Ch 3  ·  Q–
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Chapter 3 Exercise 3.1 Solutions

Matrices

Step-by-Step Solutions to the NCERT Class 12 Mathematics Chapter 3 Exercise 3.1

Class 12 Mathematics Exercise 3.1 NCERT Solutions Matrices Class 12 Mathematics Chapter 3 CBSE Board Exam JEE Main CUET Order of Matrix Elements of Matrix Matrix Equality Square Matrix Matrix Construction Matrix Elements Counting Matrices
10 Questions
20–30 min Ideal time
Q1 Now at
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Q1
NUMERIC3 marks
In the matrix \[A=\begin{bmatrix}2 & 5 & 19 & -7\\35 & -2 & \dfrac{5}{2} & 12\\\sqrt{3} & 1 & -5 & 17\end{bmatrix}\] Write
  1. The order of the matrix
  2. The number of elements
  3. The elements \(a_{13},a_{21},a_{33},a_{24},a_{23}\)
📘 Concept & Theory
Concept/Theory
›

A matrix is a rectangular arrangement of numbers, symbols, or expressions written in rows and columns. The entries of a matrix are called its elements.

If a matrix has \(m\) rows and \(n\) columns, its order is written as

\[ m\times n. \]

Thus, the order of a matrix is determined by:

  • the number of rows, followed by
  • the number of columns.

If \(A=[a_{ij}]\), then \(a_{ij}\) denotes the element situated at the \(i\)-th row and \(j\)-th column.

The first subscript indicates the row, while the second subscript indicates the column.

For example,

\[ a_{23} \]
means the element in the 2nd row and 3rd column.

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

\[ m\times n. \]

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Count the horizontal rows in the given matrix.

  2. Count the vertical columns in the given matrix.

  3. Write the order as rows \(\times\) columns.

  4. Calculate the total number of elements using

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

  5. For each \(a_{ij}\), identify the \(i\)-th row and \(j\)-th column.

  6. Read the corresponding element directly from the matrix.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  27 steps
  1. Given —
    \[A=\begin{bmatrix}2 & 5 & 19 & -7\\35 & -2 & \dfrac{5}{2} & 12\\\sqrt{3} & 1 & -5 & 17\end{bmatrix}\]
  2. (i) Order of the Matrix
  3. First, count the number of horizontal rows. The matrix has:
    \[3\text{ rows}\]
  4. Next, count the number of vertical columns. The matrix has:
    \[4\text{ columns}\]
  5. Therefore, the order of the matrix is
    \[\boxed{3\times4}\]
  6. (ii) Number of Elements
  7. A matrix with \(m\) rows and \(n\) columns contains \(m\times n\) elements. Here,
    \[m=3,\qquad n=4\]
  8. Therefore,
    \[\text{Number of elements}=3\times4\]
  9. Hence,
    \[\boxed{\text{Number of elements}=12}\]
  10. (iii) Required Elements
  11. We use the notation \(a_{ij}\), where \(i\) represents the row number and \(j\) represents the column number.
  12. 1. Finding \(a_{13}\)
  13. The notation \(a_{13}\) means the element in the 1st row and 3rd column.
  14. The first row is
    \[\begin{bmatrix}2 & 5 & 19 & -7\end{bmatrix}\]
  15. The third element of the first row is \(19\).
  16. Therefore,
    \[\boxed{a_{13}=19}\]
  17. 2. Finding \(a_{21}\)
  18. The notation \(a_{21}\) means the element in the 2nd row and 1st column.
  19. The second row is
    \[\begin{bmatrix}35 & -2 & \dfrac{5}{2} & 12\end{bmatrix}\]
  20. The first element of the second row is \(35\).
  21. Therefore,
    \[\boxed{a_{21}=35}\]
  22. 3. Finding \(a_{33}\)
  23. The notation \(a_{33}\) means the element in the 3rd row and 3rd column.
  24. The third row is
    \[\begin{bmatrix}\sqrt{3} & 1 & -5 & 17\end{bmatrix}\]
  25. The third element of the third row is \(-5\).
  26. Therefore,
    \[\boxed{a_{33}=-5}\]
  27. 4. Finding \(a_{24}\)
  28. The notation \(a_{24}\) means the element in the 2nd row and 4th column.
  29. The second row is
    \[\begin{bmatrix}35 & -2 & \dfrac{5}{2} & 12\end{bmatrix}\]
  30. The fourth element of the second row is \(12\).
  31. Therefore,
    \[\boxed{a_{24}=12}\]
  32. 5. Finding \(a_{23}\)
  33. The notation \(a_{23}\) means the element in the 2nd row and 3rd column.
  34. The second row is
    \[\begin{bmatrix}35 & -2 & \dfrac{5}{2} & 12\end{bmatrix}\]
  35. The third element of the second row is
    \[\dfrac{5}{2}.\]
  36. Therefore,
    \[\boxed{a_{23}=\dfrac{5}{2}}\]
💡 Answer
Final Answer
›

The given matrix has \(3\) rows and \(4\) columns. Hence,

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

The total number of elements is

\[ \boxed{12} \]

The required elements are:

\[ \boxed{ a_{13}=19,\qquad a_{21}=35,\qquad a_{33}=-5,\qquad a_{24}=12,\qquad a_{23}=\dfrac{5}{2} } \]
🎯 Exam Significance
Exam Significance
›

This question tests the fundamental concepts of order of a matrix, number of elements, and matrix notation. These concepts form the foundation for the entire chapter and are frequently used in questions involving equality of matrices, operations on matrices, transpose, symmetric and skew-symmetric matrices, and inverse matrices.

For board examinations, students should remember that the order is always written as

\[ \text{rows}\times\text{columns}, \]
not columns \(\times\) rows. They should also interpret \(a_{ij}\) carefully: the first suffix denotes the row and the second suffix denotes the column.

Significance for Competitive Entrance Examinations

In competitive examinations, matrix-element notation is often tested through direct questions as well as through more advanced matrix operations. A small error in interpreting \(a_{ij}\) can lead to an incorrect answer in questions involving matrix equations, determinants, inverse matrices, or transformations.

A quick and reliable approach is to remember:

\[ a_{ij}\longrightarrow\text{\(i\)-th row, \(j\)-th column}. \]
This notation should become automatic before moving to more advanced matrix problems.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. The order of a matrix is written as rows \(\times\) columns.

  2. A matrix with \(m\) rows and \(n\) columns has \(m\times n\) elements.

  3. In \(a_{ij}\), the first subscript \(i\) represents the row.

  4. In \(a_{ij}\), the second subscript \(j\) represents the column.

  5. \(a_{13}\) means the element in the 1st row and 3rd column.

  6. \(a_{24}\) means the element in the 2nd row and 4th column.

  7. The given matrix is of order \(3\times4\) and therefore contains \(12\) elements.

  8. Careful identification of row and column positions prevents common matrix-notation errors.
↑ Top
1 / 10  ·  10%
Q2 →
Q2
NUMERIC3 marks
If a matrix has 24 elements, what are the possible orders it can have? What, if it has 13 elements?
📘 Concept & Theory
Concept/Theory
›

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

\[ \text{Number of elements}=m\times n. \]

Therefore, when the number of elements is given, we find all possible factor pairs of that number. Each factor pair gives a possible order of the matrix.

For example, if

\[ mn=24, \]
then every positive factor pair \((m,n)\) of \(24\) represents a possible matrix order.

The order \((m,n)\) is different from \((n,m)\) unless \(m=n\), because the first number represents the number of rows and the second represents the number of columns.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let the matrix have \(m\) rows and \(n\) columns.

  2. Use the relation \(mn=\text{number of elements}\).

  3. Find all positive factor pairs of the given number.

  4. Write each factor pair as a possible order \(m\times n\).

  5. Include both orientations, such as \(2\times12\) and \(12\times2\), because rows and columns can be interchanged.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  11 steps
  1. Case 1: Matrix has 24 elements
  2. Suppose the matrix has \(m\) rows and \(n\) columns. Then
    \[mn=24\]
  3. We find the positive factor pairs of \(24\):
    \[24=\begin{cases}1\times 24\\ 2\times 12\\ 3\times 8\\4\times 6 \end{cases}\]
  4. Therefore, the corresponding possible orders are
    \[1\times24,\quad2\times12,\quad3\times8,\quad4\times6\]
  5. Since the number of rows and columns can be interchanged, we also have
    \[6\times4,\quad8\times3,\quad12\times2,\quad24\times1\]
  6. Hence, the complete set of possible orders is
    \[\boxed{1\times24,\;2\times12,\;3\times8,\;4\times6,\;6\times4,\;8\times3,\;12\times2,\;24\times1}\]
  7. Thus, a matrix containing \(24\) elements can have
    \[\boxed{8\text{ possible orders}}\]
  8. Case 2: Matrix has 13 elements
  9. Suppose the matrix has \(m\) rows and \(n\) columns. Then
    \[mn=13\]
  10. Since \(13\) is a prime number, its only positive factors are \(1\) and \(13\)
    \[13=1\times13\]
  11. Interchanging the number of rows and columns gives
    \[13=13\times1.\]
  12. Therefore, the possible orders are
    \[\boxed{1\times13\quad\text{and}\quad13\times1}\]
  13. Thus, a matrix containing \(13\) elements can have
    \[\boxed{2\text{ possible orders}}.\]
💡 Answer
Final Answer
›

For 24 elements:

\[\boxed{1\times24,\;2\times12,\;3\times8,\;4\times6,\;6\times4,\;8\times3,\;12\times2,\;24\times1}\]

For 13 elements:

\[\boxed{1\times13,\quad13\times1}\]
🎯 Exam Significance
Exam Significance
›

This question tests the relationship between the number of elements and the order of a matrix. It is an important basic concept because the same idea is used when determining whether two matrices can be equal or whether matrix operations such as addition and multiplication are defined.

For board examinations, remember that if a matrix contains \(N\) elements, its possible orders are obtained from all positive factor pairs of \(N\). The reversed factor pair must also be considered because \(m\times n\) and \(n\times m\) represent different orders when \(m\ne n\).

Significance for Competitive Entrance Examinations

This concept is useful in objective-type questions where the number of possible matrix orders is asked directly. Prime numbers are especially important because a matrix with a prime number \(p\) of elements can have only two possible orders:

\[ \boxed{1\times p\quad\text{and}\quad p\times1} \]

For a composite number, the number of possible orders depends on its factor pairs. Recognising these factor pairs quickly can save considerable time in competitive examinations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. If a matrix has \(m\) rows and \(n\) columns, it contains \(mn\) elements.

  2. To find possible orders, factorise the given number of elements.

  3. Every positive factor pair gives a possible matrix order.

  4. The orders \(m\times n\) and \(n\times m\) are both possible when \(m\ne n\).

  5. Since \(24\) has eight ordered positive factor pairs, it gives eight possible matrix orders.

  6. Since \(13\) is prime, it has only two possible orders: \(1\times13\) and \(13\times1\).

  7. A prime number of elements always gives exactly two possible matrix orders.
← Q1
2 / 10  ·  20%
Q3 →
Q3
NUMERIC3 marks
If a matrix has 18 elements, what are the possible orders it can have? What if it has 5 elements?
📘 Concept & Theory
Concept/Theory
›

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

\[ \text{Number of elements}=m\times n. \]

Therefore, when the number of elements of a matrix is given, we find all possible positive factor pairs of that number. Each factor pair represents a possible order of the matrix.

The order \(m\times n\) indicates that the matrix has \(m\) rows and \(n\) columns. Hence, \(m\times n\) and \(n\times m\) are generally different orders.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let the matrix have \(m\) rows and \(n\) columns.

  2. Use the relation \(mn=\text{number of elements}\).

  3. Factorise the given number into positive factor pairs.

  4. Write every factor pair as a possible order of the matrix.

  5. Include the reversed factor pair because rows and columns may be interchanged.

  6. For a prime number of elements, only \(1\times p\) and \(p\times1\) are possible.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  11 steps
  1. Case 1: Matrix has 18 elements
  2. Let the matrix have \(m\) rows and \(n\) columns. Then
    \[mn=18\]
  3. We find all positive factor pairs of \(18\):
    \[18=1\times18\]
    \[18=2\times9\]
    \[18=3\times6\]
  4. Interchanging the factors gives the corresponding reverse orders:
  5. \[18=6\times3\]
    \[18=9\times2\]
    \[18=18\times1\]
  6. Therefore, the possible orders of a matrix having \(18\) elements are
    \[\boxed{1\times18,\;2\times9,\;3\times6,\;6\times3,\;9\times2,\;18\times1}\]
  7. Hence, there are
    \[\boxed{6\text{ possible orders}}\]
    for a matrix containing \(18\) elements.
  8. Case 2: Matrix has 5 elements
  9. Let the matrix have \(m\) rows and \(n\) columns. Then
    \[mn=5\]
  10. Since \(5\) is a prime number, its only positive factors are \(1\) and \(5\).
    \[5=1\times5\]
  11. Interchanging the factors gives
    \[5=5\times1\]
  12. Therefore, the only possible orders are
    \[\boxed{1\times5\quad\text{and}\quad5\times1}\]
  13. Hence, there are
    \[\boxed{2\text{ possible orders}}\]
    for a matrix containing \(5\) elements.
💡 Answer
Final Answer
›

For 18 elements:

\[\boxed{1\times18,\;2\times9,\;3\times6,\;6\times3,\;9\times2,\;18\times1}\]

For 5 elements:

\[\boxed{1\times5,\quad5\times1}\]
🎯 Exam Significance
Exam Significance
›

This question reinforces the fundamental relationship between the order of a matrix and its number of elements. It is important for understanding the conditions required for matrix addition, subtraction, multiplication, and equality.

In board examinations, students should systematically list all factor pairs rather than stopping after finding only \(1\times18\), \(2\times9\), and \(3\times6\). The reverse orders must also be included.

Significance for Competitive Entrance Examinations

This is a useful speed-based concept for competitive examinations. Questions may ask for the number of possible orders rather than asking students to list them explicitly.

Since \(18\) has the positive factor pairs

\[ (1,18),\;(2,9),\;(3,6), \]
and their reverses, it has \(6\) possible orders.

Since \(5\) is prime, it has only

\[ (1,5)\quad\text{and}\quad(5,1), \]
giving exactly two possible orders.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. A matrix with \(m\) rows and \(n\) columns contains \(m\times n\) elements.

  2. Possible matrix orders are obtained from the positive factor pairs of the number of elements.

  3. For \(18\) elements, there are six possible orders.

  4. The possible orders for \(18\) elements are \(1\times18\), \(2\times9\), \(3\times6\), \(6\times3\), \(9\times2\), and \(18\times1\).

  5. Since \(5\) is prime, it has only two positive factor pairs when order is considered: \(1\times5\) and \(5\times1\).

  6. The order of a matrix specifies rows first and columns second.
← Q2
3 / 10  ·  30%
Q4 →
Q4
NUMERIC3 marks
Construct a \(2\times2\) matrix \[ A=[a_{ij}] \] whose elements are given by:
  1. \[ a_{ij}=\dfrac{(i+j)^2}{2} \]
  2. \[ a_{ij}=\dfrac{i}{j} \]
  3. \[ a_{ij}=\dfrac{(i+2j)^2}{2} \]
📘 Concept & Theory
Concept/Theory
›

A matrix \(A=[a_{ij}]\) of order \(2\times2\) contains two rows and two columns. Therefore, its elements are

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

Here, the subscripts \(i\) and \(j\) indicate the row and column, respectively. Since the matrix is of order \(2\times2\),

\[i=1,2\qquad\text{and}\qquad j=1,2.\]

Thus, to construct the matrix, we must calculate the four elements \(a_{11},a_{12},a_{21}\), and \(a_{22}\) using the given formula.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the general form of a \(2\times2\) matrix.

  2. Identify the four required elements: \(a_{11},a_{12},a_{21},a_{22}\).

  3. Substitute \(i=1,2\) and \(j=1,2\) into the given formula.

  4. Calculate each element separately.

  5. Place the four calculated elements in their correct positions.

  6. Write the resulting \(2\times2\) matrix.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  29 steps
  1. (i) Given \(a_{ij}=\dfrac{(i+j)^2}{2}\)
  2. We have
    \[a_{ij}=\dfrac{(i+j)^2}{2}\]
  3. Since \(A\) is a \(2\times2\) matrix,
    \[A=\begin{bmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{bmatrix}\]
  5. Put \(i=1\) and \(j=1\):
    \[a_{11}=\dfrac{(1+1)^2}{2}\]
    \[a_{11}=\dfrac{2^2}{2}=\dfrac{4}{2}=2\]
  6. Therefore,
    \[\boxed{a_{11}=2}\]
  8. Put \(i=1\) and \(j=2\):
    \[a_{12}=\dfrac{(1+2)^2}{2}\]
    \[a_{12}=\dfrac{3^2}{2}=\dfrac{9}{2}\]
  9. Therefore,
    \[\boxed{a_{12}=\dfrac{9}{2}}\]
  11. Put \(i=2\) and \(j=1\):
    \[a_{21}=\dfrac{(2+1)^2}{2}\]
    \[a_{21}=\dfrac{3^2}{2}=\dfrac{9}{2}\]
  12. Therefore,
    \[\boxed{a_{21}=\dfrac{9}{2}}\]
  14. Put \(i=2\) and \(j=2\):
    \[a_{22}=\dfrac{(2+2)^2}{2}\]
    \[a_{22}=\dfrac{4^2}{2}=\dfrac{16}{2}=8\]
  15. Therefore,
    \[\boxed{a_{22}=8}\]
  16. Constructing the Matrix
  17. Substituting the four elements into
    \[A=\begin{bmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{bmatrix},\]
    we get
    \[A=\begin{bmatrix}2 & \dfrac{9}{2}\\\dfrac{9}{2} & 8\end{bmatrix}\]
  18. (ii) Given \(a_{ij}=\dfrac{i}{j}\)
  19. We have
    \[a_{ij}=\dfrac{i}{j}\]
  20. The general form of the required \(2\times2\) matrix is
    \[A=\begin{bmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{bmatrix}\]
  21. Finding \(a_{11}\)
  22. Put \(i=1\) and \(j=1\):
    \[a_{11}=\dfrac{1}{1}=1\]
    \[\boxed{a_{11}=1}\]
  23. Finding \(a_{12}\)
  24. Put \(i=1\) and \(j=2\):
    \[a_{12}=\dfrac{1}{2}\]
    \[\boxed{a_{12}=\dfrac{1}{2}}\]
  25. Finding \(a_{21}\)
  26. Put \(i=2\) and \(j=1\):
    \[a_{21}=\dfrac{2}{1}=2\]
    \[\boxed{a_{21}=2}\]
  27. Finding \(a_{22}\)
  28. Put \(i=2\) and \(j=2\):
    \[a_{22}=\dfrac{2}{2}=1\]
    \[\boxed{a_{22}=1}\]
  29. Constructing the Matrix
  30. Therefore,
    \[\begin{aligned}A&=\begin{bmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{bmatrix}\\[8pt] &=\begin{bmatrix}1 & \dfrac{1}{2}\\2 & 1\end{bmatrix}\end{aligned}\]
  31. (iii) Given \(a_{ij}=\dfrac{(i+2j)^2}{2}\)
  32. We have
    \[a_{ij}=\dfrac{(i+2j)^2}{2}\]
  33. The general form of the required \(2\times2\) matrix is
    \[A=\begin{bmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{bmatrix}\]
  34. Finding \(a_{11}\)
  35. Put \(i=1\) and \(j=1\):
    \[\begin{aligned}a_{11}&=\dfrac{(1+2\times1)^2}{2}\\ a_{11}&=\dfrac{(1+2)^2}{2}\\&=\dfrac{3^2}{2}\\&=\dfrac{9}{2}\end{aligned}\]
    \[\boxed{a_{11}=\dfrac{9}{2}}\]
  36. Finding \(a_{12}\)
  37. Put \(i=1\) and \(j=2\):
    \[\begin{aligned}a_{12}&=\dfrac{(1+2\times2)^2}{2}\\a_{12} &=\dfrac{(1+4)^2}{2}\\ &=\dfrac{5^2}{2}\\ &=\dfrac{25}{2}\end{aligned}\]
    \[\boxed{a_{12}=\dfrac{25}{2}}\]
  38. Finding \(a_{21}\)
  39. Put \(i=2\) and \(j=1\):
    \[\begin{aligned}a_{21}&=\dfrac{(2+2\times1)^2}{2}\\ &=\dfrac{(2+2)^2}{2}\\ &=\dfrac{4^2}{2}\\ &=\dfrac{16}{2}\\ &=8\end{aligned}\]
    \[\boxed{a_{21}=8}\]
  40. Finding \(a_{22}\)
  41. Put \(i=2\) and \(j=2\):
    \[\begin{aligned} a_{22} &=\dfrac{(2+2\times2)^2}{2}\\ &=\dfrac{(2+4)^2}{2}\\ &=\dfrac{6^2}{2}\\ &=\dfrac{36}{2}\\ &=18\end{aligned}\]
    \[\boxed{a_{22}=18}\]
  42. Constructing the Matrix
  43. Therefore,
    \[\begin{aligned} A&=\begin{bmatrix}a_{11} & a_{12}\\a_{21} & a_{22}\end{bmatrix}\\[8pt] &=\begin{bmatrix}\dfrac{9}{2} & \dfrac{25}{2}\\8 & 18\end{bmatrix} \end{aligned} \]
💡 Answer
Final Answer
›

(i) For

\[ a_{ij}=\dfrac{(i+j)^2}{2}, \]
\[A=\begin{bmatrix}2 & \dfrac{9}{2}\\\dfrac{9}{2} & 8\end{bmatrix}\]

(ii) For

\[a_{ij}=\dfrac{i}{j},\]
\[A=\begin{bmatrix}1 & \dfrac{1}{2}\\2 & 1\end{bmatrix}\]

(iii) For

\[a_{ij}=\dfrac{(i+2j)^2}{2},\]
\[A=\begin{bmatrix}\dfrac{9}{2} & \dfrac{25}{2}\\8 & 18\end{bmatrix}\]
🎯 Exam Significance
Exam Significance
›

This question tests the ability to construct a matrix from a prescribed formula for its general element \(a_{ij}\). It is important because questions based on matrix elements frequently require students to substitute specific values of \(i\) and \(j\) correctly.

For a \(2\times2\) matrix, students should immediately identify the four positions:

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

A common error is to interchange the meanings of \(i\) and \(j\). Remember that \(i\) denotes the row number and \(j\) denotes the column number.

Significance for Competitive Entrance Examinations

Matrix-element formulas are frequently used as the starting point for questions involving matrix properties, equality of matrices, symmetric matrices, skew-symmetric matrices, determinants, and matrix operations.

For objective examinations, calculating the four elements systematically prevents positional errors. It is particularly important to distinguish between \(a_{12}\) and \(a_{21}\), because they correspond to different positions in a matrix.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. A \(2\times2\) matrix has exactly four elements.

  2. The four elements are \(a_{11},a_{12},a_{21}\), and \(a_{22}\).

  3. In \(a_{ij}\), \(i\) represents the row and \(j\) represents the column.

  4. To construct a matrix from a formula, substitute the appropriate values of \(i\) and \(j\) systematically.

  5. For a \(2\times2\) matrix, \(i,j\in\{1,2\}\).

  6. Do not confuse \(a_{12}\) with \(a_{21}\); they represent different matrix elements.

  7. Writing the general matrix form before substitution reduces calculation and placement errors.
← Q3
4 / 10  ·  40%
Q5 →
Q5
NUMERIC3 marks
Construct a \(3\times4\) matrix whose elements are given by:
  1. \[ a_{ij}=\dfrac{1}{2}\left|-3i+j\right| \]
  2. \[ a_{ij}=2i-j \]
📘 Concept & Theory
Concept/Theory
›

A matrix of order \(3\times4\) has \(3\) rows and \(4\) columns. Therefore, it contains

\[ 3\times4=12 \]

elements. The general form of such a matrix is

\[ A= \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14}\\ a_{21} & a_{22} & a_{23} & a_{24}\\ a_{31} & a_{32} & a_{33} & a_{34} \end{bmatrix}. \]

Since the matrix is \(3\times4\), the row index \(i\) takes the values \(1,2,3\), while the column index \(j\) takes the values \(1,2,3,4\):

\[ i\in\{1,2,3\}, \qquad j\in\{1,2,3,4\}. \]

To construct the matrix, each of the twelve elements must be calculated by substituting the corresponding values of \(i\) and \(j\) into the given formula.

In part (i), the expression contains an absolute value:

\[ \left|-3i+j\right|. \]
Recall that the absolute value of a real number is always non-negative. Thus,
\[ |-5|=5,\qquad |-2|=2,\qquad |0|=0. \]

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the general form of a \(3\times4\) matrix.

  2. Identify all twelve elements \(a_{11}\) through \(a_{34}\).

  3. For each element, substitute the appropriate row index \(i\) and column index \(j\).

  4. In part (i), evaluate the absolute value before multiplying by \(\frac12\).

  5. In part (ii), substitute \(i\) and \(j\) directly into \(2i-j\).

  6. Arrange the calculated values in their correct row and column positions.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  42 steps
  1. (i) Given \(a_{ij}=\dfrac{1}{2}\left|-3i+j\right|\)
  2. We have
    \[a_{ij}=\dfrac{1}{2}\left|-3i+j\right|\]
  3. Since \(A\) is a \(3\times4\) matrix,
    \[A=\begin{bmatrix}a_{11} & a_{12} & a_{13} & a_{14}\\ a_{21} & a_{22} & a_{23} & a_{24}\\ a_{31} & a_{32} & a_{33} & a_{34}\end{bmatrix}\]
  4. First Row: \(i=1\)
  5. For the first row, \(i=1\) and \(j=1,2,3,4\)
  6. Finding \(a_{11}\)
  7. \[\begin{aligned} a_{11}&= \dfrac{1}{2}\left|-3(1)+1\right|\\ &=\dfrac{1}{2}|-3+1|\\ &=\dfrac{1}{2}|-2|\\ &=\dfrac{2}{2}\\ &=1\end{aligned} \]
    \[\boxed{a_{11}=1}\]
  8. Finding \(a_{12}\)
  9. \[\begin{aligned} a_{12}&=\dfrac{1}{2}\left|-3(1)+2\right|\\ &=\dfrac{1}{2}|-3+2|\\ &=\dfrac{1}{2}|-1|\\ &=\dfrac{1}{2}\end{aligned}\]
    \[\boxed{a_{12}=\dfrac{1}{2}}\]
  10. Finding \(a_{13}\)
  11. \[\begin{aligned} a_{13} &=\dfrac{1}{2}\left|-3(1)+3\right|\\ &=\dfrac{1}{2}|-3+3|\\ &=\dfrac{1}{2}|0|\\ &=0\end{aligned} \]
    \[\boxed{a_{13}=0}\]
  12. Finding \(a_{14}\)
  13. \[\begin{aligned} a_{14}&=\dfrac{1}{2}\left|-3(1)+4\right|\\ &=\dfrac{1}{2}|-3+4|\\ &=\dfrac{1}{2}|1|\\ &=\dfrac{1}{2}\end{aligned} \]
    \[\boxed{a_{14}=\dfrac{1}{2}}\]
  14. Therefore, the first row is
    \[\begin{bmatrix}1 & \dfrac{1}{2} & 0 & \dfrac{1}{2}\end{bmatrix}\]
  15. Second Row: \(i=2\)
  16. For the second row, \(i=2\) and \(j=1,2,3,4\)
  17. Finding \(a_{21}\)
  18. \[\begin{aligned} a_{21}&=\dfrac{1}{2}\left|-3(2)+1\right|\\ &=\dfrac{1}{2}|-6+1|\\ &=\dfrac{1}{2}|-5|\\ &=\dfrac{5}{2}\end{aligned}\]
    \[\boxed{a_{21}=\dfrac{5}{2}}\]
  19. Finding \(a_{22}\)
  20. \[\begin{aligned} a_{22}&=\dfrac{1}{2}\left|-3(2)+2\right|\\ &=\dfrac{1}{2}|-6+2|\\ &=\dfrac{1}{2}|-4|\\ &=\dfrac{4}{2}\\ &=2\end{aligned} \]
    \[\boxed{a_{22}=2}\]
  21. Finding \(a_{23}\)
  22. \[\begin{aligned} a_{23}&=\dfrac{1}{2}\left|-3(2)+3\right|\\ &=\dfrac{1}{2}|-6+3|\\ &=\dfrac{1}{2}|-3|\\ &=\dfrac{3}{2}\end{aligned} \]
    \[\boxed{a_{23}=\dfrac{3}{2}}\]
  23. Finding \(a_{24}\)
  24. \[\begin{aligned} a_{24}&= \dfrac{1}{2}\left|-3(2)+4\right|\\ &=\dfrac{1}{2}|-6+4|\\ &=\dfrac{1}{2}|-2|\\ &=\dfrac{2}{2}\\ &=1\end{aligned} \]
    \[\boxed{a_{24}=1}\]
  25. Therefore, the second row is
    \[\begin{bmatrix}\dfrac{5}{2} & 2 & \dfrac{3}{2} & 1\end{bmatrix}\]
  26. Third Row: \(i=3\)
  27. For the third row, \(i=3\) and \(j=1,2,3,4\)
  28. Finding \(a_{31}\)
  29. \[\begin{aligned} a_{31}&=\dfrac{1}{2}\left|-3(3)+1\right|\\ &=\dfrac{1}{2}|-9+1|\\ &=\dfrac{1}{2}|-8|\\ &=\dfrac{8}{2}\\ &=4\end{aligned} \]
    \[\boxed{a_{31}=4}\]
  30. Finding \(a_{32}\)
  31. \[\begin{aligned} a_{32}&=\dfrac{1}{2}\left|-3(3)+2\right|\\ &=\dfrac{1}{2}|-9+2|\\ &=\dfrac{1}{2}|-7|\\ &=\dfrac{7}{2}\end{aligned} \]
    \[\boxed{a_{32}=\dfrac{7}{2}}\]
  32. Finding \(a_{33}\)
  33. \[\begin{aligned} a_{33}&=\dfrac{1}{2}\left|-3(3)+3\right|\\ &=\dfrac{1}{2}|-9+3|\\ &=\dfrac{1}{2}|-6|\\ &=\dfrac{6}{2}\\ &=3\end{aligned} \]
    \[\boxed{a_{33}=3}\]
  34. Finding \(a_{34}\)
  35. \[\begin{aligned} a_{34}&=\dfrac{1}{2}\left|-3(3)+4\right|\\ &=\dfrac{1}{2}|-9+4|\\ &=\dfrac{1}{2}|-5|\\ &=\dfrac{5}{2}\end{aligned} \]
    \[\boxed{a_{34}=\dfrac{5}{2}}\]
  36. Therefore, the third row is
  37. \[\begin{bmatrix}4 & \dfrac{7}{2} & 3 & \dfrac{5}{2}\end{bmatrix}\]
  38. Constructing the Matrix
  39. Combining the three rows, we obtain
    \[ A=\begin{bmatrix}1 & \dfrac{1}{2} & 0 & \dfrac{1}{2}\\ \dfrac{5}{2} & 2 & \dfrac{3}{2} & 1\\ 4 & \dfrac{7}{2} & 3 & \dfrac{5}{2}\end{bmatrix}\]
  40. (ii) Given \(a_{ij}=2i-j\)
  41. We have
    \[a_{ij}=2i-j\]
  42. The required \(3\times4\) matrix is
    \[ A= \begin{bmatrix} a_{11} & a_{12} & a_{13} & a_{14}\\ a_{21} & a_{22} & a_{23} & a_{24}\\ a_{31} & a_{32} & a_{33} & a_{34} \end{bmatrix} \]
  43. First Row: \(i=1\)
  44. For the first row, \(i=1\) and \(j=1,2,3,4\)
  45. Finding \(a_{11}\)
  46. \[a_{11}=2(1)-1=2-1=1\]
    \[\boxed{a_{11}=1}\]
  47. Finding \(a_{12}\)
  48. \[a_{12}=2(1)-2=2-2=0\]
    \[\boxed{a_{12}=0}\]
  49. Finding \(a_{13}\)
  50. \[a_{13}=2(1)-3=2-3=-1\]
    \[\boxed{a_{13}=-1}\]
  51. Finding \(a_{14}\)
  52. \[a_{14}=2(1)-4=2-4=-2\]
    \[\boxed{a_{14}=-2}\]
  53. Therefore, the first row is
  54. \[\begin{bmatrix}1 & 0 & -1 & -2\end{bmatrix}\]
  55. First Row: \(i=2\)
  56. Finding \(a_{21}\)
  57. \[a_{21}=2(2)-1=4-1=3\]
    \[\boxed{a_{21}=3}\]
  58. Finding \(a_{22}\)
  59. \[a_{22}=2(2)-2=4-2=2\]
    \[\boxed{a_{22}=2}\]
  60. Finding \(a_{23}\)
  61. \[a_{23}=2(2)-3=4-3=1\]
    \[\boxed{a_{23}=1}\]
  62. Finding \(a_{24}\)
  63. \[a_{24}=2(2)-4=4-4=0\]
    \[\boxed{a_{24}=0}\]
  64. Therefore, the second row is
    \[\begin{bmatrix}3 & 2 & 1 & 0\end{bmatrix}\]
  65. First Row: \(i=3\)
  66. Finding \(a_{31}\)
  67. \[a_{31}=2(3)-1=6-1=5\]
    \[\boxed{a_{31}=5}\]
  68. Finding \(a_{32}\)
  69. \[a_{32}=2(3)-2=6-2=4\]
    \[\boxed{a_{32}=4}\]
  70. Finding \(a_{33}\)
  71. \[a_{33}=2(3)-3=6-3=3\]
    \[\boxed{a_{33}=3}\]
  72. Finding \(a_{34}\)
  73. \[a_{34}=2(3)-4=6-4=2\]
    \[\boxed{a_{34}=2}\]
  74. Therefore, the third row is
    \[\begin{bmatrix}5 & 4 & 3 & 2\end{bmatrix}\]
  75. Constructing the Matrix
  76. Combining all three rows, we obtain
    \[ A= \begin{bmatrix} 1 & 0 & -1 & -2\\ 3 & 2 & 1 & 0\\ 5 & 4 & 3 & 2 \end{bmatrix} \]
💡 Answer
Final Answer
›

(i) For

\[ a_{ij}=\dfrac{1}{2}\left|-3i+j\right|, \]
\[ A= \begin{bmatrix} 1 & \dfrac{1}{2} & 0 & \dfrac{1}{2}\\ \dfrac{5}{2} & 2 & \dfrac{3}{2} & 1\\ 4 & \dfrac{7}{2} & 3 & \dfrac{5}{2} \end{bmatrix} \]

(ii) For

\[ a_{ij}=2i-j, \]
\[ A= \begin{bmatrix} 1 & 0 & -1 & -2\\ 3 & 2 & 1 & 0\\ 5 & 4 & 3 & 2 \end{bmatrix} \]
🎯 Exam Significance
Exam Significance
›

This question is important because it requires the complete construction of a matrix from a formula for its general element. It tests whether the student can correctly interpret row and column indices and perform substitution without missing any element.

In part (i), careful handling of the absolute value is essential. A common error is to retain the negative sign after applying the modulus. For example,

\[ |-5|=5, \]

not \(-5\).

Students should also check that the final matrix contains exactly \(3\times4=12\) elements.

Significance for Competitive Entrance Examinations

Formula-based matrix construction develops the substitution accuracy needed for more advanced matrix questions. In competitive examinations, a seemingly simple indexing error can change the entire answer.

The pattern in part (ii) can also be recognised directly:

\[ a_{ij}=2i-j. \]
For a fixed row \(i\), increasing \(j\) by \(1\) decreases the element by \(1\). For a fixed column \(j\), increasing \(i\) by \(1\) increases the element by \(2\). Recognising such patterns can speed up calculations in objective questions.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. A \(3\times4\) matrix contains exactly \(12\) elements.

  2. For a \(3\times4\) matrix, \(i=1,2,3\) and \(j=1,2,3,4\).

  3. The first subscript of \(a_{ij}\) identifies the row.

  4. The second subscript of \(a_{ij}\) identifies the column.

  5. Each matrix element must be calculated by substituting its corresponding \(i\) and \(j\) values.

  6. Absolute value always produces a non-negative result.

  7. After calculating all elements, they must be placed in their correct positions.

  8. Always verify the final matrix order and the total number of elements.
← Q4
5 / 10  ·  50%
Q6 →
Q6
NUMERIC3 marks
Find the values of \(x\), \(y\), and \(z\) from the following equations:
📘 Concept & Theory
Concept/Theory
›
  1. \[ \begin{bmatrix} 4 & 3\\ x & 5 \end{bmatrix} = \begin{bmatrix} y & z\\ 1 & 5 \end{bmatrix} \]
  2. \[ \begin{bmatrix} x+y & 2\\ 5+z & x \end{bmatrix} = \begin{bmatrix} 6 & 2\\ 5 & 8 \end{bmatrix} \]
  3. \[ \begin{bmatrix} x+y+z\\ x+z\\ y+z \end{bmatrix} = \begin{bmatrix} 9\\ 5\\ 7 \end{bmatrix} \]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Two matrices are said to be equal if and only if:

  2. They have the same order.

  3. Their corresponding elements are equal.

  4. Therefore, if

    \[ A=B, \]
    then every corresponding pair of elements must satisfy
    \[ a_{ij}=b_{ij}. \]

  5. This property allows us to convert a matrix equation into ordinary algebraic equations and then solve for the unknowns.

  6. Solution Roadmap

  7. Compare corresponding elements on both sides of the matrix equation.

  8. Equate entries occupying the same row and column positions.

  9. Solve the resulting algebraic equations systematically.

  10. Substitute any values already obtained into the remaining equations.

  11. Verify the values by substituting them back into the original matrix equation.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  33 steps
  1. (i) Finding \(x\), \(y\), and \(z\)
  2. Given —
    \[\begin{bmatrix}4 & 3\\x & 5\end{bmatrix}=\begin{bmatrix}y & z\\1 & 5\end{bmatrix}\]
  3. Since the two matrices are equal, their corresponding elements must be equal.
  4. Comparing the first-row, first-column elements:
    \[4=y\]
    \[\boxed{y=4}\]
  5. Comparing the first-row, second-column elements:
    \[3=z\]
    \[\boxed{z=3}\]
  6. Comparing the second-row, first-column elements:
    \[x=1\]
    \[\boxed{x=1}\]
  7. Comparing the second-row, second-column elements gives
    \[5=5\]
  8. which is already true and gives no additional condition.
  9. Verification
  10. Substituting \(x=1\), \(y=4\), and \(z=3\), the left-hand side becomes
    \[\begin{bmatrix}4 & 3\\1 & 5\end{bmatrix}\]
  11. The right-hand side becomes
    \[\begin{bmatrix}4 & 3\\1 & 5\end{bmatrix}\]
  12. Hence, both matrices are equal.
  13. \[\boxed{x=1,\quad y=4,\quad z=3}\]
  14. (ii) Finding \(x\), \(y\), and \(z\)
  15. Given —
    \[\begin{bmatrix}x+y & 2\\5+z & x\end{bmatrix}=\begin{bmatrix}6 & 2\\5 & 8\end{bmatrix}\]
  16. Since the matrices are equal, corresponding elements must be equal.
  17. Step 1: Compare the first-row, first-column elements
  18. \[x+y=6\]
  19. This gives
    \[\boxed{x+y=6}\]
  20. Step 2: Compare the first-row, second-column elements
  21. \[2=2\]
    This is an identity and gives no additional condition.
  22. Step 3: Compare the second-row, first-column elements
  23. \[5+z=5\]
    \[\boxed{z=0}\]
  24. Step 4: Compare the second-row, second-column elements
  25. \[x=8\]
    \[\boxed{x=8}\]
  26. Step 5: Find \(y\)
  27. We already have
    \[x+y=6\]
    Substituting \(x=8\):
    \[8+y=6\]
  28. Subtracting \(8\) from both sides:
    \[\begin{aligned}y&=6-8\\ &=-2\end{aligned}\]
    \[\boxed{y=-2}\]
  29. Verification
  30. Substituting
    \[x=8,\qquad y=-2,\qquad z=0\]
    into the left-hand side:
    \[\begin{bmatrix}x+y & 2\\5+z & x\end{bmatrix}= \begin{bmatrix}8+(-2) & 2\\5+0 & 8\end{bmatrix}\]
    \[=\begin{bmatrix}6 & 2\\5 & 8\end{bmatrix}\]
  31. This is exactly the given right-hand side. Therefore,
    \[\boxed{x=8,\quad y=-2,\quad z=0}\]
  32. (iii) Finding \(x\), \(y\), and \(z\)
  33. Given —
    \[\begin{bmatrix}x+y+z\\x+z\\y+z\end{bmatrix}= \begin{bmatrix}9\\5\\7\end{bmatrix}\]
  34. Since the two column matrices are equal, their corresponding elements must be equal.
  35. Step 1: Compare the first elements
  36. \[x+y+z=9\]
  37. Therefore,
    \[\boxed{x+y+z=9}\]
  38. Step 2: Compare the second elements
  39. \[x+z=5\]
  40. Step 3: Compare the third elements
  41. \[y+z=7\]
  42. Step 4: Find \(y\)
  43. We have
    \[x+y+z=9\]
    and
    \[x+z=5\]
  44. Subtract the second equation from the first:
    \[(x+y+z)-(x+z)=9-5\]
  45. On simplifying,
    \[y=4\]
  46. Step 5: Find \(z\)
  47. Using
    \[y+z=7\]
    and \(y=4\):
    \[4+z=7\]
  48. Subtracting \(4\) from both sides:
    \[\begin{aligned}z&=7-4\\&=3\end{aligned}\]
  49. Step 6: Find \(x\)
  50. Using
    \[x+z=5\]
    and \(z=3\):
    \[\begin{aligned}x+3&=5\\&=2\end{aligned}\]
  51. Verification
  52. Substitute
    \[x=2,\qquad y=4,\qquad z=3\]
    into the left-hand side:
    \[\begin{bmatrix}x+y+z\\x+z\\y+z\end{bmatrix}= \begin{bmatrix}2+4+3\\2+3\\4+3\end{bmatrix}= \begin{bmatrix}9\\5\\7\end{bmatrix}\]
  53. This is exactly the given right-hand side. Hence,
    \[\boxed{x=2,\quad y=4,\quad z=3}\]
💡 Answer
Final Answer
›

(i)

\[\boxed{x=1,\quad y=4,\quad z=3}\]

(ii)

\[\boxed{x=8,\quad y=-2,\quad z=0}\]

(iii)

\[\boxed{x=2,\quad y=4,\quad z=3}\]
🎯 Exam Significance
Exam Significance
›

This question directly tests the important property of equality of matrices. Students must know that two matrices are equal only when corresponding elements are equal.

Part (i) is a straightforward application of this property. Parts (ii) and (iii) require the matrix equation to be converted into simultaneous algebraic equations and then solved systematically.

A useful examination strategy is to compare entries position by position. Do not compare elements diagonally or in a different order.

Significance for Competitive Entrance Examinations

Equality of matrices is a basic but highly useful concept in competitive examinations. Matrix equations can be transformed into systems of equations, and these systems may subsequently be used with substitution, elimination, determinants, or other algebraic techniques.

In objective questions, students should first look for entries containing only one unknown. Such entries can often be used immediately to determine one variable and simplify the remaining equations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. Two matrices are equal only when they have the same order and corresponding elements are equal.

  2. For equal matrices, compare elements occupying exactly the same positions.

  3. A matrix equation can be converted into ordinary algebraic equations.

  4. Use the simplest equation first whenever possible.

  5. Always substitute the obtained values back into the original equation for verification.

  6. In Question 6(ii), the bottom-right element gives \(x=8\) directly.

  7. In Question 6(iii), subtracting \(x+z=5\) from \(x+y+z=9\) immediately gives \(y=4\).
← Q5
6 / 10  ·  60%
Q7 →
Q7
NUMERIC3 marks
Find the values of \(a\), \(b\), \(c\), and \(d\) from the equation \[\begin{bmatrix}a-b & 2a+c\\2a-b & 3c+d\end{bmatrix}=\begin{bmatrix}-1 & 5\\0 & 13\end{bmatrix}\]
📘 Concept & Theory
Concept/Theory
›

Two matrices are equal if and only if they have the same order and their corresponding elements are equal.

Therefore, when

\[ \begin{bmatrix} p & q\ r & s \end{bmatrix} = \begin{bmatrix} u & v\ w & x \end{bmatrix}, \]
we must have

\[ p=u,\qquad q=v,\qquad r=w,\qquad s=x. \]

The same principle is applied here. The given matrix equation produces four algebraic equations involving the four unknowns \(a,b,c,d\). We solve these equations systematically.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Compare the corresponding entries of the two matrices.

  2. Use the first and third entries to determine \(a\) and \(b\).

  3. Substitute the value of \(a\) into the second equation to find \(c\).

  4. Use the value of \(c\) in the fourth equation to find \(d\).

  5. Substitute all four values back into the original matrix equation to verify the answer.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  13 steps
  1. Given —
    \[\begin{bmatrix}a-b & 2a+c\\2a-b & 3c+d\end{bmatrix}=\begin{bmatrix}-1 & 5\\0 & 13\end{bmatrix}\]
  2. Since corresponding elements of equal matrices are equal, we obtain four equations.
  3. Step 1: Compare the first elements
  4. \[\begin{align} a-b&=-1 \\ \Rightarrow a&=b-1\tag{1} \end{align}\]
  5. Step 2: Compare the third elements
  6. \[\begin{aligned} 2a-b&=0\\ 2a&=b \end{aligned}\]
  7. Substituting \(b=2a\) in equation (1)
    \[\begin{aligned} a&=2a-1\\ a-2a&=-1\\ -a&=-1\\ \Rightarrow a&=1 \end{aligned}\]
  8. Since
    \[b=2a\]
  9. we get
    \[b=2\]
  10. Step 3: Find \(c\)
  11. Compare the second elements:
    \[2a+c=5\]
  12. Substituting \(a=1\):
    \[\begin{aligned}` 2(1)+c&=5\\ c&=5-2\\ &=3 \end{aligned}\]
  13. Step 4: Find \(d\)
  14. Compare the fourth elements:
    \[3c+d=13\]
  15. Substituting \(c=3\):
    \[ \begin{aligned} 3(3)+d&=13\\ d&=13-9\\ &=4 \end{aligned} \]
  16. Step 5: Verification
  17. We have obtained
    \[a=1,\quad b=2,\quad c=3,\quad d=4.\]
  18. Substitute these values into the left-hand side:
    \[\begin{aligned}\begin{bmatrix}a-b & 2a+c\\2a-b & 3c+d\end{bmatrix}&=\begin{bmatrix}1-2 & 2(1)+3\\2(1)-2 & 3(3)+4\end{bmatrix}\\[8pt] &=\begin{bmatrix}-1 & 5\\0 & 13\end{bmatrix}\end{aligned}\]
  19. This is exactly the given right-hand side. Hence, the values are verified.
💡 Answer
Final Answer
›
\[\boxed{a=1,\quad b=2,\quad c=3,\quad d=4}\]
🎯 Exam Significance
Exam Significance
›

This question is a direct application of the equality of matrices. It is important for board examinations because students must correctly convert one matrix equation into four simultaneous algebraic equations.

The most important step is to match entries according to their exact positions:

\[ a-b=-1, \]
\[ 2a+c=5, \]
\[ 2a-b=0, \]
\[ 3c+d=13. \]

A positional mistake can lead to an entirely incorrect answer. Therefore, students should always compare corresponding entries row-wise and column-wise.

Significance for Competitive Entrance Examinations

Matrix equations of this type provide a quick application of the equality property and are useful in objective-type questions. Competitive examinations may combine this concept with simultaneous equations, determinants, matrix operations, or parameter-based questions.

An efficient approach is to identify equations involving fewer unknowns first. Here,

\[ a-b=-1 \]
and
\[ 2a-b=0 \]
immediately determine \(a\) and \(b\). The remaining two equations then determine \(c\) and \(d\).

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. Equal matrices have equal corresponding elements.

  2. A \(2\times2\) matrix equality generally gives four corresponding equations.

  3. Compare entries according to their row and column positions.

  4. Solve the equations in an efficient sequence rather than treating all equations simultaneously without structure.

  5. Substitution is useful when one equation expresses one variable in terms of another.

  6. Always verify the final values in the original matrix equation.
← Q6
7 / 10  ·  70%
Q8 →
Q8
NUMERIC3 marks
Let \[ A=\left[a_{ij}\right]_{m\times n} \] be a matrix. The matrix \(A\) is a square matrix if:
  1. \(m<n\)
  2. \(m>n\)
  3. \(m=n\)
  4. None of these
📘 Concept & Theory
Concept/Theory
›

The order of a matrix is written as

\[ m\times n, \]
where \(m\) represents the number of rows and \(n\) represents the number of columns.

A matrix is called a square matrix when its number of rows is equal to its number of columns.

\[ \boxed{m=n} \]

Therefore, if

\[ A=\left[a_{ij}\right]_{m\times n}, \]
then \(A\) is a square matrix if and only if
\[ m=n. \]

For example, a \(3\times3\) matrix is square because it has three rows and three columns. Similarly, a \(4\times4\) matrix is square. However, a \(3\times4\) matrix is not square because its number of rows and columns are different.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify \(m\) as the number of rows.

  2. Identify \(n\) as the number of columns.

  3. Recall the defining condition for a square matrix.

  4. Compare \(m\) and \(n\).

  5. Select the option satisfying \(m=n\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  8 steps
  1. Given —
    \[A=\left[a_{ij}\right]_{m\times n}\]
  2. Therefore:
    \[m=\text{number of rows}\]
    \[n=\text{number of columns}\]
  3. A matrix is square if and only if the number of rows is equal to the number of columns.
  4. \[\text{Number of rows}=\text{Number of columns}\]
    \[m=n\]
  5. Hence, the required condition is
    \[\boxed{m=n}\]
  6. Checking the Options
  7. Option (a): \(m<n\) The number of rows is less than the number of columns. Therefore, the matrix is rectangular, not square.
  8. Option (b): \(m>n\) The number of rows is greater than the number of columns. Therefore, the matrix is rectangular, not square.
  9. Option (c): \(m=n\) The number of rows is equal to the number of columns. Hence, the matrix is square.
  10. Option (d): None of these Since option (c) gives the correct condition, option (d) is incorrect.
💡 Answer
Final Answer
›
\[\boxed{\text{Option (c): }m=n}\]
🎯 Exam Significance
Exam Significance
›

This is a fundamental definition-based question from the introductory concepts of matrices. The condition for a square matrix is frequently used in later topics such as diagonal matrices, scalar matrices, identity matrices, symmetric matrices, skew-symmetric matrices, determinants, and inverse matrices.

Students should remember that the equality of rows and columns is the defining condition:

\[ \boxed{\text{Square matrix}\iff m=n} \]
Significance for Competitive Entrance Examinations

This concept is often tested through direct multiple-choice questions and can also appear indirectly in questions involving determinants and inverses. A determinant is defined only for a square matrix, so identifying whether a matrix is square is an essential preliminary step.

For quick problem solving, remember that the dimensions of a square matrix have the form

\[ 1\times1,\quad2\times2,\quad3\times3,\quad4\times4,\ldots \]
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. The order of \(A=[a_{ij}]_{m\times n}\) is \(m\times n\).

  2. \(m\) represents the number of rows.

  3. \(n\) represents the number of columns.

  4. A square matrix has an equal number of rows and columns.

  5. The necessary and sufficient condition for a square matrix is \(m=n\).

  6. A \(3\times4\) matrix is not square because \(3\neq4\).

  7. A \(4\times4\) matrix is square because \(4=4\).

  8. The correct answer is option (c).
← Q7
8 / 10  ·  80%
Q9 →
Q9
NUMERIC3 marks
Which of the given values of \(x\) and \(y\) make the following pair of matrices equal? \[\begin{bmatrix}3x+7 & 5\\y+1 & 2-3x\end{bmatrix},\qquad\begin{bmatrix}0 & y-2\\8 & 4\end{bmatrix}\] The options are:
  1. \(x=-\dfrac{1}{3},\ y=7\)
  2. Not possible to find
  3. \(y=7,\ x=-\dfrac{2}{3}\)
  4. \(x=-\dfrac{1}{3},\ y=-\dfrac{2}{3}\)
📘 Concept & Theory
Concept/Theory
›

Two matrices are equal if and only if they have the same order and all their corresponding elements are equal.

Thus, if

\[ \begin{bmatrix} a & b\\ c & d \end{bmatrix} = \begin{bmatrix} p & q\\ r & s \end{bmatrix}, \]

then

\[ a=p,\qquad b=q,\qquad c=r,\qquad d=s. \]

In this question, both matrices are of order \(2\times2\). Therefore, we must equate all four corresponding elements.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Compare the first-row, first-column elements.

  2. Compare the first-row, second-column elements.

  3. Compare the second-row, first-column elements.

  4. Compare the second-row, second-column elements.

  5. Check whether the resulting equations give a common value of \(x\) and \(y\).

  6. Compare the obtained values with the given options.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  10 steps
  1. Given —
    \[A=\begin{bmatrix}3x+7 & 5\\y+1 & 2-3x\end{bmatrix}\]

    and

    \[B=\begin{bmatrix}0 & y-2\\8 & 4\end{bmatrix}\]
  2. For \(A=B\), corresponding elements must be equal.
  3. Step 1: Compare the first-row, first-column elements
  4. \[\begin{aligned}3x+7&=0\\3x&=-7\\x&=-\dfrac73\end{aligned}\]
  5. Step 2: Compare the first-row, second-column elements
  6. \[\begin{aligned}5&=y-2\\\Rightarrow y&=7\end{aligned}\]
  7. Step 3: Compare the second-row, first-column elements
  8. \[\begin{aligned}y+1&=8\\y&=7\end{aligned}\]
  9. Step 4: Compare the second-row, second-column elements
  10. \[\begin{aligned}2-3x&=4\\-3x&=2\\x&-dfrac23\end{aligned}\]
  11. Step 5: Check for consistency
  12. From the first corresponding elements, we obtained
  13. \[x=-\dfrac{7}{3}\]
  14. However, from the fourth corresponding elements, we obtained
    \[x=-\dfrac{2}{3}\]
  15. Since
    \[-\dfrac{7}{3}\neq-\dfrac{2}{3},\]
    there is no single value of \(x\) that satisfies both conditions simultaneously.
  16. Although \(y=7\) satisfies the equations involving \(y\), the two independent conditions on \(x\) are contradictory. Therefore, no pair \((x,y)\) can make the two matrices equal.
💡 Answer
Final Answer
›
\[\boxed{\text{Option (B): Not possible}}\]
🎯 Exam Significance
Exam Significance
›

This question tests a crucial point about equality of matrices: every corresponding element must satisfy the equality simultaneously.

It is not sufficient for one or two corresponding entries to match. If even one pair of corresponding elements gives a contradictory condition, the two matrices cannot be equal.

This question is particularly useful for understanding how matrix equations can produce a system of simultaneous conditions.

Significance for Competitive Entrance Examinations

This type of question can be solved rapidly in competitive examinations by comparing the simplest corresponding entries first. However, the candidate must check all corresponding entries before selecting an option.

An important competitive-exam strategy is to detect contradiction early. Here, the first and fourth entries independently give two different values of \(x\):

\[ x=-\dfrac73 \]

and

\[ x=-\dfrac23 \]

Therefore, the matrices cannot be equal, and the correct choice is immediately identified as option (B).

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. Equal matrices must have equal corresponding elements.

  2. All corresponding-element equations must be satisfied simultaneously.

  3. One contradiction is sufficient to conclude that the matrices cannot be equal.

  4. Here, the first entries give \(x=-\dfrac73\).

  5. The fourth entries give \(x=-\dfrac23\).

  6. Since these values are different, no value of \(x\) satisfies both conditions.

  7. The equations involving \(y\) consistently give \(y=7\), but this does not rescue the matrix equality because the conditions on \(x\) are contradictory.

  8. The correct answer is option (B).
← Q8
9 / 10  ·  90%
Q10 →
Q10
NUMERIC3 marks
The number of all possible matrices of order \(3\times3\), with each entry \(0\) or \(1\), is:
  1. \(27\)
  2. \(18\)
  3. \(81\)
  4. \(512\)
📘 Concept & Theory
Concept/Theory
›

A matrix of order \(m\times n\) contains

\[mn\]

elements.

If every element can independently take any one of \(k\) possible values, then the total number of possible matrices is

\[\boxed{k^{mn}}\]

In this question, each entry can take only two possible values:

\[0\quad\text{or}\quad1\]

Therefore, \(k=2\)

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Determine the total number of entries in the matrix.

  2. Determine the number of possible values for each entry.

  3. Use the multiplication principle for independent choices.

  4. Calculate the resulting power.

  5. Compare the result with the given options.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  6 steps
  1. The given matrix is of order \(3\times3\).
  2. Therefore, the total number of elements is
    \[3\times3=9\]
  3. Thus, there are \(9\) positions in the matrix:
    \[\begin{bmatrix}a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\end{bmatrix}\]
  4. Each of these \(9\) entries can independently be either \(0\) or \(1\). Hence, each position has 2 possible choices.
  5. Therefore, the number of possible matrices is
    \[\underbrace{2\times2\times2\times2\times2\times2\times2\times2\times2}_{9\text{ entries}}\]
    \[=2^9=512\]
  6. Hence, the total number of possible matrices is
    \[\boxed{512}\]
💡 Answer
Final Answer
›
\[\boxed{\text{Option (D): }512}\]
🎯 Exam Significance
Exam Significance
›

This question tests the concept of counting matrices when restrictions are placed on the possible values of their entries. It is important to understand that the number of entries depends on the order of the matrix.

For an \(m\times n\) matrix, there are \(mn\) entries. If each entry can independently assume \(k\) possible values, then

\[ \boxed{\text{Number of matrices}=k^{mn}}. \]

This general formula can be directly applied to similar board-examination questions.

Significance for Competitive Entrance Examinations

This is a standard application of the fundamental principle of counting. Competitive examinations may present the same idea with different matrix orders or with a different number of allowed values for each entry.

The fastest approach is to calculate the number of positions first and then raise the number of choices per position to that power.

\[ \boxed{k^{mn}} \]

For example, if every entry of a \(2\times3\) matrix could be \(0,1,\) or \(2\), the number of possible matrices would be

\[ 3^6. \]
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. A matrix of order \(m\times n\) contains \(mn\) entries.

  2. A \(3\times3\) matrix contains \(9\) entries.

  3. Each entry has \(2\) possible values: \(0\) and \(1\).

  4. The choices for different entries are independent.

  5. The total number of matrices is \(2^9\).

  6. \(2^9=512\).

  7. The correct answer is option (D).
← Q9
10 / 10  ·  100%
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🎓

Chapter Complete!

All 10 solutions for Matrices covered.

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ACADEMIA AETERNUM तमसो मा ज्योतिर्गमय · Est. 2025
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NCERT Class 12 Maths Matrices Exercise 3.1 Solutions
NCERT Class 12 Maths Matrices Exercise 3.1 Solutions — Complete Notes & Solutions · academia-aeternum.com
Master NCERT Class 12 Mathematics Chapter 3 Matrices Exercise 3.1 with detailed, step-by-step solutions designed for CBSE Board and competitive entrance exam preparation. This exercise introduces the fundamental concepts of matrices, including order of a matrix, number of elements, identification of matrix elements, construction of matrices using given rules, possible orders based on the number of elements, equality of matrices, and conditions for square matrices. Each solution is explained…
🎓 Class 12 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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    Matrices — Learning Resources

    📄 Detailed Notes
    🧠 Practice MCQs
    ✔️ True / False

    Frequently Asked Questions

    The order of a matrix is written as m × n, where m is the number of rows and n is the number of columns.

    A 3 × 4 matrix has 3 × 4 = 12 elements.

    A matrix is called a square matrix when its number of rows is equal to its number of columns, that is, m = n.

    If a matrix has N elements, its possible orders are obtained from the factor pairs of N, including both orders such as m × n and n × m.

    A 3 × 3 matrix has 9 entries, and each entry has 2 choices. Therefore, the number of matrices is 2? = 512.

    Two matrices are equal if they have the same order and all their corresponding elements are equal.

    The element in the ith row and jth column is represented by a??.

    Substitute i = 1, 2, ... and j = 1, 2, ... according to the required order, calculate each element, and arrange the results in matrix form.

    For a matrix of order m × n, if every entry can independently take k possible values, the total number of matrices is k^(mn).

    Exercise 3.1 builds the fundamentals of matrices, including order, elements, matrix construction, equality, and counting, which are essential for CBSE Board exams and competitive entrance examinations.

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