ Common Mistakes ›
7 ⚡ Exam Tip ›
8 📋 CBSE Case Study / HOTS ›
9 🔑 Key Takeaway ›

❓ Question

\[A=\begin{bmatrix}1&2&3\\3&-2&1\\4&2&1\end{bmatrix}\]

show that

\[\boxed{A^3-23A-40I=0}\]

💡 Concept

This question illustrates how a polynomial expression involving a matrix can be verified through ordinary matrix addition, subtraction and multiplication.

Here,

\[ A^3=A\cdot A\cdot A \]

Since \(A\) is a \(3\times3\) square matrix, all powers \(A^2,A^3,\ldots\) are defined. Also, \(I\) denotes the identity matrix of order \(3\):

\[ I= \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1 \end{bmatrix} \]

The zero on the right-hand side is the zero matrix of the same order:

\[ 0= \begin{bmatrix} 0&0&0\\ 0&0&0\\ 0&0&0 \end{bmatrix} \]

🗺️ Roadmap

Calculate \(A^2\).
Use \(A^2\) to calculate \(A^3\).
Calculate \(23A\).
Calculate \(40I\).
Substitute the matrices into \(A^3-23A-40I\).
Show that the resulting matrix is the zero matrix.

🧩 Solution

Step 1: Calculate \(A^2\)

By definition,

\[A^2=A\cdot A\]

\[\begin{aligned}A^2&=\begin{bmatrix}1&2&3\\3&-2&1\\4&2&1\end{bmatrix} \begin{bmatrix}1&2&3\\3&-2&1\\4&2&1\end{bmatrix}\end{aligned}\]
Using the row-by-column multiplication rule:

\[\begin{aligned}A^2&=\begin{bmatrix}(1)(1)+(2)(3)+(3)(4) &(1)(2)+(2)(-2)+(3)(2)&(1)(3)+(2)(1)+(3)(1)\\[4pt](3)(1)+(-2)(3)+(1)(4) &(3)(2)+(-2)(-2)+(1)(2)&(3)(3)+(-2)(1)+(1)(1)\\[4pt](4)(1)+(2)(3)+(1)(4) &(4)(2)+(2)(-2)+(1)(2)&(4)(3)+(2)(1)+(1)(1)\end{bmatrix}\\[8pt] &=\begin{bmatrix}1+6+12&2-4+6&3+2+3\\3-6+4&6+4+2&9-2+1\\4+6+4&8-4+2&12+2+1\end{bmatrix}\\[8pt] &=\begin{bmatrix}19&4&8\\1&12&8\\14&6&15\end{bmatrix}\end{aligned}\]
Therefore,

\[A^2=\begin{bmatrix}19&4&8\\1&12&8\\14&6&15\end{bmatrix}\]

Step 2: Calculate \(A^3\)

Since

\[A^3=A\cdot A^2\]
we have

\[\begin{aligned}A^3&=\begin{bmatrix}1&2&3\\3&-2&1\\4&2&1\end{bmatrix} \begin{bmatrix}19&4&8\\1&12&8\\14&6&15\end{bmatrix}\end{aligned}\]
Calculate the entries row by column

\[\begin{aligned}A^3&=\begin{bmatrix}1(19)+2(1)+3(14)&1(4)+2(12)+3(6)&1(8)+2(8)+3(15)\\[4pt] 3(19)+(-2)(1)+1(14)&3(4)+(-2)(12)+1(6)&3(8)+(-2)(8)+1(15)\\[4pt] 4(19)+2(1)+1(14)&4(4)+2(12)+1(6)&4(8)+2(8)+1(15)\end{bmatrix}\\[8pt] &=\begin{bmatrix}19+2+42&4+24+18&8+16+45\\57-2+14&12-24+6&24-16+15\\76+2+14&16+24+6&32+16+15\end{bmatrix}\\[8pt] &=\begin{bmatrix}63&46&69\\69&-6&23\\92&46&63\end{bmatrix}\end{aligned}\]

Step 3: Calculate \(23A\)

Scalar multiplication means multiplying every element of \(A\) by \(23\):

\[\begin{aligned}23A &=23\begin{bmatrix}1&2&3\\3&-2&1\\4&2&1\end{bmatrix}\\ &=\begin{bmatrix}23&46&69\\69&-46&23\\92&46&23\end{bmatrix}\end{aligned}\]

Step 4: Calculate \(40I\)

Scalar multiplication means multiplying every element of \(A\) by \(23\):
Since \(A\) is of order \(3\times3\), the identity matrix is
\[I=I_3=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\]
Therefore,

\[\begin{aligned}40I&=40\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\\ &=\begin{bmatrix}40&0&0\\0&40&0\\0&0&40\end{bmatrix}\end{aligned}\]

Step 5: Evaluate \(A^3-23A-40I\)

Substitute the matrices obtained above:

\[\begin{aligned}A^3-23A-40I&=\begin{bmatrix}63&46&69\\69&-6&23\\92&46&63\end{bmatrix}- \begin{bmatrix}23&46&69\\69&-46&23\\92&46&23\end{bmatrix}\&\qquad- \begin{bmatrix}40&0&0\\0&40&0\\0&0&40\end{bmatrix}\end{aligned}\]
Perform the subtraction entry by entry:

\[\begin{aligned}A^3-23A-40I&=\begin{bmatrix}63-23-40&46-46-0&69-69-0\\[4pt] 69-69-0&-6-(-46)-40&23-23-0\\[4pt] 92-92-0&46-46-0&63-23-40\end{bmatrix}\\[8pt] &=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\end{aligned}\]
Hence,

\[\boxed{A^3-23A-40I=0}\]

🌟 Board Examination Significance

This type of question tests several fundamental skills from the NCERT Class 12 Matrices chapter:

multiplication of square matrices,
calculation of matrix powers,
scalar multiplication,
use of the identity matrix,
matrix subtraction, and
verification of a matrix polynomial identity.

A particularly important presentation point is that the final answer must be the zero matrix, not the scalar number \(0\).

\[ \boxed{ A^3-23A-40I=0_{3\times3} } \]

Competitive Entrance Exam Significance

Polynomial identities of matrices are useful in objective examinations because they can reduce higher powers of a matrix to lower powers.

From

\[ A^3-23A-40I=0, \]

we get

\[ \boxed{ A^3=23A+40I }. \]

Therefore, instead of directly calculating \(A^3\), one may replace it by the much simpler expression \(23A+40I\).

For example, multiplying the identity by \(A\) gives

\[ A^4=23A^2+40A. \]

Multiplying again by \(A\):

\[ A^5=23A^3+40A^2. \]

Since \(A^3=23A+40I\),

\[ \begin{aligned} A^5 &=23(23A+40I)+40A^2\\ &=40A^2+529A+920I. \end{aligned} \]

Thus, higher powers of \(A\) can be reduced to expressions involving only \(I\), \(A\) and \(A^2\).

\[ \boxed{ \text{Every power }A^n\;(n\geq3)\text{ can be reduced to a combination of }I,A,A^2. } \]

❌ Common Mistakes

Writing \(40\) instead of \(40I\): A scalar constant in a matrix polynomial is represented using the identity matrix.
Forgetting the negative sign:
\[ A^3-23A-40I \]
contains subtraction of both terms.
Incorrectly calculating \(A^2\): Every entry must be obtained using row-by-column multiplication.
Confusing scalar multiplication with matrix multiplication: \(23A\) means every element of \(A\) is multiplied by \(23\).
Writing a scalar zero as the answer: The result is the \(3\times3\) zero matrix.
Arithmetic error in \(A^3\): The third-row fourth-column entry is
\[ 4(-4)+2(1)+1(1)=-13 \]
only if multiplying the original matrices incorrectly; for the actual \(A^3\) calculation the correct entry is \(63\). Always use the correct intermediate matrix \(A^2\).

⚡ Exam Tip

For \(A^3\), calculate \(A^2\) first and then multiply by \(A\).
Use \(40I\), not \(40\), when working with a matrix polynomial.
Check the order of every matrix before addition or subtraction.
Use the zero matrix as the final verification target.
If an identity such as
\[ A^3=23A+40I \]
is established, use it to simplify higher powers rather than repeatedly multiplying matrices.

📋 CBSE Case Study / HOTS

Suppose a \(3\times3\) matrix \(A\) satisfies

\[ A^3-23A-40I=0 \]

A student is asked to find \(A^4\). Instead of calculating \(A^4\) directly, first use

\[ A^3=23A+40I \]

Multiplying by \(A\):

\[ \begin{aligned} A^4 &=A(A^3)\\ &=A(23A+40I)\\ &=23A^2+40AI\\ &=23A^2+40A \end{aligned} \]

Therefore,

\[ \boxed{ A^4=23A^2+40A } \]

HOTS Question

\[ A^3-23A-40I=0 \]

express \(A^4\) in terms of \(A^2\) and \(A\).

From the given identity:

\[ A^3=23A+40I \]

Multiplying by \(A\):

\[ A^4=23A^2+40A \]

\[ \boxed{A^4=23A^2+40A} \]

🔑 Key Takeaway

The given matrix satisfies the polynomial equation
\[ A^3-23A-40I=0. \]
The correct value of \(A^2\) is
\[ \begin{bmatrix} 19&4&8\\ 1&12&8\\ 14&6&15 \end{bmatrix}. \]
The correct value of \(A^3\) is
\[ \begin{bmatrix} 63&46&69\\ 69&-6&23\\ 92&46&63 \end{bmatrix}. \]
A scalar constant in a matrix polynomial is multiplied by the identity matrix.
The resulting matrix is the \(3\times3\) zero matrix.
The identity can be rearranged as
\[ A^3=23A+40I. \]
It also gives the useful inverse formula
\[ \boxed{A^\prime =\frac{1}{40}(A^2-23I)}. \]
Higher powers of \(A\) can be reduced to expressions involving \(I\), \(A\) and \(A^2\).

▦

Transpose of a Matrix

📋

Table of Contents

22 sections

1 📘 Definition ›
2 💡 Concept of Transposition ›
3 📄 General Form ›
4 ✏️ Example 1: Transpose of a Rectangular Matrix ›
5 🧰 How to Find a Transpose Step by Step ›
6 ✏️ Example 2: Transpose of a Matrix ›
7 ✏️ Example 3: Transpose of a Row Matrix ›
8 ✏️ Example 4: Transpose of a Column Matrix ›
9 🔢 Important Formula for Individual Elements ›
10 ✏️ Example 5: Finding a Particular Entry of a Transpose ›
11 🏷️ Properties of the Transpose of a Matrix ›
12 ✏️ Example 6: Verification of Product Property ›
13 📐 General Derivation of ›
14 🤔 Why Does the Order Reverse? ›
15 🗂️ Important Special Cases ›
16 📌 Transpose of a Difference ›
17 ✏️ Combined Transpose Properties ›
18 🌟 Board Examination Significance ›
19 ❌ Common Mistakes ›
20 ⚡ Exam Tip ›
21 📋 CBSE Case Study / HOTS ›
22 ⚡ Quick Revision ›

📘 Definition

Let \(A=[a_{ij}]\) be a matrix of order \(m\times n\). The matrix obtained by interchanging the rows and columns of \(A\) is called the transpose of \(A\).

The transpose of \(A\) is denoted by \(A'\) or \(A^T\).

\[ \boxed{ A=[a_{ij}]_{m\times n} \quad\Longrightarrow\quad A^T=[a_{ji}]_{n\times m} } \]

Thus, the element \(a_{ij}\), which occupies the \(i\)-th row and \(j\)-th column of \(A\), occupies the \(j\)-th row and \(i\)-th column of \(A^T\).

\[ \boxed{ (A^T)_{ij}=a_{ji} } \]

💡 Concept of Transposition

Transposition changes the orientation of a matrix:

\[ \boxed{ \text{Rows of }A\longrightarrow\text{Columns of }A^T } \]

\[ \boxed{ \text{Columns of }A\longrightarrow\text{Rows of }A^T } \]

If \(A\) has order \(m\times n\), then its transpose has order \(n\times m\).

\[ \boxed{ A_{m\times n}\longrightarrow A^T_{n\times m} } \]

An important observation is that transposition does not change the number of elements. Both \(A\) and \(A^T\) contain \(mn\) elements.

🗒️ General Form

Let

\[ A= \begin{bmatrix} a_{11}&a_{12}&\cdots&a_{1n}\\ a_{21}&a_{22}&\cdots&a_{2n}\\ \vdots&\vdots&\ddots&\vdots\\ a_{m1}&a_{m2}&\cdots&a_{mn} \end{bmatrix} \]

Its transpose is obtained by converting every row into the corresponding column:

\[ A^T= \begin{bmatrix} a_{11}&a_{21}&\cdots&a_{m1}\\ a_{12}&a_{22}&\cdots&a_{m2}\\ \vdots&\vdots&\ddots&\vdots\\ a_{1n}&a_{2n}&\cdots&a_{mn} \end{bmatrix} \]

Therefore,

\[ \boxed{ [a_{ij}]^T=[a_{ji}] } \]

✏️ Example 1: Transpose of a Rectangular Matrix

Consider

\[ A= \begin{bmatrix} 3&5\\ \sqrt3&1\\ 0&-\dfrac15 \end{bmatrix} \]

Matrix \(A\) has \(3\) rows and \(2\) columns, so its order is \(3\times2\).

\[ \operatorname{ord}(A)=3\times2. \]

Interchanging rows and columns gives:

\[ \boxed{ A^T= \begin{bmatrix} 3&\sqrt3&0\\ 5&1&-\dfrac15 \end{bmatrix}} \]

Therefore,

\[ \operatorname{ord}(A^T)=2\times3. \]

🧰 How to Find a Transpose Step by Step

Write the given matrix clearly.
Identify its rows and columns.
Convert the first row into the first column of the transpose.
Convert the second row into the second column.
Continue until all rows have been converted into columns.
Check that the order changes from \(m\times n\) to \(n\times m\).

✏️ Example 2: Transpose of a \(2\times3\) Matrix

\[ A= \begin{bmatrix} 1&2&3\\ 4&5&6 \end{bmatrix}. \]

The rows of \(A\) are

\[ [1\quad2\quad3] \]

\[ [4\quad5\quad6] \]

These become the columns of \(A^T\):

\[ A^T= \begin{bmatrix} 1&4\\ 2&5\\ 3&6 \end{bmatrix} \]

Notice:

\[ 2\times3\longrightarrow3\times2. \]

✏️ Example 3: Transpose of a Row Matrix

Let

\[ A= \begin{bmatrix} 2&-1&5&7 \end{bmatrix} \]

\(A\) is a \(1\times4\) row matrix. Its transpose is a \(4\times1\) column matrix:

\[A^T= \begin{bmatrix} 2\\ -1\\ 5\\ 7 \end{bmatrix}\]

\[1\times4\longrightarrow4\times1\]

✏️ Example 4: Transpose of a Column Matrix

\[ A= \begin{bmatrix} 3\\ 0\\ -2 \end{bmatrix}. \]

Then

\[ A^T= \begin{bmatrix} 3&0&-2 \end{bmatrix} \]

\[ 3\times1\longrightarrow1\times3. \]

🔢 Important Formula for Individual Elements

✏️ Example 5: Finding a Particular Entry of a Transpose

Example

If \(A=[a_{ij}]\), find the \((2,3)\)-th element of \(A^T\)

By the definition of transpose:

\[(A^T)_{ij}=a_{ji}\]

Therefore,

\[(A^T)_{23}=a_{32}\]

🏷️ Properties of the Transpose of a Matrix

Properties

Property 1: Transpose of the Transpose

The transpose of the transpose of a matrix is the original matrix:

\[\boxed{(A^T)^T=A}\]

This follows because the first transposition changes rows into columns, and the second transposition changes them back.

Example

\[A=\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}\]

\[A^T=\begin{bmatrix}1&4\\2&5\\3&6\end{bmatrix}\]

Transposing again:

\[(A^T)^T=\begin{bmatrix}1&2&3\\4&5&6\end{bmatrix}=A\]

\[\boxed{(A^T)^T=A}\]

Property 2: Transpose of a Scalar Multiple

If \(k\) is any scalar, then

\[\boxed{(kA)^T=kA^T}\]

Transposition does not affect the scalar multiplier.

Example

\[A=\begin{bmatrix}1&2\\3&4\end{bmatrix},\qquad k=5\]

\[5A=\begin{bmatrix}5&10\\15&20\end{bmatrix}\]

\[(5A)^T=\begin{bmatrix}5&15\\10&20\end{bmatrix}\]

Also,

\[A^T=\begin{bmatrix}1&3\\2&4\end{bmatrix}\]

and therefore

\[5A^T=\begin{bmatrix}5&15\\10&20\end{bmatrix}\]

Hence,

\[\boxed{(5A)^T=5A^T}\]

Property 3: Transpose of a Sum

For matrices \(A\) and \(B\) of the same order,

\[\boxed{(A+B)^T=A^T+B^T}\]

Thus, transposition distributes over matrix addition.

Example

\[A=\begin{bmatrix}1&2\\3&4\end{bmatrix}\qquad =\begin{bmatrix}5&6\\7&8\end{bmatrix}\]

\[A+B=\begin{bmatrix}6&8\\10&12\end{bmatrix}\]

Hence,

\[(A+B)^T=\begin{bmatrix}6&10\\8&12\end{bmatrix}\]

On the other hand,

\[A^T=\begin{bmatrix}1&3\\2&4\end{bmatrix}\qquad B^T=\begin{bmatrix}5&7\\6&8\end{bmatrix}\]

\[\begin{aligned}A^T+B^T&=\begin{bmatrix}1&3\\2&4\end{bmatrix}+\begin{bmatrix}5&7\\6&8\end{bmatrix}\\&=\begin{bmatrix}6&10\\8&12\end{bmatrix}\end{aligned}\]

Therefore,

\[\boxed{(A+B)^T=A^T+B^T}\]

Property 4: Transpose of a Matrix Product

For matrices \(A\) and \(B\) for which \(AB\) is defined,

\[\boxed{(AB)^T=B^TA^T}\]

This is one of the most important transpose properties for Class 12 Mathematics and competitive examinations. Notice carefully that the order of the matrices is reversed.

\[\boxed{(AB)^T=B^TA^T}\]

It is generally incorrect to write

\[(AB)^T=A^TB^T\]

✏️ Example 6: Verification of Product Property

Let

\[A=\begin{bmatrix}1&2\\3&4\end{bmatrix}\qquad =\begin{bmatrix}2&1\\0&3\end{bmatrix}\]

First calculate \(AB\):

\[\begin{aligned}AB&=\begin{bmatrix}1&2\\3&4\end{bmatrix}\begin{bmatrix}2&1\\0&3\end{bmatrix}\\ &=\begin{bmatrix}2&7\\6&15\end{bmatrix}\end{aligned}\]

Therefore,

\[(AB)^T=\begin{bmatrix}2&6\\7&15\end{bmatrix}\]

Now,

\[A^T=\begin{bmatrix}1&3\\2&4\end{bmatrix}\]

\[B^T=\begin{bmatrix}2&0\\1&3\end{bmatrix}\]

Hence,

\[\begin{aligned}B^TA^T&=\begin{bmatrix}2&0\\1&3\end{bmatrix}\begin{bmatrix}1&3\\2&4\end{bmatrix}\\ &=\begin{bmatrix}2&6\\7&15\end{bmatrix}\end{aligned}\]

Thus,

\[\boxed{(AB)^T=B^TA^T}\]

📐 General Derivation of \((AB)^T=B^TA^T\)

Let

\[A=[a_{ij}]_{m\times n},\quad B=[b_{jk}]_{n\times p}\]

Then \(AB\) is an \(m\times p\) matrix. Its \((i,k)\)-th element is

\[(AB)_{ik}=\sum_{j=1}^{n}a_{ij}b_{jk}\]

The \((k,i)\)-th element of \((AB)^T\) is therefore

\[ \begin{aligned} ((AB)^T)_{ki} &=(AB)_{ik}\\ &=\sum_{j=1}^{n}a_{ij}b_{jk} \end{aligned} \]

Now consider \(B^TA^T\). The \((k,i)\)-th element is

\[ \begin{aligned} (B^TA^T)_{ki} &=\sum_{j=1}^{n}(B^T)_{kj}(A^T)_{ji}\\ &=\sum_{j=1}^{n}b_{jk}a_{ij}\\ &=\sum_{j=1}^{n}a_{ij}b_{jk} \end{aligned} \]

Hence,

\[((AB)^T)_{ki}=(B^TA^T)_{ki}\]

Since every corresponding element is equal,

\[\boxed{(AB)^T=B^TA^T}\]

🤔 Why Does the Order Reverse?

The reversal in

\[(AB)^T=B^TA^T\]

is not accidental. Transposition reverses the row-column relationship involved in matrix multiplication.

This property is analogous to the reversal that occurs when taking the transpose of a product of several matrices:

\[(ABC)^T=C^TB^TA^T\]

More generally,

\[\boxed{(A_1A_2\cdots A_n)^T=A_n^T\cdots A_2^TA_1^T}\]

🗂️ Important Special Cases

Transpose of \(A^2\)

\[\begin{aligned}(A^2)^T&=(AA)^T\\&=A^TA^T\\&=(A^T)^2\end{aligned}\]

\[\boxed{(A^2)^T=(A^T)^2}\]

Transpose of \(A^3\)

\[\boxed{(A^3)^T=(A^T)^3}\]

More generally, for every positive integer \(n\),

\[\boxed{(A^n)^T=(A^T)^n}\]

📌 Transpose of a Difference

✏️ Combined Transpose Properties

These properties can be combined to simplify complicated expressions.

question

Find the transpose of

\[3AB+2C\]

Using linearity of transpose:

\[\begin{aligned}(3AB+2C)^T&=(3AB)^T+(2C)^T\\&=3(AB)^T+2C^T\\&=3B^TA^T+2C^T\end{aligned}\]

\[\boxed{(3AB+2C)^T=3B^TA^T+2C^T}\]

question

Find the transpose of

\[ABC\]

Applying the product rule:

\[\begin{aligned}(ABC)^T&=(AB\,C)^T\\&=C^T(AB)^T\\&=C^TB^TA^T\end{aligned}\]

\[\boxed{(ABC)^T=C^TB^TA^T}\]

🌟 Board Examination Significance

Transpose is a fundamental concept in the NCERT Class 12 Mathematics chapter Matrices. Questions may ask students to:

find the transpose of a given matrix,
determine the order of the transpose,
verify transpose identities,
find unknown elements using \(A^T=B\),
prove or use symmetric and skew-symmetric properties,
simplify expressions involving transposes, and
use the product property \((AB)^T=B^TA^T\).

Competitive Entrance Exam Significance

The transpose operation is especially useful in multiple-choice questions because many complicated-looking expressions can be simplified immediately using standard identities.

For example:

\[ (ABC)^T=C^TB^TA^T. \]

Similarly,

\[ (A^2)^T=(A^T)^2. \]

and

\[ (A+B)^T=A^T+B^T. \]

Memorising these patterns reduces unnecessary matrix multiplication.

❌ Common Mistakes

Forgetting that the order changes:
\[ m\times n\longrightarrow n\times m. \]
Writing \((AB)^T=A^TB^T\): The correct formula is
\[ \boxed{(AB)^T=B^TA^T}. \]
Confusing transpose with inverse: \(A^T\) and \(A^\prime \) are different concepts.
Assuming every matrix equals its transpose:
\[ A^T=A \]
is true only for symmetric matrices.
Forgetting that addition requires equal orders:
\[ (A+B)^T=A^T+B^T \]
requires \(A+B\) to be defined.
Not reversing the order of factors: For a product of three matrices,
\[ (ABC)^T=C^TB^TA^T. \]

⚡ Exam Tip

Remember the basic transformation:
\[ a_{ij}\longrightarrow a_{ji}. \]
Always reverse the order when taking the transpose of a product:
\[ (AB)^T=B^TA^T. \]
For several factors:
\[ (ABC)^T=C^TB^TA^T. \]
Remember:
\[ (A^T)^T=A. \]
For scalar multiplication:
\[ (kA)^T=kA^T. \]
For addition:
\[ (A+B)^T=A^T+B^T. \]

📋 CBSE Case Study / HOTS

Let \(A\) and \(B\) be matrices of suitable orders such that \(AB\) is defined. A student claims:

\[ (AB)^T=A^TB^T. \]

The claim is incorrect. The correct rule is

\[ \boxed{ (AB)^T=B^TA^T }. \]

The order of multiplication is reversed after transposition.

HOTS Question

Simplify

\[ (ABC+2D)^T. \]

Using the transpose of a sum:

\[ (ABC+2D)^T=(ABC)^T+(2D)^T. \]

Using the product and scalar-multiple properties:

\[ \boxed{ (ABC+2D)^T=C^TB^TA^T+2D^T }. \]

Higher-Level Insight: Transpose as an Operation

Transposition can be viewed as an operation that changes the position of the indices:

\[ a_{ij}\longrightarrow a_{ji}. \]

This simple index interchange explains the major properties:

\[ (A^T)^T=A \]

\[ (A+B)^T=A^T+B^T \]

\[ (kA)^T=kA^T \]

\[ (AB)^T=B^TA^T. \]

The last identity is especially important because the indices associated with the product naturally reverse their order.

⚡ Quick Revision

Definition:

\[ \boxed{ A=[a_{ij}]_{m\times n} \Rightarrow A^T=[a_{ji}]_{n\times m} } \]

Order:

\[ \boxed{ A_{m\times n}\Rightarrow A^T_{n\times m} } \]

Double transpose:

\[ \boxed{(A^T)^T=A} \]

Scalar multiple:

\[ \boxed{(kA)^T=kA^T} \]

Sum:

\[ \boxed{(A+B)^T=A^T+B^T} \]

Difference:

\[ \boxed{(A-B)^T=A^T-B^T} \]

Product:

\[ \boxed{(AB)^T=B^TA^T} \]

Three matrices:

\[ \boxed{(ABC)^T=C^TB^TA^T} \]

Power:

\[ \boxed{(A^n)^T=(A^T)^n} \]

▦

Example 15

📋

Table of Contents

10 sections

1 ❓ Question ›
2 💡 Concept ›
3 🗺️ Roadmap ›
4 🧩 Solution ›
5 🔬 General Proof of the Three Properties ›
6 ✏️ Example ›
7 ❌ Common Mistakes ›
8 ⚡ Exam Tip ›
9 📋 CBSE Case Study / HOTS ›
10 ⚡ Quick Revision ›

❓ Question

\[A=\begin{bmatrix}3&\sqrt{3}&2\\4&2&0\end{bmatrix}\]

and

\[B=\begin{bmatrix}2&-1&2\\1&2&4\end{bmatrix}\]

verify that

\[\boxed{(i)\ (A^T)^T=A}\]

\[\boxed{(ii)\ (A+B)^T=A^T+B^T}\]

\[\boxed{(iii)\ (kB)^T=kB^T}\]

where \(k\) is any scalar.

💡 Concept

🗺️ Roadmap

Find \(A^T\) and then transpose it again to verify \((A^T)^T=A\).
Calculate \(A+B\), then find \((A+B)^T\).
Separately calculate \(A^T+B^T\) and compare the two results.
Choose a convenient scalar \(k\), calculate \((kB)^T\), and compare it with \(kB^T\).

🧩 Solution

Part (i): Verify \((A^T)^T=A\)

Given

\[A=\begin{bmatrix}3&\sqrt{3}&2\\4&2&0\end{bmatrix}\]
Matrix \(A\) is of order \(2\times3\). Therefore, \(A^T\) will be of order \(3\times2\).

\[\begin{aligned}A^T&=\begin{bmatrix}3&4\\\sqrt{3}&2\\2&0\end{bmatrix}\end{aligned}\]
Taking the transpose once again:

\[\begin{aligned}(A^T)^T&=\begin{bmatrix}3&\sqrt{3}&2\\4&2&0\end{bmatrix}\\&=A\end{aligned}\]
Hence,

\[\boxed{(A^T)^T=A}\]
Conceptual Explanation

The first transpose changes rows into columns:

\[ 2\times3\longrightarrow3\times2. \]

The second transpose changes the columns back into rows:

\[ 3\times2\longrightarrow2\times3. \]

Therefore, taking the transpose twice restores the original matrix.

\[ \boxed{\text{Double transposition returns the original matrix.}} \]

Part (ii): Verify \((A+B)^T=A^T+B^T\)

Given

\[A=\begin{bmatrix}3&\sqrt{3}&2\\4&2&0\end{bmatrix},\qquad B=\begin{bmatrix}2&-1&2\\1&2&4\end{bmatrix}\]
Both matrices are of order \(2\times3\), so their sum is defined.
Step 1: Calculate \(A+B\)

\[\begin{aligned}A+B&=\begin{bmatrix}3&\sqrt3&2\\4&2&0\end{bmatrix}+\begin{bmatrix}2&-1&2\\1&2&4\end{bmatrix}\\ &=\begin{bmatrix}3+2&\sqrt3-1&2+2\\4+1&2+2&0+4\end{bmatrix}\\ &=\begin{bmatrix}5&\sqrt3-1&4\\5&4&4\end{bmatrix}\end{aligned}\]
Step 2: Calculate \((A+B)^T\)

\[\begin{aligned}(A+B)^T&=\begin{bmatrix}5&5\\\sqrt3-1&4\\4&4\end{bmatrix}\end{aligned}\]

\[\boxed{(A+B)^T=\begin{bmatrix}5&5\\\sqrt3-1&4\\4&4\end{bmatrix}}\]
Step 3: Calculate \(A^T\) and \(B^T\)

\[A^T=\begin{bmatrix}3&4\\\sqrt3&2\\2&0\end{bmatrix}\]
and
\[B^T=\begin{bmatrix}2&1\\-1&2\\2&4\end{bmatrix}\]
Step 4: Calculate \(A^T+B^T\)

\[\begin{aligned}A^T+B^T&=\begin{bmatrix}3&4\\\sqrt3&2\\2&0\end{bmatrix}+\begin{bmatrix}2&1\\-1&2\\2&4\end{bmatrix}\\ &=\begin{bmatrix}3+2&4+1\\\sqrt3-1&2+2\\2+2&0+4\end{bmatrix}\\&=\boxed{\begin{bmatrix}5&5\\\sqrt3-1&4\\4&4\end{bmatrix}}\end{aligned}\]
Comparing both sides:

\[(A+B)^T=\begin{bmatrix}5&5\\\sqrt3-1&4\\4&4\end{bmatrix}\]

\[A^T+B^T=\begin{bmatrix}5&5\\\sqrt3-1&4\\4&4\end{bmatrix}\]
Therefore,

\[\boxed{(A+B)^T=A^T+B^T}\]

Part (iii): Verify \((kB)^T=kB^T\)

The property is required to hold for any scalar \(k\). We can verify it using a convenient value, say

\[k=2\]
Given

\[B=\begin{bmatrix}2&-1&2\\1&2&4\end{bmatrix}\]
Step 1: Calculate \(2B\)

\[\begin{aligned}2B&=2\begin{bmatrix}2&-1&2\\1&2&4\end{bmatrix}\\&=\begin{bmatrix}4&-2&4\\2&4&8\end{bmatrix}\end{aligned}\]
Step 2: Calculate \((2B)^T\)

\[\boxed{(2B)^T=\begin{bmatrix}4&2\\-2&4\\4&8\end{bmatrix}}\]
Step 3: Calculate \(B^T\)

\[B^T=\begin{bmatrix}2&1\\-1&2\\2&4\end{bmatrix}\]
Therefore,

\[\begin{aligned}2B^T&=2\begin{bmatrix}2&1\\-1&2\\2&4\end{bmatrix}\\&=\begin{bmatrix}4&2\\-2&4\\4&8\end{bmatrix}\end{aligned}\]
Hence,

\[(2B)^T=\begin{bmatrix}4&2\\-2&4\\4&8\end{bmatrix}=2B^T\]

\[\boxed{(2B)^T=2B^T}\]
Since \(2\) was chosen only for verification and the same argument works for every scalar \(k\), the general property is

\[\boxed{(kB)^T=kB^T}\]

🔬 General Proof of the Three Properties

1. Proof of \((A^T)^T=A\)

Let

\[A=[a_{ij}]\]

By definition,

\[(A^T)_{ij}=a_{ji}\]

Taking the transpose again:

\[((A^T)^T)_{ij}=(A^T)_{ji}=a_{ij}\]

Hence every corresponding element is equal:

\[\boxed{(A^T)^T=A}\]

2. Proof of \((A+B)^T=A^T+B^T\)

Let \(A=[a_{ij}]\) and \(B=[b_{ij}]\) be matrices of the same order. Then

\[(A+B)_{ij}=a_{ij}+b_{ij}\]

Therefore,

\[\begin{aligned}((A+B)^T)_{ij}&=(A+B)_{ji}\\&=a_{ji}+b_{ji}\\&=(A^T)_{ij}+(B^T)_{ij}\\&=(A^T+B^T)_{ij}\end{aligned}\]

Hence,

\[\boxed{(A+B)^T=A^T+B^T}\]

3. Proof of \((kB)^T=kB^T\)

Let \(k\) be a scalar and \(B=[b_{ij}]\). Then

\[(kB)_{ij}=kb_{ij}\]

Hence,

\[\begin{aligned}((kB)^T)_{ij}&=(kB)_{ji}\\&=kb_{ji}\\&=k(B^T)_{ij}\end{aligned}\]

Therefore,

\[\boxed{(kB)^T=kB^T}\]

Important Extension: Transpose of a Product

Another fundamental property, which is frequently tested in CBSE and competitive entrance examinations, is

\[\boxed{(AB)^T=B^TA^T}\]

The order of the matrices is reversed. This should not be confused with the addition rule:

\[(A+B)^T=A^T+B^T\]

For three matrices:

\[\boxed{(ABC)^T=C^TB^TA^T}\]

✏️ Example

Question

Simplify:

\[(3AB+5C)^T\]

Using the sum, scalar and product properties:

\[\begin{aligned}(3AB+5C)^T&=(3AB)^T+(5C)^T\\&=3(AB)^T+5C^T\\&=3B^TA^T+5C^T\end{aligned}\]

\[\boxed{(3AB+5C)^T=3B^TA^T+5C^T}\]

❌ Common Mistakes

Writing \((A+B)^T=A^TB^T\): Transpose of a sum is a sum of transposes, not a product.
Forgetting the second transpose:
\[ (A^T)^T=A, \]
not \(A^T\).
Changing the scalar during transposition:
\[ (kA)^T=kA^T. \]
Confusing the product rule with the sum rule:
\[ (AB)^T=B^TA^T, \qquad (A+B)^T=A^T+B^T. \]
Ignoring matrix order: Before adding matrices, verify that they have the same order.

⚡ Exam Tip

Memorise the four fundamental rules:
\[ (A^T)^T=A, \quad (kA)^T=kA^T, \quad (A+B)^T=A^T+B^T, \quad (AB)^T=B^TA^T. \]
For products, use the memory rule: transpose each factor and reverse the order.
For sums, retain the order:
\[ (A+B)^T=A^T+B^T. \]
To verify a matrix identity, calculate both sides separately and compare the resulting matrices element by element.
In objective examinations, a complicated transpose expression can often be simplified without performing any numerical matrix multiplication.

📋 CBSE Case Study / HOTS

A \(2\times3\) matrix \(A\) and a \(2\times3\) matrix \(B\) are given. A student wants to evaluate

\[ (A+B)^T. \]

Since both matrices have the same order, \(A+B\) is defined. Instead of calculating \(A+B\) first, the student can directly use:

\[ \boxed{ (A+B)^T=A^T+B^T }. \]

The resulting matrix will have order

\[ 3\times2. \]

HOTS Question

If \(A\) and \(B\) are matrices of suitable orders, simplify

\[ (2A-3B)^T. \]

Using linearity of transpose:

\[ \begin{aligned} (2A-3B)^T &=2A^T-3B^T. \end{aligned} \]

\[ \boxed{ (2A-3B)^T=2A^T-3B^T } \]

⚡ Quick Revision

\[ \boxed{ \begin{aligned} (A^T)^T&=A\\ (kA)^T&=kA^T\\ (A+B)^T&=A^T+B^T\\ (A-B)^T&=A^T-B^T\\ (AB)^T&=B^TA^T\\ (ABC)^T&=C^TB^TA^T\\ (A^n)^T&=(A^T)^n \end{aligned} } \]

▦

Symmetric and Skew-Symmetric Matrices

📋

Table of Contents

14 sections

1 🗺️ Overview ›
2 🗂️ Symmetric and Skew Symmetric Matrices ›
3 ⚖️ Symmetric vs Skew-Symmetric Matrix ›
4 🔍 Geometric Interpretation ›
5 ✏️ Example ›
6 📌 Decomposition of Any Square Matrix ›
7 ✏️ Example ›
8 📄 Important Results ›
9 🌟 CBSE Board Significance ›
10 ⚡ Exam Tip ›
11 ❌ Common Mistakes ›
12 📋 CBSE Case Study / HOTS ›
13 ⚡ Quick Revision ›
14 🔑 Key Takeaway ›

🗺️ Overview

Symmetric and skew-symmetric matrices are important special types of square matrices. They are defined using the transpose operation and are frequently tested in CBSE Class 12 Mathematics, board examinations, JEE and other competitive entrance examinations.

🗂️ Symmetric and Skew Symmetric Matrices

⚖️ Symmetric vs Skew-Symmetric Matrix

Feature	Symmetric Matrix	Skew-Symmetric Matrix
Condition	\(A^T=A\)	\(A^T=-A\)
Element condition	\(a_{ij}=a_{ji}\)	\(a_{ij}=-a_{ji}\)
Diagonal elements	Can be arbitrary	Always zero
Order	Always square	Always square
Example	\[ \begin{bmatrix} 1&2\\ 2&3 \end{bmatrix} \]	\[ \begin{bmatrix} 0&2\\ -2&0 \end{bmatrix} \]

🔍 Geometric Interpretation

Imagine the principal diagonal of a square matrix as a mirror. In a symmetric matrix, entries reflected across this diagonal are identical:

\[a_{12}=a_{21},\qquad a_{13}=a_{31},\qquad a_{23}=a_{32}\]

In a skew-symmetric matrix, the reflected entries have opposite signs:

\[a_{12}=-a_{21},\qquad _{13}=-a_{31},\qquad a_{23}=-a_{32}\]

✏️ Example

Question

Find Unknown Values for a Symmetric Matrix. If

\[A=\begin{bmatrix}2&x&5\\3&4&y\\5&7&6\end{bmatrix}\]

is symmetric, find \(x\) and \(y\).

For a symmetric matrix:

\[a_{ij}=a_{ji}\]
Comparing the \((1,2)\) and \((2,1)\) entries:

\[x=3\]
Comparing the \((2,3)\) and \((3,2)\) entries:

\[y=7\]

\[\boxed{x=3,\qquad y=7}\]

Question

Find Unknown Values for a Skew-Symmetric Matrix. If

\[A=\begin{bmatrix}0&x&4\\-x&0&y\\-4&-y&0\end{bmatrix}\]

show that \(A\) is skew-symmetric.

Its transpose is

\[A^T=\begin{bmatrix}0&-x&-4\\x&0&-y\\4&y&0\end{bmatrix}\]

Also,

\[-A=\begin{bmatrix}0&-x&-4\\x&0&-y\\4&y&0\end{bmatrix}\]

Hence,

\[A^T=-A\]

\[\boxed{A\text{ is skew-symmetric}}\]

📌 Decomposition of Any Square Matrix

One of the most useful higher-level results is that every square matrix can be uniquely expressed as the sum of a symmetric matrix and a skew-symmetric matrix. Let \(A\) be any square matrix. Write:

\[\boxed{A=\frac{A+A^T}{2}+\frac{A-A^T}{2}}\]

Define

\[S=\frac{A+A^T}{2}\]

and

\[K=\frac{A-A^T}{2}\]

Then \(S\) is symmetric because

\[ \begin{aligned} S^T&=\left(\frac{A+A^T}{2}\right)^T\\ &=\frac{A^T+(A^T)^T}{2}\\ &=\frac{A^T+A}{2}\\ &=S\end{aligned} \]

Similarly, \(K\) is skew-symmetric:

\[ \begin{aligned} K^T&=\left(\frac{A-A^T}{2}\right)^T\\ &=\frac{A^T-A}{2}\\ &=-\frac{A-A^T}{2}\\ &=-K\end{aligned} \]

Therefore,

\[\boxed{A=\underbrace{\frac{A+A^T}{2}}_{\text{Symmetric part}}+\underbrace{\frac{A-A^T}{2}}_{\text{Skew-symmetric part}}}\]

✏️ Example

Question

Decompose a Matrix into Symmetric and Skew-Symmetric Parts

Let

\[A=\begin{bmatrix}2&3\\5&4\end{bmatrix}\]

Its transpose is

\[A^T=\begin{bmatrix}2&5\\3&4\end{bmatrix}\]

The symmetric part is

\[ \begin{aligned} S&=\frac12(A+A^T)\\ &=\frac12\begin{bmatrix} 4&8\\ 8&8\end{bmatrix}\\ &=\begin{bmatrix} 2&4\\4&4\end{bmatrix} \end{aligned} \]

The skew-symmetric part is

\[ \begin{aligned} K&=\frac12(A-A^T)\\ &=\frac12\begin{bmatrix} 0&-2\\2&0 \end{bmatrix}\\ &=\begin{bmatrix} 0&-1\\ 1&0\end{bmatrix} \end{aligned} \]

Hence,

\[A=S+K\]

\[\begin{bmatrix}2&3\\5&4\end{bmatrix}=\begin{bmatrix}2&4\\4&4\end{bmatrix}+\begin{bmatrix}0&-1\\1&0\end{bmatrix}\]

🗒️ Important Results

Diagonal of a Skew-Symmetric Matrix

If \(A\) is skew-symmetric, then:

\[\boxed{a_{11}=a_{22}=\cdots=a_{nn}=0}\]

Consequently, the trace of a skew-symmetric matrix is zero:

\[\operatorname{tr}(A)=a_{11}+a_{22}+\cdots+a_{nn}=0\]

\[\boxed{\operatorname{tr}(A)=0}\]

A Matrix Cannot Be Both Symmetric and Skew-Symmetric Except the Zero Matrix

Suppose \(A\) is both symmetric and skew-symmetric. Then

\[A^T=A\]

and

\[A^T=-A\]

Therefore,

\[\begin{aligned}A&=-A\\2A&=0\\[8pt]A&=0\end{aligned}\]

Thus, the only matrix that is both symmetric and skew-symmetric is the zero matrix.

🌟 CBSE Board Significance

Symmetric and skew-symmetric matrices are high-value concepts for board examinations. Questions commonly involve:

checking whether a matrix is symmetric or skew-symmetric,
finding unknown entries using \(A^T=A\),
finding unknown entries using \(A^T=-A\),
proving that diagonal entries of a skew-symmetric matrix are zero,
decomposing a square matrix into symmetric and skew-symmetric matrices, and
using transpose identities to simplify matrix expressions.

Competitive Examination Significance

For JEE and other entrance examinations, the fastest approach is often to use the defining identities rather than explicitly calculating a complete transpose.

The two identities to recognise immediately are:

\[ \boxed{A^T=A\quad\Rightarrow\quad\text{symmetric}} \]

\[ \boxed{A^T=-A\quad\Rightarrow\quad\text{skew-symmetric}} \]

For a skew-symmetric matrix, the zero diagonal is an immediate consequence and can eliminate incorrect options quickly.

⚡ Exam Tip

A symmetric or skew-symmetric matrix must always be square.
For symmetry, corresponding elements across the principal diagonal are equal:
\[ a_{ij}=a_{ji}. \]
For skew symmetry, corresponding elements are negatives:
\[ a_{ij}=-a_{ji}. \]
Every diagonal element of a skew-symmetric matrix is zero.
The zero matrix is both symmetric and skew-symmetric.
Any square matrix can be decomposed as
\[ A=\frac12(A+A^T)+\frac12(A-A^T). \]
Do not assume that a matrix containing zero diagonal entries is automatically skew-symmetric. The off-diagonal entries must also have opposite signs.

❌ Common Mistakes

Calling a rectangular matrix symmetric: Symmetric and skew-symmetric matrices must be square.
Forgetting the sign condition: A skew-symmetric matrix requires \(a_{ij}=-a_{ji}\), not merely unequal entries.
Assuming diagonal entries of a symmetric matrix must be zero: This is false. Only skew-symmetric matrices necessarily have zero diagonal entries.
Checking only one pair of off-diagonal elements: Every corresponding pair must satisfy the required condition.
Incorrectly copying entries while taking transpose: Carefully interchange \(a_{ij}\) and \(a_{ji}\).

📋 CBSE Case Study / HOTS

Let

\[ A= \begin{bmatrix} 2&x&3\\ y&5&4\\ 3&z&6 \end{bmatrix} \]

Suppose \(A\) is symmetric. Find \(x,y,z\).

Since \(A^T=A\),

\[ a_{12}=a_{21} \]

\[ x=y. \]

Also,

\[ a_{23}=a_{32} \]

\[ 4=z. \]

Therefore,

\[ \boxed{y=x,\qquad z=4}. \]

HOTS Extension

If the same matrix is skew-symmetric, its diagonal elements must be zero. Since the diagonal entries are \(2,5,6\), the matrix cannot be skew-symmetric.

\[ \boxed{\text{The matrix cannot be skew-symmetric.}} \]

⚡ Quick Revision

Symmetric matrix:

\[ \boxed{A^T=A} \]

\[ \boxed{a_{ij}=a_{ji}} \]

Skew-symmetric matrix:

\[ \boxed{A^T=-A} \]

\[ \boxed{a_{ij}=-a_{ji}} \]

Diagonal entries of skew-symmetric matrix:

\[ \boxed{a_{ii}=0} \]

Trace of a skew-symmetric matrix:

\[ \boxed{\operatorname{tr}(A)=0} \]

Symmetric part of any square matrix:

\[ \boxed{ S=\frac12(A+A^T) } \]

Skew-symmetric part of any square matrix:

\[ \boxed{ K=\frac12(A-A^T) } \]

Decomposition:

\[ \boxed{ A=S+K = \frac12(A+A^T)+\frac12(A-A^T) } \]

Both symmetric and skew-symmetric:

\[ \boxed{A=0} \]

🔑 Key Takeaway

A symmetric matrix satisfies \(A^T=A\).
A skew-symmetric matrix satisfies \(A^T=-A\).
Both types must be square matrices.
In a symmetric matrix, entries reflected across the principal diagonal are equal.
In a skew-symmetric matrix, entries reflected across the principal diagonal have opposite signs.
Every diagonal entry of a skew-symmetric matrix is zero.
The trace of a skew-symmetric matrix is zero.
Every square matrix can be uniquely decomposed into symmetric and skew-symmetric parts.
The zero matrix is the only matrix that is both symmetric and skew-symmetric.

▦

Theorem 1: Symmetric and Skew-Symmetric Matrices from a Square Matrix

📋

Table of Contents

12 sections

1 🧮 Theorem ›
2 📌 Element-Wise Explanation ›
3 ✏️ Example ›
4 📎 Important Corollary: Symmetric-Skew-Symmetric Decomposition ›
5 ✏️ Example ›
6 📌 Uniqueness of the Decomposition ›
7 🌟 Board Examination Significance ›
8 ⚡ Exam Tip ›
9 ❌ Common Mistakes ›
10 📋 CBSE Case Study / HOTS ›
11 ⚡ Quick Revision Formulae ›
12 🔑 Key Takeaway ›

🧮 Theorem

Statement

Let \(A\) be a square matrix with real number entries. Then:

\(A+A^T\) is a symmetric matrix.
\(A-A^T\) is a skew-symmetric matrix.

\[ \boxed{A+A^T\text{ is symmetric}} \]

\[ \boxed{A-A^T\text{ is skew-symmetric}} \]

Prerequisite

We use the following transpose properties:
\[\boxed{(A+B)^T=A^T+B^T}\]
\[\boxed{(A-B)^T=A^T-B^T}\]
\[\boxed{(A^T)^T=A}\]
We also use the commutative property of matrix addition:
\[\boxed{A+B=B+A}\]

Part I: Proof that \(A+A^T\) is Symmetric

Let
\[B=A+A^T\]
To prove that \(B\) is symmetric, we have to show that
\[B^T=B\]
Taking transpose on both sides of the definition of \(B\):
\[\begin{aligned}B^T&=(A+A^T)^T\\&=A^T+(A^T)^T\\&=A^T+A\\&=A+A^T\\&=B\end{aligned}\]
Therefore,
\[\boxed{B^T=B}\]
Hence, \(B=A+A^T\) is a symmetric matrix.
\[\boxed{A+A^T\text{ is symmetric}}\]

Part II: Proof that \(A-A^T\) is Skew-Symmetric

Let
\[C=A-A^T\]
To prove that \(C\) is skew-symmetric, we have to show that
\[C^T=-C\]
Taking transpose:
\[\begin{aligned}C^T&=(A-A^T)^T\\&=A^T-(A^T)^T\\&=A^T-A\\&=-(A-A^T)\\&=-C\end{aligned}\]
Therefore,
\[\boxed{C^T=-C}\]
Hence, \(C=A-A^T\) is a skew-symmetric matrix.
\[\boxed{A-A^T\text{ is skew-symmetric}}\]

Complete Proof at a Glance For \(A+A^T\)

\[\begin{aligned}(A+A^T)^T&=A^T+(A^T)^T\\&=A^T+A\\&=A+A^T\end{aligned}\]

\[\boxed{(A+A^T)^T=A+A^T}\]
Therefore \(A+A^T\) is symmetric.

Complete Proof at a Glance For \(A-A^T\)

\[\begin{aligned}(A-A^T)^T&=A^T-(A^T)^T\\&=A^T-A\\&=-(A-A^T)\end{aligned}\]

\[\boxed{(A-A^T)^T=-(A-A^T)}\]
Therefore \(A-A^T\) is skew-symmetric.

📌 Element-Wise Explanation

Let

\[A=[a_{ij}]\]

The \((i,j)\)-th element of \(A+A^T\) is

\[a_{ji}+a_{ij}\]

Since addition is commutative,

\[a_{ij}+a_{ji}=a_{ji}+a_{ij}\]

Thus,

\[(A+A^T)_{ij}=(A+A^T)_{ji}\]

Hence \(A+A^T\) is symmetric.

Similarly, the \((i,j)\)-th element of \(A-A^T\) is

\[a_{ij}-a_{ji}\]

Its \((j,i)\)-th element is

\[a_{ji}-a_{ij}=-(a_{ij}-a_{ji})\]

Therefore,

\[(A-A^T)_{ji}=-(A-A^T)_{ij}\]

Hence \(A-A^T\) is skew-symmetric.

✏️ Example

Let

\[A=\begin{bmatrix}2&3&5\\1&4&6\\7&8&9\end{bmatrix}\]

Then

\[A^T=\begin{bmatrix}2&1&7\\3&4&8\\5&6&9\end{bmatrix}\]

Finding \(A+A^T\)

\[\begin{aligned}A+A^T&=\begin{bmatrix}2&3&5\\1&4&6\\7&8&9\end{bmatrix}+ \begin{bmatrix}2&1&7\\3&4&8\\5&6&9\end{bmatrix}\\ &=\begin{bmatrix}4&4&12\\4&8&14\\12&14&18\end{bmatrix}\end{aligned}\]

Taking its transpose:

\[(A+A^T)^T=\begin{bmatrix}4&4&12\\4&8&14\\12&14&18\end{bmatrix}\]

Therefore,

\[(A+A^T)^T=A+A^T\]

\[\boxed{A+A^T\text{ is symmetric}}\]

Finding \(A-A^T\)

\[\begin{aligned}A-A^T&=\begin{bmatrix}2&3&5\\1&4&6\\7&8&9\end{bmatrix}- \begin{bmatrix}2&1&7\\3&4&8\\5&6&9\end{bmatrix}\\ &=\begin{bmatrix}0&2&-2\\-2&0&-2\\2&2&0\end{bmatrix}\end{aligned}\]

Its transpose is

\[(A-A^T)^T=\begin{bmatrix}0&-2&2\\2&0&2\\-2&-2&0\end{bmatrix}\]

Also,

\[-(A-A^T)=\begin{bmatrix}0&-2&2\\2&0&2\\-2&-2&0\end{bmatrix}\]

Hence,

\[(A-A^T)^T=-(A-A^T)\]

\[\boxed{A-A^T\text{ is skew-symmetric}}\]

📎 Important Corollary: Symmetric-Skew-Symmetric Decomposition

Since \(A+A^T\) is symmetric and \(A-A^T\) is skew-symmetric, multiplying each by \(\frac12\) gives:

\[\boxed{\frac12(A+A^T)\text{ is symmetric}}\]

\[\boxed{\frac12(A-A^T)\text{ is skew-symmetric}}\]

Adding them:

\[\begin{aligned}\frac12(A+A^T)+\frac12(A-A^T)&=\frac12(2A)\\&=A\end{aligned}\]

Therefore every square matrix \(A\) can be written as

\[\boxed{A=\frac12(A+A^T)+\frac12(A-A^T)}\]

This is called the symmetric and skew-symmetric decomposition of a square matrix.

Symmetric Part and Skew-Symmetric Part

The symmetric part of \(A\) is

\[\boxed{S=\frac12(A+A^T)}\]

The skew-symmetric part of \(A\) is

\[\boxed{K=\frac12(A-A^T)}\]

Hence,

\[\boxed{A=S+K}\]

✏️ Example

Question

Find the symmetric and skew-symmetric parts of

\[A=\begin{bmatrix}1&2&4\\3&5&6\\7&8&9\end{bmatrix}\]

Step 1: Find \(A^T\)

\[A^T=\begin{bmatrix}1&3&7\\2&5&8\\4&6&9\end{bmatrix}\]

Step 2: Symmetric Part

\[\begin{aligned}S&=\frac12(A+A^T)\\&=\frac12\begin{bmatrix}2&5&11\\5&10&14\\11&14&18\end{bmatrix}\\ &=\begin{bmatrix}1&\frac52&\frac{11}{2}\\\frac52&5&7\\\frac{11}{2}&7&9\end{bmatrix}\end{aligned}\]

Step 3: Skew-Symmetric Part

\[\begin{aligned}K&=\frac12(A-A^T)\\&=\frac12\begin{bmatrix}0&-1&-3\\1&0&-2\\3&2&0\end{bmatrix}\\ &=\begin{bmatrix}0&-\frac12&-\frac32\\\frac12&0&-1\\\frac32&1&0\end{bmatrix}\end{aligned}\]

Step 4: Verification

\[\begin{aligned}S+K&=\begin{bmatrix}1&\frac52&\frac{11}{2}\\\frac52&5&7\\\frac{11}{2}&7&9\end{bmatrix}+ \begin{bmatrix}0&-\frac12&-\frac32\\\frac12&0&-1\\\frac32&1&0\end{bmatrix}\\ &=\begin{bmatrix}1&2&4\\3&5&6\\7&8&9\end{bmatrix}\\&=A\end{aligned}\]
Thus the decomposition is verified.

📌 Uniqueness of the Decomposition

The representation

\[A=S+K\]

where \(S\) is symmetric and \(K\) is skew-symmetric, is unique.
Suppose

\[A=S_1+K_1=S_2+K_2\]

where \(S_1,S_2\) are symmetric and \(K_1,K_2\) are skew-symmetric.
Then

\[S_1-S_2=K_2-K_1\]

The left side is symmetric, while the right side is skew-symmetric. Therefore the common matrix must be both symmetric and skew-symmetric.

The only matrix having both properties is the zero matrix. Hence,

\[\begin{aligned}S_1-S_2&=0\\S_1&=S_2\end{aligned}\]

Similarly,

\[K_1=K_2\]

Therefore, the decomposition is unique.

\[\boxed{A=\frac12(A+A^T)+\frac12(A-A^T)\text{ is unique}}\]

🌟 Board Examination Significance

This theorem is highly useful in CBSE Class 12 Mathematics questions involving:

proof of symmetric and skew-symmetric matrix identities,
finding the symmetric part of a matrix,
finding the skew-symmetric part of a matrix,
expressing a matrix as the sum of symmetric and skew-symmetric matrices,
verification of transpose properties, and
case-study and application-based questions.

Competitive Entrance Examination Significance

In JEE and other competitive examinations, this theorem is particularly useful for rapid identification of matrix properties.

Whenever an expression has the form

\[ A+A^T, \]

immediately identify it as symmetric.

Whenever an expression has the form

\[ A-A^T, \]

immediately identify it as skew-symmetric.

This recognition can avoid unnecessary matrix multiplication and save valuable time in objective-type examinations.

⚡ Exam Tip

Memorise:
\[ \boxed{A+A^T\rightarrow\text{symmetric}} \]
Memorise:
\[ \boxed{A-A^T\rightarrow\text{skew-symmetric}} \]
For a proof, take the transpose of the complete expression.
Always use:
\[ (A^T)^T=A. \]
For \(A+A^T\), use commutativity of addition to obtain the original expression.
For \(A-A^T\), factor out \(-1\) to establish skew symmetry.
Remember the decomposition:
\[ \boxed{ A=\frac12(A+A^T)+\frac12(A-A^T) }. \]

❌ Common Mistakes

Forgetting the square-matrix condition: Symmetric and skew-symmetric matrices are defined only for square matrices.
Writing \((A-A^T)^T=A-A^T\): The correct identity is
\[ (A-A^T)^T=-(A-A^T). \]
Forgetting the double-transpose rule:
\[ (A^T)^T=A. \]
Confusing the symmetric and skew-symmetric parts:
\[ \frac12(A+A^T) \]
is symmetric, whereas
\[ \frac12(A-A^T) \]
is skew-symmetric.
Ignoring the factor \(\frac12\) in decomposition: The sum \(A+A^T+A-A^T\) gives \(2A\), not \(A\). Therefore division by \(2\) is essential.

📋 CBSE Case Study / HOTS

Let \(A\) be a square matrix and define

\[ P=A+A^T \]

and

\[ Q=A-A^T \]

A student claims that both \(P\) and \(Q\) are symmetric. Determine whether the claim is correct.

Solution

For \(P\):

\[ \begin{aligned} P^T &=(A+A^T)^T\\ &=A^T+(A^T)^T\\ &=A^T+A\\ &=A+A^T\\ &=P \end{aligned} \]

Hence \(P\) is symmetric.

For \(Q\):

\[ \begin{aligned} Q^T &=(A-A^T)^T\\ &=A^T-(A^T)^T\\ &=A^T-A\\ &=-(A-A^T)\\ &=-Q \end{aligned} \]

Hence \(Q\) is skew-symmetric, not symmetric in general.

\[ \boxed{ P\text{ is symmetric and }Q\text{ is skew-symmetric} } \]

Therefore, the student's claim is incorrect.

⚡ Quick Revision

Quick Revision Formulae

Symmetric matrix:

\[ \boxed{A^T=A} \]

Skew-symmetric matrix:

\[ \boxed{A^T=-A} \]

Symmetric matrix obtained from \(A\):

\[ \boxed{A+A^T} \]

Skew-symmetric matrix obtained from \(A\):

\[ \boxed{A-A^T} \]

Symmetric part:

\[ \boxed{ S=\frac12(A+A^T) } \]

Skew-symmetric part:

\[ \boxed{ K=\frac12(A-A^T) } \]

Decomposition:

\[ \boxed{ A= \frac12(A+A^T)+ \frac12(A-A^T) } \]

🔑 Key Takeaway

\(A+A^T\) is always symmetric for a square matrix \(A\).
\(A-A^T\) is always skew-symmetric for a square matrix \(A\).
The proof follows directly from the transpose properties.
The symmetric part of \(A\) is
\[ \frac12(A+A^T). \]
The skew-symmetric part of \(A\) is
\[ \frac12(A-A^T). \]
Every square matrix can be uniquely decomposed as
\[ A=\frac12(A+A^T)+\frac12(A-A^T). \]

▦

Theorem 2: Expression of a Square Matrix as the Sum of a Symmetric and a Skew-Symmetric Matrix

📋

Table of Contents

12 sections

1 🧮 Theorem ›
2 🔬 Compact Proof for Board Examination ›
3 📐 Derivation of the Decomposition Formula ›
4 ✏️ Example ›
5 👁️ Important Observation About the Diagonal Entries ›
6 🧮 Important Corollary: Uniqueness ›
7 🌟 CBSE Board Examination Significance ›
8 📋 CBSE Case Study / HOTS ›
9 ⚡ Exam Tip ›
10 ❌ Common Mistakes ›
11 ⚡ Quick Revision Formulae ›
12 🔑 Key Takeaway ›

🧮 Theorem

Statement

Every square matrix can be expressed as the sum of a symmetric matrix and a skew-symmetric matrix.
If \(A\) is any square matrix, then

\[\boxed{A=\frac{1}{2}(A+A^T)+\frac{1}{2}(A-A^T)}\]

where

\[\boxed{\frac{1}{2}(A+A^T)}\]

is symmetric and

\[\boxed{\frac{1}{2}(A-A^T)}\]

is skew-symmetric.

Concept

A general square matrix need not be symmetric or skew-symmetric. However, its entries can always be separated into two parts:

a part that remains unchanged when the transpose is taken, and
a part that changes its sign when the transpose is taken.

These two parts are respectively called the symmetric part and the skew-symmetric part of the matrix.

Therefore, the theorem is not merely a method for representing a matrix; it gives a systematic decomposition of every square matrix into two structurally different components.

Prerequisite

The proof depends on Theorem 1:

\[ \boxed{A+A^T\text{ is symmetric}} \]

\[ \boxed{A-A^T\text{ is skew-symmetric}} \]

We also use the transpose property:

\[ \boxed{(kA)^T=kA^T} \]

for any scalar \(k\).

In addition,

\[ \boxed{(A^T)^T=A}. \]

Proof

Step 1: Let \(A\) be a square matrix
Let \(A\) be any square matrix. Consider the expression
\[\frac12(A+A^T)+\frac12(A-A^T)\]
Step 2: Simplify the expression
\[\begin{aligned}\frac12(A+A^T)+\frac12(A-A^T)&=\frac12\left[(A+A^T)+(A-A^T)\right]\\&=\frac12(2A)\\&=A\end{aligned}\]
Hence,
\[\boxed{A=\frac12(A+A^T)+\frac12(A-A^T)}\]
Step 3: Identify the first component
From Theorem 1,
\[A+A^T\]
is symmetric.
Multiplying a symmetric matrix by a scalar does not destroy symmetry. Therefore,
\[\frac12(A+A^T)\]
is also symmetric.
Verification
\[\begin{aligned}\left[\frac12(A+A^T)\right]^T&=\frac12(A+A^T)^T\\&=\frac12(A+A^T)\end{aligned}\]
Therefore,
\[\boxed{\frac12(A+A^T)\text{ is symmetric}}\]
Step 4: Identify the second component
From Theorem 1,
\[A-A^T\]
is skew-symmetric.
Multiplication by the scalar \(\frac12\) preserves skew symmetry. Hence,
\[\frac12(A-A^T)\]
is skew-symmetric.
Verification
\[\begin{aligned}\left[\frac12(A-A^T)\right]^T&=\frac12(A-A^T)^T\\&=\frac12\left[-(A-A^T)\right]\\&=-\frac12(A-A^T)\end{aligned}\]
Therefore,
\[\boxed{\frac12(A-A^T)\text{ is skew-symmetric}}\]
Thus,
\[\boxed{A=\underbrace{\frac12(A+A^T)}_{\text{Symmetric matrix}}+\underbrace{\frac12(A-A^T)}_{\text{Skew-symmetric matrix}}}.\]

Hence Proved.

🔬 Compact Proof for Board Examination

🔬 Proof

Let \(A\) be a square matrix. We can write
\[\begin{aligned}A&=\frac12(A+A^T)+\frac12(A-A^T).\end{aligned}\]
By Theorem 1, \(A+A^T\) is symmetric and \(A-A^T\) is skew-symmetric. Since multiplication by a scalar preserves these properties,
\[\frac12(A+A^T)\]
is symmetric and
\[\frac12(A-A^T)\]
is skew-symmetric.
Therefore,
\[\boxed{A=\frac12(A+A^T)+\frac12(A-A^T)}\]
expresses \(A\) as the sum of a symmetric and a skew-symmetric matrix.

📐 Derivation of the Decomposition Formula

The formula can also be derived by adding and subtracting \(A^T\) to \(A\).

\[\begin{aligned}A&=A+0\\&=A+\frac12A^T-\frac12A^T\\&=\frac12A+\frac12A^T+\frac12A-\frac12A^T\\&=\frac12(A+A^T)+\frac12(A-A^T)\end{aligned}\]

Thus the decomposition is obtained naturally by pairing \(A\) with its transpose.

Why the Factor \(\frac12\) Is Necessary

A common mistake is to write

\[A=(A+A^T)+(A-A^T)\]

This is incorrect because

\[\begin{aligned}(A+A^T)+(A-A^T)&=2A\end{aligned}\]

Therefore, the factor \(\frac12\) is essential:

\[\boxed{\frac12(A+A^T)+\frac12(A-A^T)=A}\]

✏️ Example

Question

Express

\[A=\begin{bmatrix}2&5\\1&4\end{bmatrix}\]

as the sum of a symmetric matrix and a skew-symmetric matrix.

Step 1: Find \(A^T\)

\[A^T=\begin{bmatrix}2&1\\5&4\end{bmatrix}\]

Step 2: Find the symmetric part

\[\begin{aligned}S&=\frac12(A+A^T)\\&=\frac12\left[\begin{bmatrix}2&5\\1&4\end{bmatrix}+ \begin{bmatrix}2&1\\5&4\end{bmatrix}\right]\\ &=\frac12\begin{bmatrix}4&6\\6&8\end{bmatrix}\\ &=\begin{bmatrix}2&3\\3&4\end{bmatrix}\end{aligned}\]
Clearly,

\[S^T=S\]

Step 3: Find the skew-symmetric part

\[\begin{aligned}K&=\frac12(A-A^T)\\&=\frac12\left[\begin{bmatrix}2&5\\1&4\end{bmatrix}- \begin{bmatrix}2&1\\5&4\end{bmatrix}\right]\\ &=\frac12\begin{bmatrix}0&4\\-4&0\end{bmatrix}\\ &=\begin{bmatrix}0&2\\-2&0\end{bmatrix}\end{aligned}\]
Clearly,

\[K^T=K\]

Step 4: Verify the decomposition

\[\begin{aligned}S+K&=\begin{bmatrix}2&3\\3&4\end{bmatrix}+ \begin{bmatrix}0&2\\-2&0\end{bmatrix}\\ &=\begin{bmatrix}2&5\\1&4\end{bmatrix}\\&=A\end{aligned}\]
Therefore,

\[\boxed{A=\begin{bmatrix}2&3\\3&4\end{bmatrix}+\begin{bmatrix}0&2\\-2&0\end{bmatrix}}\]

Question

Express

\[A=\begin{bmatrix}1&2&4\\3&5&6\\7&8&9\end{bmatrix}\]

as the sum of a symmetric and a skew-symmetric matrix.

Step 1: Transpose of \(A\)

\[A^T=\begin{bmatrix}1&3&7\\2&5&8\\4&6&9\end{bmatrix}\]

Step 2: Symmetric Part

\[\begin{aligned}S&=\frac12(A+A^T)\\&=\frac12\begin{bmatrix}2&5&11\\5&10&14\\11&14&18\end{bmatrix}\\ &=\boxed{\begin{bmatrix}1&\frac52&\frac{11}{2}\\\frac52&5&7\\\frac{11}{2}&7&9\end{bmatrix}}\end{aligned}\]

Step 3: Skew-Symmetric Part

\[\begin{aligned}K&=\frac12(A-A^T)\\&=\frac12\begin{bmatrix}0&-1&-3\\1&0&-2\\3&2&0\end{bmatrix}\\ &=\begin{bmatrix}0&-\frac12&-\frac32\\\frac12&0&-1\\\frac32&1&0\end{bmatrix}\end{aligned}\]

Step 4: Final Representation

\[A=\begin{bmatrix}1&\frac52&\frac{11}{2}\\\frac52&5&7\\\frac{11}{2}&7&9\end{bmatrix}+ \begin{bmatrix}0&-\frac12&-\frac32\\\frac12&0&-1\\\frac32&1&0\end{bmatrix}\]

📌

Note Consider a general \(3\times3\) matrix:

\[A=\begin{bmatrix}a&b&c\\d&e&f\\g&h&i\end{bmatrix}\]

Its symmetric part is

\[\frac12(A+A^T)=\begin{bmatrix}a&\frac{b+d}{2}&\frac{c+g}{2}\\\frac{b+d}{2}&e&\frac{f+h}{2}\\\frac{c+g}{2}&\frac{f+h}{2}&i\end{bmatrix}\]

Its skew-symmetric part is

\[\frac12(A-A^T)=\begin{bmatrix}0&\frac{b-d}{2}&\frac{c-g}{2}\\-\frac{b-d}{2}&0&\frac{f-h}{2}\\-\frac{c-g}{2}&-\frac{f-h}{2}&0\end{bmatrix}\]

This makes the decomposition especially clear:

The symmetric part contains the average of corresponding off-diagonal entries.
The skew-symmetric part contains half of the difference between corresponding off-diagonal entries.
The entire diagonal belongs to the symmetric part.
Every diagonal entry of the skew-symmetric part is zero.

👁️ Important Observation About the Diagonal Entries

🧮 Important Corollary: Uniqueness

The decomposition of a square matrix into a symmetric matrix and a skew-symmetric matrix is unique.

Suppose

\[A=S_1+K_1=S_2+K_2\]

where \(S_1,S_2\) are symmetric and \(K_1,K_2\) are skew-symmetric. Then

\[S_1-S_2=K_2-K_1\]

The left-hand side is symmetric, while the right-hand side is skew-symmetric. Therefore the common matrix must be both symmetric and skew-symmetric.

The only matrix having both properties is the zero matrix. Hence,

\[S_1=S_2\]

and

\[K_1=K_2\]

Therefore the decomposition is unique:

\[\boxed{A=\frac12(A+A^T)+\frac12(A-A^T)}\]

🧮 Theorem

The decomposition of a square matrix into a symmetric matrix and a skew-symmetric matrix is unique.

Suppose

\[A=S_1+K_1=S_2+K_2\]

where \(S_1,S_2\) are symmetric and \(K_1,K_2\) are skew-symmetric. Then

\[S_1-S_2=K_2-K_1\]

\[S_1=S_2\]

and

\[K_1=K_2\]

Therefore the decomposition is unique:

\[\boxed{A=\frac12(A+A^T)+\frac12(A-A^T)}\]

🌟 CBSE Board Examination Significance

This theorem is important for questions involving:

proof-based questions on symmetric and skew-symmetric matrices,
finding the symmetric part of a given matrix,
finding the skew-symmetric part of a given matrix,
expressing a matrix as the sum of two special matrices,
verification of \(A=S+K\), and
case-study and competency-based questions.

Competitive Entrance Examination Significance

For JEE and other competitive examinations, the decomposition formula is a high-value shortcut:

\[ \boxed{ S=\frac12(A+A^T),\qquad K=\frac12(A-A^T) } \]

Instead of solving for unknown entries individually, these formulas immediately give the two required matrices.

If an objective question asks which expression is symmetric, identify \(\frac12(A+A^T)\) immediately. If it asks for a skew-symmetric expression, identify \(\frac12(A-A^T)\).

📋 CBSE Case Study / HOTS

A square matrix \(A\) is represented as

\[ A=S+K, \]

where \(S\) is symmetric and \(K\) is skew-symmetric. A student claims that \(S=A+A^T\) and \(K=A-A^T\). Is the claim correct?

Solution

The matrices

\[ A+A^T \]

and

\[ A-A^T \]

are respectively symmetric and skew-symmetric, but their sum is

\[ (A+A^T)+(A-A^T)=2A. \]

Therefore they do not directly add to \(A\). Each must be multiplied by \(\frac12\).

\[ \boxed{ S=\frac12(A+A^T) } \]

\[ \boxed{ K=\frac12(A-A^T) } \]

Hence the student's claim is incorrect.

⚡ Exam Tip

Remember the master formula:
\[ \boxed{ A=\frac12(A+A^T)+\frac12(A-A^T) }. \]
The first term is always symmetric.
The second term is always skew-symmetric.
The factor \(\frac12\) cannot be omitted.
The skew-symmetric component always has zero diagonal entries.
The original diagonal entries of \(A\) appear unchanged in its symmetric part.
For a numerical problem, calculate \(A^T\) first and then use the two formulas.

❌ Common Mistakes

Writing
\[ A=(A+A^T)+(A-A^T). \]
This gives \(2A\), not \(A\).
Forgetting the factor \(\frac12\).
Interchanging the symmetric and skew-symmetric components.
Assuming that \(A\) itself must be symmetric or skew-symmetric. The theorem applies to any square matrix.
Forgetting that the skew-symmetric component has zero diagonal.
Making arithmetic errors while calculating \(A^T\).

⚡ Quick Revision Formulae

Symmetric part of \(A\):

\[ \boxed{ S=\frac12(A+A^T) } \]

Skew-symmetric part of \(A\):

\[ \boxed{ K=\frac12(A-A^T) } \]

Decomposition of \(A\):

\[ \boxed{ A=S+K } \]

Therefore,

\[ \boxed{ A= \frac12(A+A^T)+ \frac12(A-A^T) } \]

Properties:

\[ \boxed{S^T=S} \]

\[ \boxed{K^T=-K} \]

\[ \boxed{A=S+K} \]

🔑 Key Takeaway

Every square matrix can be decomposed into two parts: one symmetric and one skew-symmetric.
The symmetric part is
\[ \frac12(A+A^T). \]
The skew-symmetric part is
\[ \frac12(A-A^T). \]
The factor \(\frac12\) ensures that the two components add to exactly \(A\).
The symmetric part contains all the diagonal entries of the original matrix.
The skew-symmetric part has zero diagonal entries.
The decomposition is unique.
This theorem is a direct and important application of transpose properties and is highly useful in CBSE and competitive examinations.

▦

Example 16

📋

Table of Contents

5 sections

1 ❓ Question ›
2 💡 Concept ›
3 🧩 Solution ›
4 🧩 Exam-Oriented Shortcut ›
5 ⚡ Quick Revision ›

❓ Question

Express the matrix

\[B=\begin{bmatrix}2&-2&-4\\-1&3&4\\&-2&-3\end{bmatrix}\]

as the sum of a symmetric matrix and a skew-symmetric matrix.

💡 Concept

By Theorem 2, every square matrix \(B\) can be uniquely expressed in the form

\[ \boxed{B=P+Q} \]

where \(P\) is symmetric and \(Q\) is skew-symmetric.

The required matrices are obtained directly from:

\[ \boxed{ P=\frac12(B+B^T) } \]

\[ \boxed{ Q=\frac12(B-B^T) } \]

Therefore, the solution strategy is:

Find \(B^T\).
Calculate \(P=\frac12(B+B^T)\).
Calculate \(Q=\frac12(B-B^T)\).
Verify that \(P^T=P\).
Verify that \(Q^T=-Q\).
Finally verify that \(P+Q=B\).

🧩 Solution

Step 1: Find the Transpose of \(B\)

Given

\[B=\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix}\]
Interchanging rows and columns,

\[B^T=\begin{bmatrix}2&-1&1\\-2&3&-2\\-4&4&-3\end{bmatrix}\]

Step 2: Find the Symmetric Matrix \(P\)

We use

\[P=\frac12(B+B^T)\]
Therefore,

\[\begin{aligned}P&=\frac12\left[\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix}+ \begin{bmatrix}2&-1&1\\-2&3&-2\\-4&4&-3\end{bmatrix}\right]\\ &=\frac12\begin{bmatrix}2+2&-2-1&-4+1\\-1-2&3+3&4-2\\1-4&-2+4&-3-3\end{bmatrix}\\ &=\frac12\begin{bmatrix}4&-3&-3\\-3&6&2\\-3&2&-6\end{bmatrix}\\ &=\begin{bmatrix}2&-\frac32&-\frac32\\-\frac32&3&1\\-\frac32&1&-3\end{bmatrix}\end{aligned}\]
Verification that \(P\) is Symmetric

\[P^T=\begin{bmatrix}2&-\frac32&-\frac32\\-\frac32&3&1\\-\frac32&1&-3\end{bmatrix}\]
Thus,

\[\boxed{P^T=P}\]

Hence \(P\) is a symmetric matrix.

Step 3: Find the Skew-Symmetric Matrix \(Q\)

We use

\[Q=\frac12(B-B^T)\]
Therefore,

\[\begin{aligned}Q&=\frac12\left[\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix}- \begin{bmatrix}2&-1&1\\-2&3&-2\\-4&4&-3\end{bmatrix}\right]\\ &=\frac12\begin{bmatrix}2-2&-2-(-1)&-4-1\\-1-(-2)&3-3&4-(-2)\\1-(-4)&-2-4&-3-(-3)\end{bmatrix}\\ &=\frac12\begin{bmatrix}0&-1&-5\\1&0&6\\5&-6&0\end{bmatrix}\\ &=\begin{bmatrix}0&-\frac12&-\frac52\\\frac12&0&3\\\frac52&-3&0\end{bmatrix}\end{aligned}\]
Verification that \(Q\) is Skew-Symmetric

\[Q^T=\begin{bmatrix}0&\frac12&\frac52\\-\frac12&0&-3\\-\frac52&3&0\end{bmatrix}\]
On the other hand,

\[-Q=\begin{bmatrix}0&\frac12&\frac52\\-\frac12&0&-3\\-\frac52&3&0\end{bmatrix}\]
Hence, \[\boxed{Q^T=-Q}\]
Therefore, \(Q\) is a skew-symmetric matrix.

Step 4: Verify that \(P+Q=B\)

\[\begin{aligned}P+Q&=\begin{bmatrix}2&-\frac32&-\frac32\\-\frac32&3&1\\-\frac32&1&-3\end{bmatrix}+ \begin{bmatrix}0&-\frac12&-\frac52\\\frac12&0&3\\\frac52&-3&0\end{bmatrix}\\ &=\begin{bmatrix}2+0&-\frac32-\frac12&-\frac32-\frac52\\-\frac32+\frac12&3+0&1+3\\-\frac32+\frac52&1-3&-3+0\end{bmatrix}\\ &=\begin{bmatrix}2&-2&-4\\-1&3&4\\1&-2&-3\end{bmatrix}\\&=B\end{aligned}\]
Hence,

\[\boxed{P+Q=B}\]

🧩 Exam-Oriented Shortcut

For any square matrix \(B\), there is no need to derive the decomposition from scratch. Directly use:

\[ \boxed{ \text{Symmetric part}=\frac12(B+B^T) } \]

\[ \boxed{ \text{Skew-symmetric part}=\frac12(B-B^T) } \]

In a board examination, showing these two formulas followed by the calculation and a brief verification is a clean and efficient method.

⚡ Quick Revision

\(P^T=P\), so \(P\) is symmetric.
\(Q^T=-Q\), so \(Q\) is skew-symmetric.
\(P+Q=B\).
Every square matrix has a unique symmetric-skew-symmetric decomposition.

\[ \boxed{ B= \frac12(B+B^T)+ \frac12(B-B^T) } \]

▦

Invertible Matrix

📋

Table of Contents

22 sections

1 📘 Definition ›
2 📄 Important Terminology ›
3 💡 Concept of an Inverse Matrix ›
4 🧭 Existence Condition for an Inverse ›
5 🤔 Why Must the Matrix Be Square? ›
6 ✏️ Example ›
7 🔢 Finding the Inverse of a Matrix ›
8 📐 Derivation of the Inverse Formula ›
9 ✏️ Example ›
10 📎 Inverse Matrix and Matrix Equation ›
11 🏷️ Important Properties of the Inverse Matrix ›
12 📐 Derivation of ›
13 📄 Uniqueness Of The Inverse ›
14 🔗 Connection Between Inverse and Determinant ›
15 🌟 Geometrical Significance ›
16 🛠️ Applications and Significance ›
17 🌟 CBSE Board Examination Significance ›
18 ⚡ Exam Tip ›
19 ❌ Common Mistakes ›
20 📋 CBSE Case Study / HOTS ›
21 ⚡ Quick Revision Formulae ›
22 🔑 Key Takeaway ›

📘 Definition

Let \(A\) be a square matrix of order \(n\). If there exists a square matrix \(B\) of the same order such that

\[AB=BA=I_n\]

where \(I_n\) is the identity matrix of order \(n\), then \(B\) is called the inverse of \(A\). The inverse of \(A\) is denoted by \(A^\prime \). Thus,

\[\boxed{AA^\prime =A^\prime A=I_n}\]

In this case, \(A\) is called an invertible matrix or non-singular matrix.

🗒️ Important Terminology

\(A\) is called the invertible matrix.
\(A^\prime \) is called the inverse matrix of \(A\).
\(I_n\) is the identity matrix of the same order as \(A\).
If \(A^\prime \) exists, then \(A\) is also called non-singular.
If no inverse exists, \(A\) is called a singular matrix.

💡 Concept of an Inverse Matrix

🧭 Existence Condition for an Inverse

🧭

Condition A matrix inverse exists only for a square matrix satisfying the appropriate invertibility condition.

For a square matrix \(A\),

\[\boxed{A\text{ is invertible}\iff |A|\neq0}\]

Equivalently,

\[\boxed{A\text{ is invertible}\iff \det(A)\neq0}\]

and

\[\boxed{A\text{ is singular}\iff \det(A)=0}\]

The determinant criterion is particularly important for Class 12 board examinations and competitive entrance examinations.

🤔 Why Must the Matrix Be Square?

If \(A\) has an inverse \(A^\prime\), then both products

\[AA^\prime\]

and

\[A^\prime A\]

must be identity matrices. This requires the dimensions to be compatible in both directions. Consequently, an invertible matrix must be square.

Therefore:

\[\boxed{\text{Only square matrices can have a two-sided matrix inverse.}}\]

✏️ Example

Question

Let

\[A=\begin{bmatrix}2&3\\1&2\end{bmatrix}\]

and

\[B=\begin{bmatrix}2&-3\\-1&2\end{bmatrix}\]

We verify whether \(B=A^\prime \).

Step 1: Calculate \(AB\)

\[\begin{aligned}AB&=\begin{bmatrix}2&3\\1&2\end{bmatrix}\begin{bmatrix}2&-3\\-1&2\end{bmatrix}\\ &=\begin{bmatrix}2(2)+3(-1)&2(-3)+3(2)\\1(2)+2(-1)&1(-3)+2(2)\end{bmatrix}\\ &=\begin{bmatrix}4-3&-6+6\\2-2&-3+4\end{bmatrix}\\ &=\begin{bmatrix}1&0\\0&1\end{bmatrix}\end{aligned}\]
Thus,

\[AB=I_2\]

Step 2: Calculate \(BA\)

\[\begin{aligned}BA&=\begin{bmatrix}2&-3\\-1&2\end{bmatrix} \begin{bmatrix}2&3\\1&2\end{bmatrix}\\ &=\begin{bmatrix}2(2)+(-3)(1)&2(3)+(-3)(2)\\(-1)(2)+2(1)&(-1)(3)+2(2)\end{bmatrix}\\ &=\begin{bmatrix}4-3&6-6\\-2+2&-3+4\end{bmatrix}\\ &=\begin{bmatrix}1&0\\0&1\end{bmatrix}\end{aligned}\]

🔢 Finding the Inverse of a \(2\times2\) Matrix

📐 Derivation of the \(2\times2\) Inverse Formula

Let

\[A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\]

Consider the matrix

\[C=\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\]

Multiplying \(A\) and \(C\):

\[\begin{aligned}AC&=\begin{bmatrix}a&b\\c&d\end{bmatrix} \begin{bmatrix}d&-b\\-c&a\end{bmatrix}\\ &=\begin{bmatrix}ad-bc&-ab+ab\\cd-dc&-bc+ad\end{bmatrix}\\ &=\begin{bmatrix}ad-bc&0\\0&ad-bc\end{bmatrix}\\ &=(ad-bc)\begin{bmatrix}1&0\\0&1\end{bmatrix}\end{aligned}\]

Therefore,

\[AC=|A|I\]

If \(|A|\neq0\), divide both sides by \(|A|\):

\[A\left(\frac{1}{|A|}C\right)=I\]

Hence,

\[A^\prime =\frac{1}{|A|}\begin{bmatrix}d&-b\\-c&a\end{bmatrix}\]

✏️ Example

Question

Find the inverse of

\[A=\begin{bmatrix}3&2\\1&1\end{bmatrix}\]

Step 1: Find the determinant

\[|A|=(3)(1)-(2)(1)=1\]
Since

\[|A|=1\neq0\]
the matrix is invertible.

Step 2: Apply the inverse formula

\[\begin{aligned}A^\prime &=\frac{1}{1}\begin{bmatrix}1&-2\\-1&3\end{bmatrix}\\&=\boxed{\begin{bmatrix}1&-2\\-1&3\end{bmatrix}}\end{aligned}\]

Question

Determine whether

\[A=\begin{bmatrix}2&4\\1&2\end{bmatrix}\]

\[|A|=(2)(2)-(4)(1)=4-4=0\]
Since

\[\boxed{|A|=0}\]
the matrix is singular. Therefore, its inverse does not exist.
\[\boxed{A^\prime\text{ does not exist}}\]

📎 Inverse Matrix and Matrix Equation

\[AX=B\]

and \(A\) is invertible, multiply both sides on the left by \(A^\prime \):

\[\begin{aligned}A^\prime AX&=A^\prime B\\IX&=A^\prime B\\X&=A^\prime B\end{aligned}\]

Hence,

\[\boxed{X=A^\prime B}\]

Similarly, if

\[XA=B\]

then

\[\boxed{X=BA^\prime}\]

The order of multiplication is important because matrix multiplication is generally not commutative.

🏷️ Important Properties of the Inverse Matrix

📐 Derivation of \((AB)^\prime =B^\prime A^\prime \)

Consider

\[(AB)(B^\prime A^\prime )\]

Using associativity:

\[\begin{aligned}(AB)(B^\prime A^\prime )&=A(BB^\prime )A^\prime \\&=AIA^\prime \\&=AA^\prime \\&=I\end{aligned}\]

Similarly,

\[\begin{aligned}(B^\prime A^\prime )(AB)&=B^\prime (A^\prime A)B\\&=B^\prime IB\\&=B^\prime B\\&=I\end{aligned}\]

Therefore,

\[\boxed{(AB)^\prime =B^\prime A^\prime }\]

Notice the reversal of order. This is an important point in matrix algebra.

🗒️ Uniqueness Of The Inverse

A matrix cannot have two different inverses. Suppose \(B\) and \(C\) are both inverses of \(A\). Then

\[AB=BA=I\]

and

\[AC=CA=I\]

Now,

\[\begin{aligned}B&=BI\\&=B(AC)\\&=(BA)C\\&=IC\\&=C\end{aligned}\]

Hence,

\[\boxed{B=C}\]

Therefore, the inverse of a matrix, whenever it exists, is unique.

🔗 Connection Between Inverse and Determinant

For a square matrix \(A\), the following statements are equivalent:

\[ \boxed{ \begin{aligned} A&\text{ is invertible}\\ &\Longleftrightarrow \det(A)\neq0\\ &\Longleftrightarrow A\text{ is non-singular}\\ &\Longleftrightarrow \operatorname{rank}(A)=n \end{aligned} } \]

For a \(2\times2\) matrix

\[A=\begin{bmatrix}a&b\\c&d\end{bmatrix}\]

this becomes

\[\boxed{A\text{ is invertible}\iff ad-bc\neq0}\]

🌟 Geometrical Significance

A matrix can represent a linear transformation of vectors. If a matrix is invertible, the transformation can be reversed.

\[\mathbf{x}\xrightarrow{\ A\ }A\mathbf{x}\xrightarrow{\ A^\prime \ }A^\prime A\mathbf{x}=\mathbf{x}\]

Thus, \(A^\prime \) represents the reverse transformation corresponding to \(A\).

A singular matrix does not possess such a reverse transformation because it maps different vectors into a lower-dimensional configuration.

🛠️ Applications and Significance

solving simultaneous linear equations,
linear transformations and coordinate transformations,
computer graphics and geometric transformations,
engineering and physical modelling,
economics and input-output models,
statistics and data analysis,
optimization and numerical methods, and
systems of linear equations in competitive mathematics.

🌟 CBSE Board Examination Significance

In Class 12 Mathematics, inverse matrices are particularly important for:

finding the inverse of a \(2\times2\) or \(3\times3\) matrix,
checking whether a matrix is invertible,
solving matrix equations,
solving simultaneous linear equations using matrix methods,
proving identities involving inverses, and
case-study and application-based questions.

Competitive Entrance Examination Significance

For JEE and other entrance examinations, the following results are particularly useful for quick problem solving:

\[ \boxed{AA^\prime =A^\prime A=I} \]

\[ \boxed{(AB)^\prime =B^\prime A^\prime } \]

\[ \boxed{(A^T)^\prime =(A^\prime )^T} \]

\[ \boxed{\det(A)\neq0\iff A^\prime \text{ exists}} \]

The reversal of order in \((AB)^\prime \) is a frequent source of multiple-choice questions.

⚡ Exam Tip

Always check whether the matrix is square before discussing its inverse.
For a \(2\times2\) matrix, calculate the determinant first.
If \(\det(A)=0\), stop: the inverse does not exist.
Do not confuse \(A^\prime \) with \(\frac{1}{A}\). Matrix inverse is a distinct algebraic concept.
Remember that
\[ (AB)^\prime =B^\prime A^\prime , \]
not \(A^\prime B^\prime \).
To verify a proposed inverse \(B\), check
\[ AB=BA=I. \]
When solving \(AX=B\), multiply on the left:
\[ X=A^\prime B. \]

❌ Common Mistakes

Trying to find the inverse of a non-square matrix.
Forgetting to check that \(\det(A)\neq0\).
Using the wrong signs in the \(2\times2\) inverse formula.
Writing
\[ (AB)^\prime =A^\prime B^\prime . \]
The correct order is reversed.
Assuming \(AB=I\) automatically means \(A\) and \(B\) are arbitrary matrices. In the standard Class 12 context, \(A\) and \(B\) must be square matrices of the same order for a two-sided inverse.
Confusing the inverse \(A^\prime \) with the transpose \(A^T\). They are generally different matrices.

📋 CBSE Case Study / HOTS

A school uses the matrix

\[ A= \begin{bmatrix} 2&3\\ 1&2 \end{bmatrix} \]

to represent a linear transformation. A student claims that

\[ B= \begin{bmatrix} 2&-3\\ -1&2 \end{bmatrix} \]

is the reverse transformation matrix. Verify the claim.

Solution

\[ \begin{aligned} AB &= \begin{bmatrix} 2&3\\ 1&2 \end{bmatrix} \begin{bmatrix} 2&-3\\ -1&2 \end{bmatrix}\\ &= \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix} \end{aligned} \]

Also,

\[ \begin{aligned} BA &= \begin{bmatrix} 2&-3\\ -1&2 \end{bmatrix} \begin{bmatrix} 2&3\\ 1&2 \end{bmatrix}\\ &= \begin{bmatrix} 1&0\\ 0&1 \end{bmatrix} \end{aligned} \]

Therefore,

\[ AB=BA=I_2 \]

Hence,

\[ \boxed{B=A^\prime } \]

⚡ Quick Revision Formulae

Definition of inverse:

\[ \boxed{AA^\prime =A^\prime A=I} \]

Invertibility condition:

\[ \boxed{\det(A)\neq0} \]

Singularity condition:

\[ \boxed{\det(A)=0} \]

Inverse of a \(2\times2\) matrix:

\[ \boxed{ \begin{bmatrix} a&b\\ c&d \end{bmatrix}^\prime = \frac{1}{ad-bc} \begin{bmatrix} d&-b\\ -c&a \end{bmatrix} } \]

provided

\[ ad-bc\neq0 \]

Inverse of a product:

\[ \boxed{(AB)^\prime =B^\prime A^\prime } \]

Inverse of a transpose:

\[ \boxed{(A^T)^\prime =(A^\prime )^T} \]

Inverse of an inverse:

\[ \boxed{(A^\prime )^\prime =A} \]

Identity matrix:

\[ \boxed{I^\prime =I} \]

🔑 Key Takeaway

An inverse matrix exists only when a square matrix is non-singular.
The defining relationship is
\[ AA^\prime =A^\prime A=I. \]
For a square matrix,
\[ \det(A)\neq0 \]
is the necessary and sufficient condition for invertibility.
For a \(2\times2\) matrix, the inverse can be obtained directly using the standard formula.
The inverse of a product reverses the order:
\[ (AB)^\prime =B^\prime A^\prime . \]
The inverse of a matrix, when it exists, is unique.
Matrix inverses provide an efficient method for solving systems of linear equations.

▦

Theorem 3: Uniqueness of the Inverse of a Matrix

📋

Table of Contents

13 sections

1 🧮 Statement of the Theorem ›
2 🤔 Why the Proof Works ›
3 📌 Proof in One Continuous Chain ›
4 👁️ Important Observation ›
5 ✏️ Example ›
6 📄 Connection with Ordinary Algebra ›
7 📌 Logical Form of the Theorem ›
8 🧮 Important Corollary ›
9 🌟 Board Examination Significance ›
10 ⚡ Exam Tip ›
11 ❌ Common Mistakes ›
12 📋 CBSE Case Study / HOTS ›
13 🔑 Key Takeaway ›

🧮 Theorem

Statement of the Theorem

🧮 Statement of the Theorem

Statement

The inverse of a square matrix, if it exists, is unique. In other words, if a square matrix \(A\) has an inverse, there cannot be two different matrices that both serve as its inverse.

\[\boxed{A^\prime \text{ exists }\Longrightarrow A^\prime \text{ is unique}}\]

Concept

Let \(A\) be a square matrix. Suppose two matrices \(B\) and \(C\) both claim to be the inverse of \(A\). By definition of an inverse,

\[AB=BA=I\]

and

\[AC=CA=I\]

The theorem states that these two supposed inverses must actually be the same matrix:

\[\boxed{B=C}\]

This proves that a matrix has at most one inverse. If an inverse exists, it is therefore unique.

Proof

Let \(A=[a_{ij}]\) be a square matrix of order \(m\)
Suppose, if possible, that \(B\) and \(C\) are two inverses of \(A\).
Since \(B\) is the inverse of \(A\),
\[AB=BA=I\]
Since \(C\) is also the inverse of \(A\),
\[AC=CA=I\]
Now,
\[\begin{aligned}B&=BI\\&=B(AC)\\&=(BA)C\\&=IC\\&=C\end{aligned}\]
Therefore,
\[\boxed{B=C}\]
Hence, the inverse of a square matrix, if it exists, is unique.

Hence Proved

🤔 Why the Proof Works

The key idea is to insert the identity matrix in a suitable way:

\[B=BI\]

Since \(C\) is an inverse of \(A\),

\[I=AC\]

Therefore,

\[B=B(AC)\]

Using associativity of matrix multiplication,

\[B(AC)=(BA)C\]

Since \(B\) is also an inverse of \(A\),

\[BA=I\]

Hence,

\[(BA)C=IC=C\]

Thus \(B=C\)

📌 Proof in One Continuous Chain

👁️ Important Observation

The proof uses only three fundamental matrix properties:

The definition of inverse:
\[ AB=BA=I. \]
The definition of the second inverse:
\[ AC=CA=I. \]
Associativity of matrix multiplication:
\[ B(AC)=(BA)C. \]

Notice that commutativity of matrix multiplication is not used. In general,

\[AB\neq BA\]

The proof relies on associativity, not commutativity.

✏️ Example

Consider

\[A=\begin{bmatrix}2&3\\1&2\end{bmatrix}\]

We know that

\[A^\prime =\begin{bmatrix}2&-3\\-1&2\end{bmatrix}\]

Suppose another matrix \(C\) were also an inverse of \(A\). Then

\[ AC=CA=I\]

But the uniqueness theorem gives

\[C=A^\prime =\begin{bmatrix}2&-3\\-1&2\end{bmatrix}\]

Thus no different matrix can serve as another inverse of \(A\).

🗒️ Connection with Ordinary Algebra

The result is analogous to the uniqueness of the reciprocal of a non-zero real number. For a non-zero number \(a\), if \(b\) and \(c\) both satisfy

\[ab=ba=1\]

and

\[ac=ca=1\]

then

\[b=b(1)=b(ac)=(ba)c=c\]

The matrix proof follows the same fundamental idea, with the identity matrix \(I\) playing the role of the number \(1\).

📌 Logical Form of the Theorem

🧮 Important Corollary

If \(B\) is an inverse of \(A\), then \(A\) is automatically the inverse of \(B\).

Since

\[AB=BA=I,\]

\(A\) satisfies the defining condition for being the inverse of \(B\). Therefore,

\[\boxed{B^\prime =A}\]

If \(B=A^\prime \), then

\[\boxed{(A^\prime )^\prime =A}\]

🧮 Theorem

If \(B\) is an inverse of \(A\), then \(A\) is automatically the inverse of \(B\).

Since

\[AB=BA=I,\]

\(A\) satisfies the defining condition for being the inverse of \(B\). Therefore,

\[\boxed{B^\prime =A}\]

If \(B=A^\prime \), then

\[\boxed{(A^\prime )^\prime =A}\]

🌟 Board Examination Significance

This theorem is useful in Class 12 Mathematics questions involving:

proof of uniqueness of the inverse,
verification of a proposed inverse,
matrix identities involving \(A^\prime \),
solving matrix equations,
properties of inverse matrices, and
case-study and competency-based questions.

For a proof question, the chain

\[B=BI=B(AC)=(BA)C=IC=C\]

is the key result to remember.

Competitive Examination Significance

The uniqueness property is frequently useful when a problem gives an equation involving matrices and asks for an unknown matrix. If two matrices are shown to satisfy the inverse condition, they must be identical.

A useful recognition rule is:

\[\boxed{AB=BA=I\Longrightarrow B=A^\prime }\]

and if another matrix \(C\) also satisfies

\[AC=CA=I\]

then immediately,

\[\boxed{B=C}\]

⚡ Exam Tip

Start with \(B=BI\), not \(B=B\).
Replace \(I\) by \(AC\), because \(C\) is an inverse of \(A\).
Use associativity:
\[ B(AC)=(BA)C. \]
Replace \(BA\) by \(I\), because \(B\) is an inverse of \(A\).
Finally use \(IC=C\).
Do not write \(B(AC)=B(IA)\) or interchange factors arbitrarily.
Remember that matrix multiplication is generally non-commutative.

❌ Common Mistakes

Assuming \(AB=BA\) for arbitrary matrices.
Using commutativity instead of associativity in the proof.
Forgetting that both \(B\) and \(C\) must satisfy the inverse condition.
Writing \(B=AC\) directly without explaining the insertion of the identity matrix.
Confusing uniqueness with existence. The theorem says that an inverse is unique if it exists; it does not say that every square matrix has an inverse.

📋 CBSE Case Study / HOTS

Let \(A\) be a square matrix. Suppose matrices \(B\) and \(C\) satisfy

\[ AB=BA=I \]

and

\[ AC=CA=I. \]

A student says that \(B\) and \(C\) may be different because matrix multiplication is not commutative. Is the student correct?

Solution

No. Although matrix multiplication is generally not commutative, the uniqueness proof requires only associativity.

\[ \begin{aligned} B &=BI\\ &=B(AC)\\ &=(BA)C\\ &=IC\\ &=C \end{aligned} \]

Therefore,

\[ \boxed{B=C} \]

Hence the student's claim is incorrect.

⚡ Quick Revision

Definition of inverse:

\[ \boxed{AA^\prime =A^\prime A=I} \]

Uniqueness:

\[ \boxed{ AB=BA=I,\quad AC=CA=I \Longrightarrow B=C } \]

Inverse of the inverse:

\[ \boxed{(A^\prime )^\prime =A} \]

Inverse relationship:

\[ \boxed{B=A^\prime \Longrightarrow B^\prime =A} \]

🔑 Key Takeaway

A square matrix can have at most one inverse.
If \(B\) and \(C\) are both inverses of \(A\), then \(B=C\).
The proof is based on
\[ B=BI=B(AC)=(BA)C=IC=C. \]
Associativity of matrix multiplication is used; commutativity is not required.
The theorem establishes uniqueness, not existence.
If \(A^\prime \) exists, then
\[ (A^\prime )^\prime =A. \]

▦

Theorem 4: Inverse of the Product of Two Invertible Matrices

📋

Table of Contents

12 sections

1 🧮 Inverse of the Product of Two Invertible Matrices ›
2 🔬 Alternative Proof Using the Definition of Inverse ›
3 🤔 Why Does the Order Reverse? ›
4 ✏️ Example ›
5 🔗 Connection with Matrix Transformations ›
6 📎 Important Related Results ›
7 🌟 Board Examination Significance ›
8 ⚡ Exam Tip ›
9 ❌ Common Mistakes ›
10 📋 CBSE Case Study / HOTS ›
11 ⚡ Quick Revision ›
12 🔑 Key Takeaway ›

🧮 Theorem

Inverse of the Product of Two Invertible Matrices

🧮 Inverse of the Product of Two Invertible Matrices

Statement

If \(A\) and \(B\) are invertible matrices of the same order, then the product \(AB\) is also invertible and its inverse is obtained by taking the inverses of the individual matrices in the reverse order.

\[\boxed{(AB)^\prime =B^\prime A^\prime }\]

Concept

For ordinary non-zero numbers, the reciprocal of a product is

\[(ab)^\prime =a^\prime b^\prime \]

For matrices, multiplication is generally not commutative. Therefore, the corresponding matrix rule contains an important reversal of order: \boxed{(AB)^\prime =B^\prime A^\prime } Thus, while finding the inverse of a matrix product, the factors appear in the opposite order.

Important Note

Do not write

\[(AB)^\prime =A^\prime B^\prime \]

This is generally incorrect. The correct identity is

\[\boxed{(AB)^\prime =B^\prime A^\prime }\]

The reversal is necessary because matrix multiplication is generally non-commutative:

\[AB\neq BA\]

Proof

Let \(A\) and \(B\) be invertible matrices of the same order. We need to prove that
\[\boxed{(AB)^\prime =B^\prime A^\prime }\]
Let
\[C=B^\prime A^\prime \]
To prove that \(C\) is the inverse of \(AB\), it is sufficient to show that
\[(AB)C=C(AB)=I\]
First Product
\[\begin{aligned}(AB)(B^\prime A^\prime )&=A(BB^\prime )A^\prime \\&=AIA^\prime \\&=AA^\prime \\&=I\end{aligned}\]
Second Product
\[\begin{aligned}(B^\prime A^\prime )(AB)&=B^\prime (A^\prime A)B\\&=B^\prime IB\\&=B^\prime B\\&=I\end{aligned}\]
Therefore,
\[(AB)(B^\prime A^\prime )=(B^\prime A^\prime )(AB)=I\]
Hence \(B^\prime A^\prime \) is the inverse of \(AB\). By the uniqueness of the inverse,
\[\boxed{(AB)^\prime =B^\prime A^\prime }\]

Hence Proved

🔬 Alternative Proof Using the Definition of Inverse

🔬 Proof

Since \(A\) and \(B\) are invertible,
\[AA^\prime =A^\prime A=I\]
and
\[BB^\prime =B^\prime B=I\]
Now consider
\[(AB)(B^\prime A^\prime )\]

\[\begin{aligned}(AB)(B^\prime A^\prime )&=A(BB^\prime )A^\prime \\&=AIA^\prime \\&=AA^\prime \\&=I\end{aligned}\]
Similarly,
\[\begin{aligned}(B^\prime A^\prime )(AB)&=B^\prime (A^\prime A)B\\&=B^\prime IB\\&=B^\prime B\\&=I\end{aligned}\]
Thus \(B^\prime A^\prime \) satisfies the defining condition for the inverse of \(AB\). Therefore,
\[\boxed{(AB)^\prime =B^\prime A^\prime }\]

🤔 Why Does the Order Reverse?

This is the central conceptual point of the theorem.

Starting with \(AB\) we first have to cancel \(B\) and then \(A\). Therefore the inverse operation must act as

\[B^\prime A^\prime \]

Symbolically:

\[\boxed{A\longrightarrow B\quad\text{is reversed as}\quad B^\prime \longrightarrow A^\prime }\]

This is analogous to reversing the order of operations when undoing a sequence of transformations.

✏️ Example

Question

Verification with Two \(2\times2\) Matrices

Let

\[A=\begin{bmatrix}1&1\\0&1\end{bmatrix},\qquad B=\begin{bmatrix}1&0\\1&1\end{bmatrix}\]
Their inverses are

\[A^\prime =\begin{bmatrix}1&-1\\0&1\end{bmatrix},\qquad B^\prime =\begin{bmatrix}1&0\\-1&1\end{bmatrix}\]
First calculate \(AB\):

\[\begin{aligned}AB&=\begin{bmatrix}1&1\\0&1\end{bmatrix}\begin{bmatrix}1&0\\1&1\end{bmatrix}\\ &=\begin{bmatrix}2&1\\1&1\end{bmatrix}\end{aligned}\]
The determinant is

\[|AB|=(2)(1)-(1)(1)=1.\]
Hence,

\[\begin{aligned}(AB)^\prime &=\begin{bmatrix}1&-1\\-1&2\end{bmatrix}\end{aligned}\]
Now calculate \(B^\prime A^\prime \):

\[\begin{aligned}B^\prime A^\prime &=\begin{bmatrix}1&0\\-1&1\end{bmatrix} \begin{bmatrix}1&-1\\0&1\end{bmatrix}\\ &=\begin{bmatrix}1&-1\\-1&2\end{bmatrix}\end{aligned}\]
Therefore,

\[\boxed{(AB)^\prime =B^\prime A^\prime }\]

Question

If \(A\) and \(B\) are invertible matrices and

\[A^\prime =P,\qquad B^\prime =Q,\]

then

\[\boxed{(AB)^\prime =QP}\]

Notice that \(P\) and \(Q\) occur in reverse order.
Extension to Three Matrices

The theorem can be extended to three invertible matrices:
\[\boxed{(ABC)^\prime =C^\prime B^\prime A^\prime }\]
To see this, write

\[(ABC)^\prime =\left((AB)C\right)^\prime \]
Applying the theorem,

\[\begin{aligned}(ABC)^\prime &=C^\prime (AB)^\prime \\&=C^\prime B^\prime A^\prime \end{aligned}\]
General Result for \(n\) Invertible Matrices
If \(A_1,A_2,\ldots,A_n\) are invertible matrices of compatible orders, then
\[\boxed{(A_1A_2\cdots A_n)^\prime =A_n^\prime \cdots A_2^\prime A_1^\prime }\]
Thus, the inverse of a product is obtained by:
1. taking the inverse of every factor, and
2. reversing the order of the factors.

🔗 Connection with Matrix Transformations

Suppose \(A\) and \(B\) represent two successive linear transformations:

\[\mathbf{x}\xrightarrow{\ B\ }B\mathbf{x}\xrightarrow{\ A\ }A(B\mathbf{x})=AB\mathbf{x}\]

To reverse the complete transformation, \(B\) must be undone first, followed by \(A\). Therefore,

\[\mathbf{y}\xrightarrow{\ A^\prime \ }A^\prime \mathbf{y}\xrightarrow{\ B^\prime \ }B^\prime A^\prime \mathbf{y}\]

Hence the reverse transformation is represented by

\[\boxed{B^\prime A^\prime }\]

This gives an intuitive explanation for the reversal of order in the theorem.

📎 Important Related Results

The following identities should be learned together:

Inverse of a product:

\[ \boxed{(AB)^\prime =B^\prime A^\prime } \]

Inverse of an inverse:

\[ \boxed{(A^\prime )^\prime =A} \]

Inverse of a transpose:

\[ \boxed{(A^T)^\prime =(A^\prime )^T} \]

Inverse of a scalar multiple:

\[ \boxed{(kA)^\prime =\frac{1}{k}A^\prime },\qquad k\neq0 \]

Inverse of identity:

\[ \boxed{I^\prime =I} \]

🌟 Board Examination Significance

This theorem is important for Class 12 Mathematics because it can appear directly as a proof question or indirectly in problems involving matrix identities, inverse matrices and matrix equations.

A typical board question may ask:

prove that \((AB)^\prime =B^\prime A^\prime \),
find the inverse of a matrix product,
verify an inverse identity,
simplify an expression containing \(A^\prime \) and \(B^\prime \), or
use the result to solve a matrix equation.

Competitive Entrance Examination Significance

For JEE and other competitive examinations, this theorem is particularly useful because it can reduce lengthy matrix calculations to a short identity.

If a question gives

\[ A^\prime =P,\qquad B^\prime =Q, \]

then immediately,

\[ \boxed{(AB)^\prime =QP}. \]

There is no need to calculate \(AB\) first and then find its inverse.

⚡ Exam Tip

Memorise the reversal:
\[ \boxed{(AB)^\prime =B^\prime A^\prime }. \]
Do not confuse this with
\[ (AB)^\prime =A^\prime B^\prime . \]
Remember that the theorem requires \(A\) and \(B\) to be invertible.
In a proof, verify both:
\[ (AB)(B^\prime A^\prime )=I \]
and
\[ (B^\prime A^\prime )(AB)=I. \]
Use associativity to group the factors correctly.
For three or more matrices, reverse the complete order.

❌ Common Mistakes

Writing \((AB)^\prime =A^\prime B^\prime \).
Assuming \(AB=BA\).
Forgetting that the matrices must be invertible.
Reversing the order of the original matrices but not the order of their inverses.
Applying the theorem to a product whose inverse does not exist.
Using commutativity instead of associativity during the proof.

📋 CBSE Case Study / HOTS

Let \(A\) and \(B\) be invertible matrices of the same order. Suppose

\[ A^\prime = \begin{bmatrix} 1&0\\ -1&1 \end{bmatrix}, \qquad B^\prime = \begin{bmatrix} 1&-1\\ 0&1 \end{bmatrix} \]

Find \((AB)^\prime \).

Solution

Using Theorem 4,

\[ (AB)^\prime =B^\prime A^\prime . \]

Therefore,

\[ \begin{aligned} (AB)^\prime &= \begin{bmatrix} 1&-1\\ 0&1 \end{bmatrix} \begin{bmatrix} 1&0\\ -1&1 \end{bmatrix}\\ &= \begin{bmatrix} 1+1&-1\\ -1&1 \end{bmatrix}\\ &= \boxed{ \begin{bmatrix} 2&-1\\ -1&1 \end{bmatrix} }. \end{aligned} \]

⚡ Quick Revision

\[ \boxed{(AB)^\prime =B^\prime A^\prime } \]

\[ \boxed{(ABC)^\prime =C^\prime B^\prime A^\prime } \]

\[ \boxed{ (A_1A_2\cdots A_n)^\prime = A_n^\prime \cdots A_2^\prime A_1^\prime } \]

Memory rule: Take the inverse of every factor and reverse their order.

🔑 Key Takeaway

If \(A\) and \(B\) are invertible, then \(AB\) is also invertible.
The inverse of a product is the product of the inverses in reverse order.
\[ (AB)^\prime =B^\prime A^\prime . \]
The reversal of order is essential because matrix multiplication is generally non-commutative.
For three matrices:
\[ (ABC)^\prime =C^\prime B^\prime A^\prime . \]
For \(n\) matrices, reverse the entire sequence while taking the inverse of each factor.

Class XII · Mathematics · Chapter 3

Matrices

A complete, self-contained learning engine — concepts, formulas, a rule-based step-by-step solver, curated practice questions and interactive labs, built for JEE Main, JEE Advanced, BITSAT & CBSE Boards.

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Core Concepts

Everything the NCERT syllabus covers in Matrices, rebuilt with clean explanations and worked micro-examples.

What is a Matrix?

Definition & Order

A matrix is a rectangular arrangement of numbers (or functions) into rows and columns, enclosed in brackets. If a matrix has m rows and n columns, it is said to be of order m × n (read "m by n"), and it contains exactly m×n elements.

2-15 073

This is a matrix of order 2 × 3 — 2 rows, 3 columns. The element in row i and column j is written a_ij. So here, a₁₂ = -1 and a₂₃ = 3.

Rows are counted first, columns second — remembered by "RC Cola" (Row × Column).

Types of Matrices

Classifying by shape and pattern

Row matrix: only 1 row, order 1×n.
Column matrix: only 1 column, order m×1.
Square matrix: rows = columns (m = n).
Diagonal matrix: square, all non-diagonal entries are 0.
Scalar matrix: diagonal matrix with all diagonal entries equal.

Identity matrix (I): scalar matrix with diagonal entries = 1.
Zero / Null matrix (O): every entry is 0, any order.
Symmetric matrix: A' = A (mirror image across the diagonal).
Skew-symmetric matrix: A' = −A (diagonal entries must be 0).
Upper/Lower triangular: zeros below/above the main diagonal.

A matrix is called square of order n when m = n — square matrices are the only ones that can be diagonal, scalar, identity, symmetric, or skew-symmetric.

Equality of Matrices

When are two matrices "the same"?

Two matrices A and B are equal only when both conditions hold:

They have the same order (same number of rows and columns).
Every corresponding element is equal: a_ij = b_ij for all i, j.

This is exactly how you solve "find x and y" problems — equate corresponding entries to build simultaneous equations.

Matrix Algebra: Addition, Subtraction & Scalar Multiplication

Element-wise operations

Addition/Subtraction is only defined when two matrices have the same order — corresponding elements are added or subtracted. Scalar multiplication multiplies every single element by the constant k, and works for any matrix.

Addition of matrices is commutative (A+B = B+A) and associative, exactly like ordinary numbers — but multiplication is not (see below)!

Matrix Multiplication

The compatibility rule & the "row-into-column" method

The product AB is defined only when the number of columns of A equals the number of rows of B. If A is m×n and B is n×p, then AB exists and has order m × p.

Each entry (AB)ᵢⱼ is found by taking the dot product of row i of A with column j of B: multiply corresponding entries and add them up.

Matrix multiplication is NOT commutative in general — AB ≠ BA, and often one of AB or BA won't even be defined. It IS associative: (AB)C = A(BC), and distributive over addition.

Transpose of a Matrix

Flipping rows into columns

The transpose of A, written A' (or Aᵀ), is obtained by interchanging the rows and columns of A. If A is m×n, then A' is n×m, and (A')ᵢⱼ = Aⱼᵢ.

Transpose gives rise to two special families of square matrices:

Symmetric: A' = A → the matrix is a mirror image of itself about the main diagonal.
Skew-symmetric: A' = −A → every diagonal entry must be 0 (since aᵢᵢ = −aᵢᵢ forces aᵢᵢ = 0).

A powerful theorem: every square matrix A can be written uniquely as the sum of a symmetric and a skew-symmetric matrix:

A = ½(A + A') + ½(A − A')

Elementary Operations & Invertible Matrices

Row/column transformations and finding A⁻¹

An elementary operation (or transformation) on a matrix is one of three moves applied to rows (Rᵢ) or columns (Cᵢ):

Interchange two rows/columns: Rᵢ ↔ Rⱼ
Multiply a row/column by a non-zero scalar: Rᵢ → kRᵢ (k ≠ 0)
Add a scalar multiple of one row/column to another: Rᵢ → Rᵢ + kRⱼ

A square matrix A is invertible if there exists a matrix B such that AB = BA = I. This B is called A⁻¹, and it is unique when it exists. Elementary row (or column) operations — applied simultaneously to A and an identity matrix of the same order — reduce A to I while transforming I into A⁻¹. This is the Gauss–Jordan method, used throughout the AI Solver and Interactive Lab below.

Only a square matrix can possibly be invertible — and even then, not every square matrix has an inverse (it must be "non-singular").

Formula Reference Sheet

Every identity you need for Matrices, grouped by operation — perfect for last-minute revision.

Addition & Scalar Multiplication

(A+B)ᵢⱼ = Aᵢⱼ + Bᵢⱼ — defined only when A, B have the same order.

(kA)ᵢⱼ = k·Aᵢⱼ

A+B = B+A (commutative)

(A+B)+C = A+(B+C) (associative)

k(A+B) = kA+kB (k+l)A = kA+lA

Matrix Multiplication

A(m×n) · B(n×p) = C(m×p)

Cᵢⱼ = Σₖ Aᵢₖ·Bₖⱼ (sum over k = 1 to n)

A(BC) = (AB)C (associative)

A(B+C) = AB+AC (distributive)

AB ≠ BA (generally — non-commutative)

AI = IA = A

Transpose Properties

(A')' = A

(kA)' = kA'

(A+B)' = A'+B'

(AB)' = B'A' (order reverses!)

Symmetric / Skew-Symmetric

Symmetric: A' = A

Skew-symmetric: A' = −A (diagonal entries = 0)

Unique decomposition of any square A:

A = ½(A+A') + ½(A−A')

where ½(A+A') is symmetric and ½(A−A') is skew-symmetric.

⁻¹

Inverse of a Matrix

A⁻¹ exists ⇔ A is a square, non-singular matrix (|A| ≠ 0).

A·A⁻¹ = A⁻¹·A = I

(A⁻¹)⁻¹ = A

(AB)⁻¹ = B⁻¹A⁻¹ (order reverses)

The inverse, when it exists, is unique.

Elementary Operations

Rᵢ ↔ Rⱼ — swap two rows

Rᵢ → kRᵢ, k ≠ 0 — scale a row

Rᵢ → Rᵢ + kRⱼ — add a multiple of one row to another

Same rules apply to columns (Cᵢ). Used in Gauss–Jordan reduction: [A | I] → [I | A⁻¹].

Tricks & Tips

Exam-tested shortcuts that save minutes on the paper.

✦

Check compatibility before you multiply

Write the two orders side by side, e.g. (2×3)(3×2). If the inner numbers match, the product exists, and the outer numbers give its order — here, 2×2. Do this mental check first; it instantly eliminates half the traps in "which product is defined" questions.

✦

Spot symmetric/skew-symmetric instantly

Don't compute the full transpose if you don't need to — just scan the matrix visually. If it looks identical when flipped about the diagonal, it's symmetric. If every off-diagonal pair is a negative of its mirror AND the diagonal is all zeros, it's skew-symmetric. Trust the visual check, then verify algebraically only if unsure.

✦

The A + A' / A − A' trick works every time

Whenever asked to express a square matrix as symmetric + skew-symmetric, don't think — just apply the formula: P = ½(A+A'), Q = ½(A−A'). Compute A' first, then it's pure addition/subtraction and halving. No cleverness required, just speed.

✦

For (AB)' problems, reverse and transpose

Whenever you see (AB)', immediately rewrite it as B'A' — reverse the order, then transpose each. This single reflex resolves a huge chunk of "prove that..." questions in one line.

✦

Inverse by elementary operations: stick to ONE side

When finding A⁻¹ using [A | I], apply every operation to both matrices in the same step — never operate on A alone and forget I. Also, always use only row operations if you started with rows (don't mix rows and columns) — mixing is the single most common way marks are lost here.

✦

Diagonal entries of skew-symmetric matrices are free marks

The moment a question says "A is skew-symmetric," write "all diagonal entries = 0" immediately — this is guaranteed and often unlocks the rest of the problem (e.g., finding unknowns on the diagonal) in one step.

Common Mistakes

The exact errors examiners see year after year — recognise them so you never make them.

Pitfall

Assuming AB = BA

Students often multiply matrices in whichever order is "easier" and assume it won't matter. Matrix multiplication is not commutative — AB and BA can be completely different matrices, or one may not even be defined. Always multiply in the exact order the question specifies.

Pitfall

Adding matrices of different orders

A frequent slip in "find A+B" questions is forcing addition even when the orders don't match, by pairing up whatever elements happen to line up. If the orders differ, addition is undefined — say so, don't force it.

Pitfall

Forgetting to reverse order while transposing a product

Writing (AB)' = A'B' instead of the correct B'A' is one of the most repeated errors on board papers. The reversal rule applies to transpose and inverse of a product alike: (AB)' = B'A', and (AB)⁻¹ = B⁻¹A⁻¹.

Pitfall

Calling every "diagonal-looking" matrix scalar or identity

A diagonal matrix only becomes a scalar matrix if all diagonal entries are equal, and only becomes the identity matrix if that common value is exactly 1. Don't skip the check — a matrix with diagonal (2,2,2) is scalar, not identity.

Pitfall

Ignoring that skew-symmetric diagonal entries must be 0

When solving for unknowns and told a matrix is skew-symmetric, students sometimes forget to also set the diagonal entries (which may contain a variable) to zero, missing an equation and get stuck or produce a wrong answer.

Pitfall

Mixing row and column operations while finding the inverse

Once you begin reducing [A | I] using row operations, switching mid-way to a column operation invalidates the whole process — the augmented identity no longer tracks correctly. Pick rows or columns at the start and stay consistent throughout.

Pitfall

Treating "equal matrices" as "equal totals"

Some students check if the sum of all elements matches instead of comparing element-by-element. Matrix equality requires every single corresponding entry to match, not just an overall total or trace.

AI Solver — Step-by-Step Matrix Engine

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Operation

Order of A (rows × cols)

Order of B (rows × cols)

Concept-Building Practice Questions

Original questions, organised by concept, each with a full step-by-step solution — click a question to reveal it.

Q1Foundational

A matrix has 12 elements. List all possible orders it could have, and state which of these orders could give a square matrix.

Show full solution

The order m×n must satisfy m·n = 12. Listing all factor pairs of 12: 1×12, 2×6, 3×4, 4×3, 6×2, 12×1.

A matrix is square only when the number of rows equals the number of columns (m = n). None of the pairs above have m = n, since 12 has no perfect-square factor pair — so no order among these gives a square matrix.

Q2Standard

Find x, y, z, w if [[x+y, 2z+w],[x−y, z−w]] = [[9, 5],[1, 3]].

Show full solution

Both matrices are of order 2×2, so equate corresponding entries:

x + y = 9 …(i) x − y = 1 …(ii) 2z + w = 5 …(iii) z − w = 3 …(iv)

Adding (i) and (ii): 2x = 10 ⟹ x = 5, and substituting back gives y = 4.

Adding (iii) and (iv) after making coefficients of w match: from (iv), w = z − 3. Substitute into (iii): 2z + (z−3) = 5 ⟹ 3z = 8 ⟹ z = 8/3, then w = 8/3 − 3 = −1/3.

Answer: x = 5, y = 4, z = 8/3, w = −1/3.

Q3Standard

Construct a 3×3 matrix A = [aᵢⱼ] where aᵢⱼ = i² − j². Identify its type (if any special type applies).

Show full solution

Compute each entry using aᵢⱼ = i² − j² for i, j = 1, 2, 3:

a₁₁=0, a₁₂=1−4=−3, a₁₃=1−9=−8
a₂₁=4−1=3, a₂₂=0, a₂₃=4−9=−5
a₃₁=9−1=8, a₃₂=9−4=5, a₃₃=0

So A = [[0,−3,−8],[3,0,−5],[8,5,0]]. Notice aᵢⱼ = −aⱼᵢ for every pair, and all diagonal entries are 0 — so A is skew-symmetric. This makes sense structurally since i²−j² = −(j²−i²).

Q1Foundational

If A = [[3,−1],[2,4]] and B = [[1,5],[−2,0]], find 2A − 3B.

Show full solution

First scale each matrix: 2A = [[6,−2],[4,8]], and 3B = [[3,15],[−6,0]].

Now subtract element-wise: 2A − 3B = [[6−3, −2−15],[4−(−6), 8−0]] = [[3,−17],[10,8]].

Answer: 2A − 3B = [[3,−17],[10,8]].

Q2Standard

Find matrix X such that 2X + [[1,2],[3,4]] = [[7,4],[1,0]].

Show full solution

Isolate 2X by subtracting: 2X = [[7,4],[1,0]] − [[1,2],[3,4]] = [[6,2],[−2,−4]].

Divide every entry by 2: X = [[3,1],[−1,−2]].

Q3Challenging

If A is a 2×2 matrix such that A + Aᵀ = [[4,2],[2,6]], and A is known to be symmetric, find A.

Show full solution

Since A is symmetric, A' = A, so A + A' = A + A = 2A.

Therefore 2A = [[4,2],[2,6]], which gives A = [[2,1],[1,3]].

Check: this A is indeed symmetric (equal to its own transpose), confirming the answer is consistent.

Q1Foundational

If A is of order 2×3 and B is of order 3×4, state which of AB and BA are defined, and give the order of whichever exists.

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For AB: A is (2×3), B is (3×4) — inner numbers 3 and 3 match, so AB is defined with order 2×4.

For BA: B is (3×4), A is (2×3) — inner numbers 4 and 2 don't match, so BA is not defined.

Q2Standard

If A = [[2,0],[1,3]] and B = [[1,−1],[0,2]], compute AB and BA, and verify that AB ≠ BA.

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AB: row1·col1 = 2·1+0·0=2, row1·col2 = 2·(−1)+0·2=−2, row2·col1 = 1·1+3·0=1, row2·col2 = 1·(−1)+3·2=5.
So AB = [[2,−2],[1,5]].

BA: row1·col1 = 1·2+(−1)·1=1, row1·col2 = 1·0+(−1)·3=−3, row2·col1 = 0·2+2·1=2, row2·col2 = 0·0+2·3=6.
So BA = [[1,−3],[2,6]].

Since AB = [[2,−2],[1,5]] and BA = [[1,−3],[2,6]] are different, this confirms AB ≠ BA — matrix multiplication is not commutative.

Q3Challenging

If A = [[0,1],[1,0]], show that A² = I, and use this fact to quickly find A⁴ and A⁵.

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Compute A² = A·A: row1·col1 = 0·0+1·1=1, row1·col2 = 0·1+1·0=0, row2·col1 = 1·0+0·1=0, row2·col2 = 1·1+0·0=1.

So A² = [[1,0],[0,1]] = I, confirming the claim.

Since A² = I: A⁴ = (A²)² = I² = I. And A⁵ = A⁴·A = I·A = A = [[0,1],[1,0]]. Recognising the pattern A² = I lets you skip repeated multiplication entirely for any power of A.

Q1Foundational

If A = [[2,3,1],[0,−1,4]], find Aᵀ and state its order.

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A is 2×3, so Aᵀ will be 3×2. Row 1 of A becomes column 1 of Aᵀ, and row 2 of A becomes column 2 of Aᵀ.

Aᵀ = [[2,0],[3,−1],[1,4]], order 3×2.

Q2Standard

Express A = [[4,2],[6,8]] as the sum of a symmetric and a skew-symmetric matrix.

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First find A' = [[4,6],[2,8]].

P = ½(A+A') = ½[[8,8],[8,16]] = [[4,4],[4,8]] — check: P' = P, so P is symmetric. ✓

Q = ½(A−A') = ½[[0,−4],[4,0]] = [[0,−2],[2,0]] — check: Q' = −Q, so Q is skew-symmetric. ✓

Answer: A = P + Q = [[4,4],[4,8]] + [[0,−2],[2,0]].

Q3Challenging

A = [[0,a,−2],[−3,0,b],[2,−4,0]] is skew-symmetric. Find a and b.

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For a skew-symmetric matrix, aᵢⱼ = −aⱼᵢ for every pair. Compare the (1,2) and (2,1) entries: a and −3 must satisfy a = −(−3) = 3.

Compare the (2,3) and (3,2) entries: b and −4 must satisfy b = −(−4) = 4.

Check the (1,3)/(3,1) pair for consistency: −2 and 2 — indeed −2 = −(2). ✓ Consistent.

Answer: a = 3, b = 4.

Q1Standard

Using elementary row operations, find the inverse of A = [[2,1],[1,1]].

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Write [A | I] = [[2,1 | 1,0],[1,1 | 0,1]].

Apply R₁ → R₁ − R₂: [[1,0 | 1,−1],[1,1 | 0,1]].

Apply R₂ → R₂ − R₁: [[1,0 | 1,−1],[0,1 | −1,2]].

The left block is now I, so the right block is A⁻¹: A⁻¹ = [[1,−1],[−1,2]]. (Verify: A·A⁻¹ should give I.)

Q2Standard

Show that A = [[1,2],[2,4]] is NOT invertible, using elementary row operations.

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Write [A | I] = [[1,2 | 1,0],[2,4 | 0,1]].

Apply R₂ → R₂ − 2R₁: [[1,2 | 1,0],[0,0 | −2,1]].

Row 2 of the left block has become entirely zero — it is impossible to continue reducing the left side to the identity matrix. This means A has no inverse (A is singular).

Q3Challenging

Find the inverse of A = [[1,0,2],[2,1,0],[0,1,1]] using elementary row operations.

Show full solution

Start with [A | I]: [[1,0,2 | 1,0,0],[2,1,0 | 0,1,0],[0,1,1 | 0,0,1]].

R₂ → R₂ − 2R₁: [[1,0,2 | 1,0,0],[0,1,−4 | −2,1,0],[0,1,1 | 0,0,1]].

R₃ → R₃ − R₂: [[1,0,2 | 1,0,0],[0,1,−4 | −2,1,0],[0,0,5 | 2,−1,1]].

R₃ → R₃ ÷ 5: [[1,0,2 | 1,0,0],[0,1,−4 | −2,1,0],[0,0,1 | 0.4,−0.2,0.2]].

R₁ → R₁ − 2R₃, and R₂ → R₂ + 4R₃: [[1,0,0 | 0.2,0.4,−0.4],[0,1,0 | −0.4,0.2,0.8],[0,0,1 | 0.4,−0.2,0.2]].

A⁻¹ = [[0.2,0.4,−0.4],[−0.4,0.2,0.8],[0.4,−0.2,0.2]] — equivalently, multiply through by 5 to write with a common denominator of 5: (1/5)[[1,2,−2],[−2,1,4],[2,−1,1]].

Interactive Lab

Four hands-on modules for building intuition — no lectures, just play and discover.

Matrix Type Identifier

Classify the randomly generated matrix

A fresh matrix appears below. Decide which type best describes it, then check your answer.

Multiplication Dot-Product Visualizer

Click a cell in the result to see how it's built

A (2×3) is multiplied by B (3×2). Click any cell of the result matrix C to highlight the row of A and column of B that produce it.

A (2×3)

B (3×2)

C = AB (2×2) — click a cell

Click a result cell above to see its dot-product calculation here.

Symmetric / Skew-Symmetric Decomposer

Enter any 3×3 matrix and watch it split live

Type any values into matrix A. The symmetric part P and skew-symmetric part Q update automatically, and we verify P + Q = A.

A (editable)

P = ½(A+Aᵀ) — symmetric

Q = ½(A−Aᵀ) — skew-symmetric

Waiting for input…

Elementary Row-Operation Trainer

Reduce [A | I] to [I | A⁻¹] yourself, one move at a time

Apply row operations to steer the left block toward the identity matrix. Your operations are also applied to the right block, which becomes A⁻¹ once you succeed.

with row k =

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Class 12 Matrices Notes | NCERT Maths Chapter 3

Class 12 Matrices Notes | NCERT Maths Chapter 3 — Complete Notes & Solutions · academia-aeternum.com

NCERT Class 12 Mathematics Chapter 3: Matrices is a fundamental chapter that introduces an efficient algebraic framework for organising numbers, solving equations, representing transformations, and handling mathematical data systematically. These Class 12 Matrices Notes cover the complete range of concepts required for CBSE Board examinations as well as competitive entrance examinations, including the definition and order of a matrix, types of matrices, equality of matrices, matrix operations,…

🎓 Class 12 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE

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Matrices

Chapter Snapshot

Why This Chapter Matters for Entrance Exams

Key Concept Highlights

Important Formula Capsules

What You Will Learn

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🏆 Exam Strategy & Preparation Tips

For example

For example

General Form of a Matrix

For example

Importance for JEE and Competitive Entrance Examinations

Questions

Answers

Higher-Order Thinking Question

Solution

Key Concepts

For example, consider

Meaning of the Subscripts in \(a_{ij}\)

For example, in

The \(i\)-th Row

For example, if

The \(j\)-th Column

For example,

Row-Wise Coordinate Representation

Solution

Answer

Definition

For example,

General Form

Example

Geometrical Interpretation

Definition

For example,

General Form

Example

Connection with Coordinate Geometry

Definition

Example

General Form

Important Features of a Square Matrix

Definition

General Form

Example

Important Note

Definition

General Form

Example

Definition

General Form

Example

Identity Matrix as a Scalar Matrix

Why Is the Identity Matrix Important?

Definition

General Form

Example

Important Observation

Significance for Competitive Entrance Examinations

Solution

Condition 1: Same Order

Condition 2: Equal Corresponding Elements

Example 1: Equal Matrices

Example 2: Different Orders

Example 3: Same Elements but Different Positions

Why This Method Works

Same Elements Does Not Mean Equal Matrices

For example,

For example,

Significance for Competitive Entrance Examinations

Solution

Competitive Exam Significance

General Formula

For example,

Solution

Competitive Entrance Exam Significance

Competitive Entrance Exam Significance

General Formula

Competitive Entrance Exam Significance

Competitive Entrance Exam Significance