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Chapter 3 Exercise 3.2 Solutions

Matrices

Step-by-Step Solutions to the NCERT Class 12 Mathematics Chapter 3 Exercise 3.2

Class 12 Mathematics Exercise 3.2 NCERT Solutions Matrices Class 12 Mathematics Chapter 3 CBSE Board Exam JEE Main CUET Types of Matrices Matrix Operations Addition of Matrices Subtraction of Matrices Scalar Multiplication Matrix Multiplication Matrix Equations Properties of Matrices
22 Questions
50–70 min Ideal time
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Q1
NUMERIC3 marks
Let \[A=\begin{bmatrix}2 & 4\\3 & 2\end{bmatrix},\quad B=\begin{bmatrix}1 & 3\\-2 & 5\end{bmatrix},\quad C=\begin{bmatrix}-2 & 5\\3 & 4\end{bmatrix}\]

Find:

  1. \(A+B\)
  2. \(A-B\)
  3. \(3A-C\)
  4. \(AB\)
  5. \(BA\)
📘 Concept & Theory
Concept/Theory
›

This question tests the fundamental operations on matrices: matrix addition, matrix subtraction, scalar multiplication and matrix multiplication. These operations follow specific rules, so the order and dimensions of the matrices must be checked before performing any calculation.

1. Addition of Matrices

Two matrices can be added only when they have the same order. The corresponding elements are added.

\[ A+B=\left[a_{ij}+b_{ij}\right] \]
2. Subtraction of Matrices

Two matrices can be subtracted only when they have the same order. The corresponding elements are subtracted.

\[ A-B=\left[a_{ij}-b_{ij}\right] \]
3. Scalar Multiplication

When a matrix is multiplied by a scalar, every element of the matrix is multiplied by that scalar.

\[ kA=\left[ka_{ij}\right] \]
4. Matrix Multiplication

If \(A\) is of order \(m\times n\) and \(B\) is of order \(n\times p\), then the product \(AB\) is defined and has order \(m\times p\).

\[ (AB)_{ij}=\sum_{k=1}^{n}a_{ik}b_{kj} \]

In matrix multiplication, each element is obtained by multiplying the corresponding elements of a row of the first matrix by a column of the second matrix and then adding the products.

A crucial property is that matrix multiplication is generally not commutative.

\[ AB\neq BA \]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Verify that \(A\), \(B\) and \(C\) have compatible orders.

  2. For \(A+B\), add corresponding elements.

  3. For \(A-B\), subtract corresponding elements carefully, particularly when negative entries occur.

  4. For \(3A-C\), first multiply every element of \(A\) by \(3\), then subtract \(C\).

  5. For \(AB\), multiply each row of \(A\) by each column of \(B\).

  6. For \(BA\), reverse the order and multiply each row of \(B\) by each column of \(A\).

  7. Compare \(AB\) and \(BA\) to observe whether the matrices commute.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  18 steps
  1. Given — Matrices
    \[A=\begin{bmatrix}2 & 4\\3 & 2\end{bmatrix},\quad B=\begin{bmatrix}1 & 3\\-2 & 5\end{bmatrix},\quad C=\begin{bmatrix}-2 & 5\\3 & 4\end{bmatrix}\]
  2. All three matrices are of order \(2\times2\). Therefore, addition, subtraction and scalar multiplication are defined. Also, both products \(AB\) and \(BA\) are defined and will each be of order \(2\times2\).
  3. (i) Find \(A+B\)
  4. We add the corresponding elements of \(A\) and \(B\).
    \[\begin{aligned}A+B&=\begin{bmatrix}2 & 4\\3 & 2\end{bmatrix}+ \begin{bmatrix}1 & 3\\-2 & 5\end{bmatrix}\\[8pt] &=\begin{bmatrix}2+1 & 4+3\\3+(-2) & 2+5\end{bmatrix}\\[8pt] &=\begin{bmatrix}3 & 7\\1 & 7\end{bmatrix}\end{aligned}\]
  5. (ii) Find \(A-B\)
  6. To subtract \(B\) from \(A\), subtract each corresponding element of \(B\) from the corresponding element of \(A\).
    \[\begin{aligned}A-B&=\begin{bmatrix}2 & 4\\3 & 2\end{bmatrix}-\begin{bmatrix}1 & 3\\-2 & 5\end{bmatrix}\\[8pt] &=\begin{bmatrix}2-1 & 4-3\\3-(-2) & 2-5\end{bmatrix}\\[8pt] &=\begin{bmatrix}1 & 1\\5 & -3\end{bmatrix}\end{aligned}\]
  7. (iii) Find \(3A-C\)
  8. First multiply every element of \(A\) by \(3\).
    \[\begin{aligned}3A&=3\begin{bmatrix}2 & 4\\3 & 2\end{bmatrix}\\[8pt] &=\begin{bmatrix}3(2) & 3(4)\\3(3) & 3(2)\end{bmatrix}\\[8pt] &=\begin{bmatrix}6 & 12\\9 & 6\end{bmatrix}\end{aligned}\]
  9. Now subtract \(C\)
    \[\begin{aligned}3A-C&=\begin{bmatrix}6 & 12\\9 & 6\end{bmatrix}-\begin{bmatrix}-2 & 5\\3 & 4\end{bmatrix}\\[8pt] &=\begin{bmatrix}6-(-2) & 12-5\\9-3 & 6-4\end{bmatrix}\\[8pt] &=\begin{bmatrix}8 & 7\\6 & 2\end{bmatrix}\end{aligned}\]
  10. (iv) Find \(AB\)
  11. Since both \(A\) and \(B\) are \(2\times2\) matrices, the product \(AB\) is defined and will be a \(2\times2\) matrix
  12. \[\begin{aligned}AB&=\begin{bmatrix}2 & 4\\3 & 2\end{bmatrix}\begin{bmatrix}1 & 3\\-2 & 5\end{bmatrix}\end{aligned}\]
  13. The \((1,1)\)-entry is obtained by multiplying the first row of \(A\) by the first column of \(B\):
    \[(AB)_{11}=2(1)+4(-2)=2-8=-6\]
  14. The \((1,2)\)-entry is obtained by multiplying the first row of \(A\) by the second column of \(B\):
    \[(AB)_{12}=2(3)+4(5)=6+20=26\]
  15. The \((2,1)\)-entry is obtained by multiplying the second row of \(A\) by the first column of \(B\):
    \[(AB)_{21}=3(1)+2(-2)=3-4=-1\]
  16. The \((2,2)\)-entry is obtained by multiplying the second row of \(A\) by the second column of \(B\):
    \[(AB)_{22}=3(3)+2(5)=9+10=19\]
  17. Therefore,
    \[\begin{aligned}AB&=\begin{bmatrix}-6 & 26\\-1 & 19\end{bmatrix}\end{aligned}\]
  18. (v) Find \(BA\)
  19. Matrix multiplication is order-sensitive, so \(BA\) must be calculated separately. We cannot assume that \(BA=AB\).
  20. \[\begin{aligned}BA&=\begin{bmatrix}1 & 3\\-2 & 5\end{bmatrix}\begin{bmatrix}2 & 4\\3 & 2\end{bmatrix}\end{aligned}\]
  21. The \((1,1)\)-entry is obtained by multiplying the first row of \(B\) by the first column of \(A\):
    \[(BA)_{11}=1(2)+3(3)=2+9=11\]
  22. The \((1,2)\)-entry is obtained by multiplying the first row of \(B\) by the second column of \(A\):
    \[(BA)_{12}=1(4)+3(2)=4+6=10\]
  23. The \((2,1)\)-entry is obtained by multiplying the second row of \(B\) by the first column of \(A\):
    \[(BA)_{21}=(-2)(2)+5(3)=-4+15=11\]
  24. The \((2,2)\)-entry is obtained by multiplying the second row of \(B\) by the second column of \(A\):
    \[(BA)_{22}=(-2)(4)+5(2)=-8+10=2\]
  25. Therefore,
    \[\begin{aligned}BA&=\begin{bmatrix}11 & 10\\11 & 2\end{bmatrix}\end{aligned}\]
  26. Comparison of \(AB\) and \(BA\)
  27. \[AB=\begin{bmatrix}-6 & 26\\-1 & 19\end{bmatrix}\]
    \[BA=\begin{bmatrix}11 & 10\\11 & 2\end{bmatrix}\]
  28. Clearly,
    \[\boxed{AB\neq BA}\]
  29. Thus, this example demonstrates that matrix multiplication is generally not commutative.
🎯 Exam Significance
Exam Significance
›

This problem covers the basic matrix operations that form the foundation of the Class 12 Mathematics chapter on Matrices. Questions involving addition, subtraction, scalar multiplication and matrix multiplication are frequently used as direct-mark questions as well as intermediate steps in longer problems.

  • Always check the order of matrices before performing an operation.
  • For addition and subtraction, corresponding elements are operated upon.
  • For scalar multiplication, the scalar multiplies every element.
  • For matrix multiplication, use the row-by-column rule.
  • Do not interchange \(AB\) and \(BA\).
  • Write intermediate calculations clearly to avoid sign and multiplication errors.
  • Negative entries require particular attention during subtraction and multiplication.
Significance for Competitive Entrance Examinations

Matrix operations are fundamental for competitive examinations such as JEE and other engineering entrance examinations. Problems may test not only direct multiplication but also order conditions, matrix identities, non-commutativity and algebraic manipulation involving several matrices.

A common competitive-exam trap is to treat matrices like ordinary numbers. In particular,

\[ AB\neq BA \]

in general. Therefore, the order of multiplication must always be preserved. Careful row-by-column multiplication is essential for avoiding calculation errors in time-bound examinations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  11 points

  1. Two matrices can be added or subtracted only when they have the same order.

  2. Matrix addition is performed element by element.

  3. Matrix subtraction is also performed element by element.

  4. A scalar multiplies every element of a matrix.

  5. If \(A\) is \(m\times n\) and \(B\) is \(n\times p\), then \(AB\) exists and is of order \(m\times p\).

  6. Every entry of a matrix product is obtained using the row-by-column rule.

  7. Matrix multiplication is generally not commutative.

  8. For the given matrices,

    \[ AB= \begin{bmatrix} -6 & 26\\ -1 & 19 \end{bmatrix} \]

  9. For the given matrices,

    \[ BA= \begin{bmatrix} 11 & 10\\ 11 & 2 \end{bmatrix} \]

  10. Hence,

    \[ AB\neq BA. \]

  11. Sign errors and incorrect row-column pairing are among the most common mistakes in matrix multiplication.
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1 / 22  ·  5%
Q2 →
Q2
NUMERIC3 marks
Compute the following:
  1. \[\begin{bmatrix}a & b\\-b & a\end{bmatrix}+\begin{bmatrix}a & b\\b & a\end{bmatrix}\]
  2. \[\begin{bmatrix}a^{2}+b^{2} & b^{2}+c^{2}\\a^{2}+c^{2} & a^{2}+b^{2}\end{bmatrix}+\begin{bmatrix}2ab & 2bc\\-2ac & -2ba\end{bmatrix}\]
  3. \[\begin{bmatrix}-1 & 4 & -6\\8 & 5 & 16\\2 & 8 & 5\end{bmatrix}+\begin{bmatrix}12 & 7 & 6\\8 & 0 & 5\\3 & 2 & 4\end{bmatrix}\]
  4. \[\begin{bmatrix}\cos^{2}x & \sin^{2}x\\\sin^{2}x & \cos^{2}x\end{bmatrix}+\begin{bmatrix}\sin^{2}x & \cos^{2}x\\\cos^{2}x & \sin^{2}x\end{bmatrix}\]
📘 Concept & Theory
Concept/Theory
›

Matrix addition is performed by adding corresponding entries of the two matrices. Therefore, two matrices can be added only when they have the same order.

If

\[A=[a_{ij}]\quad\text{and}\quad B=[b_{ij}]\]
are matrices of the same order, then
\[A+B=[a_{ij}+b_{ij}]\]

Thus, the entry in the \(i\)-th row and \(j\)-th column of \(A+B\) is obtained by adding the entries in the \(i\)-th row and \(j\)-th column of \(A\) and \(B\), respectively.

In algebraic matrix problems, simplification of individual entries may require standard algebraic identities such as

\[a^{2}+2ab+b^{2}=(a+b)^{2},\]
\[a^{2}-2ab+b^{2}=(a-b)^{2},\]
and
\[\cos^{2}x+\sin^{2}x=1.\]

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Check that the two matrices in each part have the same order.

  2. Add the corresponding entries position by position.

  3. Retain algebraic expressions in their expanded form initially so that no term is missed.

  4. For part (ii), simplify the resulting expressions using standard identities involving squares.

  5. For part (iv), use the fundamental trigonometric identity

    \[\sin^{2}x+\cos^{2}x=1\]

  6. Present the final matrix in simplified form.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  15 steps
  1. (i) Compute
    \[\begin{bmatrix}a & b\\-b & a\end{bmatrix}+\begin{bmatrix}a & b\\b & a\end{bmatrix}\]
  2. Both matrices are of order \(2\times2\), so their sum is defined. We add corresponding entries.
    \[\begin{aligned}&\begin{bmatrix}a & b\\-b & a\end{bmatrix}+\begin{bmatrix}a & b\\b & a\end{bmatrix}\\[8pt] &=\begin{bmatrix}a+a & b+b\\-b+b & a+a\end{bmatrix}\end{aligned}\]
  3. Now simplify each entry:
    \[a+a=2a\]
    \[b+b=2b\]
    \[-b+b=0\]
    \[a+a=2a\]
  4. Therefore,
    \[\begin{aligned}&\begin{bmatrix}a & b\\-b & a\end{bmatrix}+\begin{bmatrix}a & b\\b & a\end{bmatrix}\\[8pt] &=\begin{bmatrix}2a & 2b\\0 & 2a\end{bmatrix}\end{aligned}\]
  5. Taking \(2\) as a common factor from every entry,
    \[\begin{bmatrix}2a & 2b\\0 & 2a\end{bmatrix}=2\begin{bmatrix}a & b\\0 & a\end{bmatrix}\]
  6. (ii) Compute
    \[\begin{bmatrix}a^{2}+b^{2} & b^{2}+c^{2}\\a^{2}+c^{2} & a^{2}+b^{2}\end{bmatrix}+\begin{bmatrix}2ab & 2bc\\-2ac & -2ba\end{bmatrix}\]
  7. Both matrices are of order \(2\times2\). Hence, matrix addition is defined. Add corresponding entries.
  8. \[\begin{aligned}&\begin{bmatrix}a^{2}+b^{2} & b^{2}+c^{2}\\a^{2}+c^{2} & a^{2}+b^{2}\end{bmatrix}+\begin{bmatrix}2ab & 2bc\\-2ac & -2ba\end{bmatrix}\\[8pt] &=\begin{bmatrix}a^{2}+b^{2}+2ab & b^{2}+c^{2}+2bc\\a^{2}+c^{2}-2ac & a^{2}+b^{2}-2ba\end{bmatrix}\end{aligned}\]
  9. Since multiplication of algebraic quantities is commutative,
    \[ba=ab\]
    Therefore,
    \[-2ba=-2ab\]
  10. Hence,
    \[\begin{aligned}&=\begin{bmatrix}a^{2}+b^{2}+2ab & b^{2}+c^{2}+2bc\\a^{2}+c^{2}-2ac & a^{2}+b^{2}-2ab\end{bmatrix}\end{aligned}\]
  11. Now simplify each entry using the identities
    \[a^{2}+2ab+b^{2}=(a+b)^{2},\]
    \[b^{2}+2bc+c^{2}=(b+c)^{2},\]
    \[a^{2}-2ac+c^{2}=(a-c)^{2},\]
    and
    \[a^{2}-2ab+b^{2}=(a-b)^{2}.\]
  12. \[\begin{aligned}&\begin{bmatrix}a^{2}+b^{2}+2ab & b^{2}+c^{2}+2bc\\a^{2}+c^{2}-2ac & a^{2}+b^{2}-2ab\end{bmatrix}\\[8pt] &=\begin{bmatrix}(a+b)^{2} & (b+c)^{2}\\(a-c)^{2} & (a-b)^{2}\end{bmatrix}\end{aligned}\]
  13. (iii) Compute
  14. We are given two matrices of order \(3\times3\), so their sum is defined.
  15. \[\begin{bmatrix}-1 & 4 & -6\\8 & 5 & 16\\2 & 8 & 5\end{bmatrix}+\begin{bmatrix}12 & 7 & 6\\8 & 0 & 5\\3 & 2 & 4\end{bmatrix}\]
  16. Add corresponding entries:
    \[\begin{aligned}&\begin{bmatrix}-1 & 4 & -6\\8 & 5 & 16\\2 & 8 & 5\end{bmatrix}+\begin{bmatrix}12 & 7 & 6\\8 & 0 & 5\\3 & 2 & 4\end{bmatrix}\\[8pt] &=\begin{bmatrix}-1+12 & 4+7 & -6+6\\8+8 & 5+0 & 16+5\\2+3 & 8+2 & 5+4\end{bmatrix}\end{aligned}\]
  17. Now simplify each entry:
    \[-1+12=11\]
    \[4+7=11\]
    \[-6+6=0\]
    \[8+8=16\]
    \[5+0=5\]
    \[16+5=21\]
    \[2+3=5\]
    \[8+2=10\]
    \[5+4=9\]
  18. Therefore,
    \[\begin{bmatrix}11 & 11 & 0\\16 & 5 & 21\\5 & 10 & 9\end{bmatrix}\]
  19. (iv) Compute
    \[\begin{bmatrix}\cos^{2}x & \sin^{2}x\\\sin^{2}x & \cos^{2}x\end{bmatrix}+\begin{bmatrix}\sin^{2}x & \cos^{2}x\\\cos^{2}x & \sin^{2}x\end{bmatrix}\]
  20. Both matrices are of order \(2\times2\). Therefore, we add their corresponding entries.
  21. \[\begin{aligned}&\begin{bmatrix}\cos^{2}x & \sin^{2}x\\\sin^{2}x & \cos^{2}x\end{bmatrix}+\begin{bmatrix}\sin^{2}x & \cos^{2}x\\\cos^{2}x & \sin^{2}x\end{bmatrix}\\[8pt] &=\begin{bmatrix}\cos^{2}x+\sin^{2}x &\sin^{2}x+\cos^{2}x\\\sin^{2}x+\cos^{2}x &\cos^{2}x+\sin^{2}x\end{bmatrix}\end{aligned}\]
  22. Using the fundamental trigonometric identity
    \[\sin^{2}x+\cos^{2}x=1,\]
    each entry becomes \(1\)
    \[\begin{aligned}&=\begin{bmatrix}1 & 1\\1 & 1\end{bmatrix}\end{aligned}\]
🎯 Exam Significance
Exam Significance
›

This question reinforces the basic operation of matrix addition and also tests whether the student can simplify algebraic and trigonometric expressions correctly. Such questions are useful for securing marks in direct computational questions and for building accuracy before attempting more advanced matrix identities.

  • Always check that the matrices have the same order before adding them.
  • Add only corresponding entries.
  • Do not change the position of any matrix element while copying the matrices.
  • Use algebraic identities only after the corresponding entries have been added correctly.
  • For trigonometric entries, remember
    \[ \sin^{2}x+\cos^{2}x=1. \]
  • Negative signs must be retained carefully while adding numerical matrices.
Significance for Competitive Entrance Examinations

Competitive examinations frequently combine elementary matrix operations with algebraic identities, trigonometric identities and simplification. The main challenge is often not the matrix addition itself but identifying the correct algebraic structure quickly and avoiding transcription errors.

Part (ii), for example, tests recognition of perfect-square identities, while part (iv) tests immediate application of the Pythagorean identity. Such pattern recognition reduces calculation time in objective examinations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. Matrix addition is defined only for matrices of the same order.

  2. Corresponding entries are added directly.

  3. Algebraic identities can significantly simplify the resulting entries.

  4. The identities

    \[ (a+b)^{2}=a^{2}+2ab+b^{2} \]
    and
    \[ (a-b)^{2}=a^{2}-2ab+b^{2} \]
    are particularly useful.

  5. The trigonometric identity

    \[ \sin^{2}x+\cos^{2}x=1 \]
    simplifies every entry in part (iv).

  6. Careful transcription of the original matrices is essential.

  7. The correct answer to part (iii) is

    \[ \begin{bmatrix} 11 & 11 & 0\\ 16 & 5 & 21\\ 5 & 10 & 9 \end{bmatrix}. \]

  8. The correct answer to part (ii) is

    \[ \begin{bmatrix} (a+b)^{2} & (b+c)^{2}\\ (a-c)^{2} & (a-b)^{2} \end{bmatrix}. \]
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2 / 22  ·  9%
Q3 →
Q3
NUMERIC3 marks
Compute as indicated products
  1. \(\quad\begin{bmatrix} a & b \\ -b & a \end{bmatrix}\begin{bmatrix} a & -b \\ b & a \end{bmatrix}\)

  2. \(\quad\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\times \begin{bmatrix} 2 & 3 & 4 \end{bmatrix}\)

  3. \(\quad\begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix}\)

  4. \(\quad\begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix}\begin{bmatrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{bmatrix}\)

  5. \(\quad\begin{bmatrix} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1 \\ -1 & 2 & 1 \end{bmatrix}\)

  6. \(\quad\begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{bmatrix}\)
📘 Concept & Theory
Concept/Theory
›

Matrix multiplication is defined when the number of columns of the first matrix is equal to the number of rows of the second matrix. If A is of order \(m\times n\) and B is of order \(n\times p\), then the product \(AB\) is defined and is of order \(m\times p\).

Each element of the product matrix is obtained by multiplying the corresponding entries of a row of the first matrix by the corresponding entries of a column of the second matrix and then adding the products. Thus,

\[ (AB)_{ij}=\sum_{k=1}^{n}a_{ik}b_{kj}. \]

A particularly important point is that matrix multiplication is generally not commutative; in general,

\[ AB\neq BA. \]
Therefore, the order of the matrices must be preserved throughout the calculation.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Check the orders of the two matrices to ensure that the product is defined.

  2. Determine the order of the resulting matrix.

  3. Multiply each row of the first matrix by each column of the second matrix.

  4. Write every element of the product matrix explicitly.

  5. Simplify the resulting entries carefully.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  14 steps
  1. (i)\(\quad\begin{bmatrix} a & b \\ -b & a \end{bmatrix}\begin{bmatrix} a & -b \\ b & a \end{bmatrix}\)
  2. Both matrices are of order \(2\times2\). Therefore, their product is defined and the resulting matrix will also be of order \(2\times2\)
  3. Using row-by-column multiplication,
    \[\begin{aligned}&\begin{bmatrix}a & b\\-b & a\end{bmatrix}\begin{bmatrix}a & -b\\b & a\end{bmatrix}\\ &=\begin{bmatrix}a(a)+b(b) & a(-b)+b(a)\\(-b)(a)+a(b) & (-b)(-b)+a(a)\end{bmatrix}\end{aligned}\]
  4. Now simplify each entry:
    \[\begin{aligned}a_{11}&=a^2+b^2,\\a_{12}&=-ab+ab=0,\\a_{21}&=-ab+ab=0,\\a_{22}&=b^2+a^2=a^2+b^2\end{aligned}\]
  5. Hence,
    \[\begin{bmatrix}a & b\\-b & a\end{bmatrix}\begin{bmatrix}a & -b\\b & a\end{bmatrix}=\begin{bmatrix}a^2+b^2 & 0\\0 & a^2+b^2\end{bmatrix}\]
  6. or equivalently,
    \[(a^2+b^2)I_2\]
  7. (ii)\(\quad\begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}\times \begin{bmatrix} 2 & 3 & 4 \end{bmatrix}\)
  8. The first matrix is of order \(3\times1\), while the second matrix is of order \(1\times3\). Since the inner dimensions are equal, the product is defined.
    \[(3\times1)(1\times3)=3\times3\]
    Therefore, the answer must be a \(3\times3\) matrix.
  9. \[\begin{aligned}&\begin{bmatrix}1\\2\\3\end{bmatrix}\begin{bmatrix}2 & 3 & 4\end{bmatrix}\\[8pt] &=\begin{bmatrix}1(2) & 1(3) & 1(4)\\2(2) & 2(3) & 2(4)\\3(2) & 3(3) & 3(4)\end{bmatrix}\\[8pt] &=\begin{bmatrix}2 & 3 & 4\\4 & 6 & 8\\6 & 9 & 12\end{bmatrix}\end{aligned}\]
  10. Hence,
    \[\begin{bmatrix}2 & 3 & 4\\4 & 6 & 8\\6 & 9 & 12\end{bmatrix}\]
  11. (iii)\(\quad\begin{bmatrix} 1 & -2 \\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 & 2 & 3 \\ 2 & 3 & 1 \end{bmatrix}\)
  12. The first matrix is \(2\times2\) and the second is \(2\times3\). Hence,
    \[(2\times2)(2\times3)=2\times3\]
    Thus, the product will be a \(2\times3\) matrix.
  13. \[\begin{aligned}&\begin{bmatrix}1 & -2\\2 & 3\end{bmatrix}\begin{bmatrix}1 & 2 & 3\\2 & 3 & 1\end{bmatrix}\\[8pt] &=\begin{bmatrix}1(1)+(-2)(2) & 1(2)+(-2)(3) & 1(3)+(-2)(1)\\2(1)+3(2) & 2(2)+3(3) & 2(3)+3(1)\end{bmatrix}\\[8pt] &=\begin{bmatrix}1-4 & 2-6 & 3-2\\2+6 & 4+9 & 6+3\end{bmatrix}\\[8pt] &=\boxed{\begin{bmatrix}-3 & -4 & 1\\8 & 13 & 9\end{bmatrix}}\end{aligned}\]
  14. (iv)\(\quad\begin{bmatrix} 2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6 \end{bmatrix}\begin{bmatrix} 1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5 \end{bmatrix}\)
  15. Both matrices are of order \(3\times3\). Hence, the product is defined and the result is of order \(3\times3\).
  16. Let
    \[AB=\begin{bmatrix}a_{11} & a_{12} & a_{13}\\a_{21} & a_{22} & a_{23}\\a_{31} & a_{32} & a_{33}\end{bmatrix}\]
  17. First row:
  18. \[\begin{aligned}a_{11}&=\begin{bmatrix}2&3&4\end{bmatrix}\begin{bmatrix}1\\0\\3\end{bmatrix}\\&=2(1)+3(0)+4(3)\\&=2+0+12\\&=14\end{aligned}\]
    \[\begin{aligned}a_{12}&=\begin{bmatrix}2&3&4\end{bmatrix}\begin{bmatrix}-3\\2\\0\end{bmatrix}\\&=2(-3)+3(2)+4(0)\\&=-6+6+0\\&=0\end{aligned}\]
    \[\begin{aligned}a_{13}&=\begin{bmatrix}2&3&4\end{bmatrix}\begin{bmatrix}5\\4\\5\end{bmatrix}\\&=2(5)+3(4)+4(5)\\&=10+12+20\\&=42\end{aligned}\]
  19. Second row:
  20. \[\begin{aligned}a_{21}&=\begin{bmatrix}3&4&5\end{bmatrix}\begin{bmatrix}1\\0\\3\end{bmatrix}\\&=3(1)+4(0)+5(3)\\&=3+0+15\\&=18\end{aligned}\]
    \[\begin{aligned}a_{22}&=\begin{bmatrix}3&4&5\end{bmatrix}\begin{bmatrix}-3\\2\\0\end{bmatrix}\\&=3(-3)+4(2)+5(0)\\&=-9+8+0\\&=-1\end{aligned}\]
    \[\begin{aligned}a_{23}&=\begin{bmatrix}3&4&5\end{bmatrix}\begin{bmatrix}5\\4\\5\end{bmatrix}\\&=3(5)+4(4)+5(5)\\&=15+16+25\\&=56\end{aligned}\]
  21. Third row:
  22. \[\begin{aligned}a_{31}&=\begin{bmatrix}4&5&6\end{bmatrix}\begin{bmatrix}1\\0\\3\end{bmatrix}\\&=4(1)+5(0)+6(3)\\&=4+0+18\\&=22\end{aligned}\]
    \[\begin{aligned}a_{32}&=\begin{bmatrix}4&5&6\end{bmatrix}\begin{bmatrix}-3\\2\\0\end{bmatrix}\\&=4(-3)+5(2)+6(0)\\&=-12+10+0\\&=-2\end{aligned}\]
    \[\begin{aligned}a_{33}&=\begin{bmatrix}4&5&6\end{bmatrix}\begin{bmatrix}5\\4\\5\end{bmatrix}\\&=4(5)+5(4)+6(5)\\&=20+20+30\\&=70\end{aligned}\]
  23. Therefore,
    \[\begin{bmatrix}2&3&4\\3&4&5\\4&5&6\end{bmatrix}\begin{bmatrix}1&-3&5\\0&2&4\\3&0&5\end{bmatrix}=\begin{bmatrix}14&0&42\\18&-1&56\\22&-2&70\end{bmatrix}\]
  24. (v)\(\quad\begin{bmatrix} 2 & 1 \\ 3 & 2 \\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1 \\ -1 & 2 & 1 \end{bmatrix}\)
  25. The first matrix is \(3\times2\) and the second matrix is \(2\times3\). Therefore,
    \[(3\times2)(2\times3)=3\times3\]
  26. \[\begin{aligned}&\begin{bmatrix}2&1\\3&2\\-1&1\end{bmatrix}\begin{bmatrix}1&0&1\\-1&2&1\end{bmatrix}\\[8pt] &=\begin{bmatrix}2(1)+1(-1) & 2(0)+1(2) & 2(1)+1(1)\\3(1)+2(-1) & 3(0)+2(2) & 3(1)+2(1)\\(-1)(1)+1(-1) & (-1)(0)+1(2) & (-1)(1)+1(1)\end{bmatrix}\\[8pt] &=\begin{bmatrix}2-1&2&2+1\\3-2&4&3+2\\-1-1&2&-1+1\end{bmatrix}\\[8pt] &=\begin{bmatrix}1&2&3\\1&4&5\\-2&2&0\end{bmatrix}\end{aligned}\]
  27. (vi)\(\quad\begin{bmatrix} 3 & -1 & 3 \\ -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3 \\ 1 & 0 \\ 3 & 1 \end{bmatrix}\)
  28. The first matrix is \(2\times3\), while the second matrix is \(3\times2\). Therefore,
    \[(2\times3)(3\times2)=2\times2\]
  29. \[\begin{aligned}&\begin{bmatrix}3&-1&3\\-1&0&2\end{bmatrix}\begin{bmatrix}2&-3\\1&0\\3&1\end{bmatrix}\\[8pt] &=\begin{bmatrix}3(2)+(-1)(1)+3(3)&3(-3)+(-1)(0)+3(1)\\(-1)(2)+0(1)+2(3)&(-1)(-3)+0(0)+2(1)\end{bmatrix}\\[8pt] &=\begin{bmatrix}6-1+9&-9+0+3\\-2+0+6&3+0+2\end{bmatrix}\\[8pt] &=\begin{bmatrix}14&-6\\4&5\end{bmatrix}\end{aligned}\]
🎯 Exam Significance
Exam Significance
›
This question is fundamental for mastering matrix multiplication, which is frequently tested in Class 12 Mathematics board examinations. It develops accuracy in row-column multiplication, order checking and algebraic simplification. The same skills are directly useful in competitive examinations such as JEE and other entrance tests, where matrix products, matrix equations, transformations and higher-order matrix expressions often require rapid and error-free calculations.
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. Matrix multiplication is possible only when the number of columns of the first matrix equals the number of rows of the second matrix.

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

  3. Every entry of a matrix product is obtained by the row-by-column rule.

  4. A \(3\times1\) matrix multiplied by a \(1\times3\) matrix produces a \(3\times3\) matrix.

  5. Always preserve the signs of entries, particularly negative elements, during multiplication.

  6. Matrix multiplication is generally not commutative, so the order of multiplication matters.

  7. Before performing lengthy calculations, determine the order of the product. This is an effective error-checking technique in examinations.

  8. For competitive examinations, careful order checking and systematic row-column multiplication can significantly reduce calculation errors.
← Q2
3 / 22  ·  14%
Q4 →
Q4
NUMERIC3 marks
If \[A=\begin{bmatrix}1 & 2 & -3\\5 & 0 & 2\\1 & -1 & 1\end{bmatrix},\quad B=\begin{bmatrix}3 & -1 & 2\\4 & 2 & 5\\2 & 0 & 3\end{bmatrix},\quad C=\begin{bmatrix}4 & 1 & 2\\0 & 3 & 2\\1 & -2 & 3\end{bmatrix}\] then compute \(A+B\) and \(B-C\). Also verify that \[A+(B-C)=(A+B)-C.\]
📘 Concept & Theory
Concept/Theory
›

This question demonstrates the distributive property of subtraction over matrix addition, expressed as

\[ A+(B-C)=(A+B)-C. \]

Matrix addition and subtraction are performed element by element. Therefore, if \(A\), \(B\) and \(C\) are matrices of the same order, then

\[ A+(B-C) \]
and
\[ (A+B)-C \]
are both defined and must produce the same matrix.

The identity can also be understood algebraically:

\[ A+(B-C)=A+B-C \]
and
\[ (A+B)-C=A+B-C. \]
Hence,
\[ \boxed{A+(B-C)=(A+B)-C}. \]

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Observe that \(A\), \(B\) and \(C\) are all \(3\times3\) matrices.

  2. Compute \(A+B\) by adding corresponding entries.

  3. Compute \(B-C\) by subtracting corresponding entries.

  4. Use the calculated value of \(B-C\) to determine \(A+(B-C)\).

  5. Use the calculated value of \(A+B\) to determine \((A+B)-C\).

  6. Compare the two resulting matrices entry by entry.

  7. If the two matrices are identical, the required identity is verified.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  17 steps
  1. Given — Matrices
    \[A=\begin{bmatrix}1 & 2 & -3\\5 & 0 & 2\\1 & -1 & 1\end{bmatrix}\]
    \[B=\begin{bmatrix}3 & -1 & 2\\4 & 2 & 5\\2 & 0 & 3\end{bmatrix}\]
    \[C=\begin{bmatrix}4 & 1 & 2\\0 & 3 & 2\\1 & -2 & 3\end{bmatrix}\]
  2. Each matrix is of order \(3\times3\). Therefore, all the required additions and subtractions are defined.
  3. Step 1: Compute \(A+B\)
  4. Add the corresponding entries of \(A\) and \(B\).
    \[\begin{aligned}A+B&=\begin{bmatrix}1 & 2 & -3\\5 & 0 & 2\\1 & -1 & 1\end{bmatrix}+\begin{bmatrix}3 & -1 & 2\\4 & 2 & 5\\2 & 0 & 3\end{bmatrix}\\[8pt] &=\begin{bmatrix}1+3 & 2+(-1) & -3+2\\5+4 & 0+2 & 2+5\\1+2 & -1+0 & 1+3\end{bmatrix}\end{aligned}\]
  5. Now simplify each entry:
    \[1+3=4\]
    \[2+(-1)=1\]
    \[-3+2=-1\]
    \[5+4=9\]
    \[0+2=2\]
    \[2+5=7\]
    \[1+2=3\]
    \[-1+0=-1\]
    \[1+3=4\]
  6. Therefore,
    \[A+B=\begin{bmatrix}4 & 1 & -1\\9 & 2 & 7\\3 & -1 & 4\end{bmatrix}\]
  7. Step 2: Compute \(B-C\)
  8. Subtract the corresponding entries of \(C\) from \(B\).
    \[\begin{aligned}B-C&=\begin{bmatrix}3 & -1 & 2\\4 & 2 & 5\\2 & 0 & 3\end{bmatrix}-\begin{bmatrix}4 & 1 & 2\\0 & 3 & 2\\1 & -2 & 3\end{bmatrix}\\[8pt] &=\begin{bmatrix}3-4 & -1-1 & 2-2\\4-0 & 2-3 & 5-2\\2-1 & 0-(-2) & 3-3\end{bmatrix}\end{aligned}\]
  9. Now simplify the entries one by one:
    \[3-4=-1\]
    \[-1-1=-2\]
    \[2-2=0\]
    \[4-0=4\]
    \[2-3=-1\]
    \[5-2=3\]
    \[2-1=1\]
    \[0-(-2)=0+2=2\]
    \[3-3=0\]
  10. Therefore,
    \[B-C=\begin{bmatrix}-1 & -2 & 0\\4 & -1 & 3\\1 & 2 & 0\end{bmatrix}\]
  11. Step 3: Compute \(A+(B-C)\)
  12. From the previous step,
    \[B-C=\begin{bmatrix}-1 & -2 & 0\\4 & -1 & 3\\1 & 2 & 0\end{bmatrix}\]
  13. Therefore,
    \[\begin{aligned}A+(B-C)&=\begin{bmatrix}1 & 2 & -3\\5 & 0 & 2\\1 & -1 & 1\end{bmatrix}+\begin{bmatrix}-1 & -2 & 0\\4 & -1 & 3\\1 & 2 & 0\end{bmatrix}\\[8pt] &=\begin{bmatrix}1+(-1) & 2+(-2) & -3+0\\5+4 & 0+(-1) & 2+3\\1+1 & -1+2 & 1+0\end{bmatrix}\end{aligned}\]
  14. Now simplify:
    \[1+(-1)=0\]
    \[2+(-2)=0\]
    \[-3+0=-3\]
    \[5+4=9\]
    \[0+(-1)=-1\]
    \[2+3=5\]
    \[1+1=2\]
    \[-1+2=1\]
    \[1+0=1\]
  15. Hence,
    \[A+(B-C)=\begin{bmatrix}0 & 0 & -3\\9 & -1 & 5\\2 & 1 & 1\end{bmatrix}\]
  16. Step 4: Compute \((A+B)-C\)
  17. From Step 1,
    \[A+B=\begin{bmatrix}4 & 1 & -1\\9 & 2 & 7\\3 & -1 & 4\end{bmatrix}\]
  18. Therefore,
    \[\begin{aligned}(A+B)-C&=\begin{bmatrix}4 & 1 & -1\\9 & 2 & 7\\3 & -1 & 4\end{bmatrix}-\begin{bmatrix}4 & 1 & 2\\0 & 3 & 2\\1 & -2 & 3\end{bmatrix}\\[8pt] &=\begin{bmatrix}4-4 & 1-1 & -1-2\\9-0 & 2-3 & 7-2\\3-1 & -1-(-2) & 4-3\end{bmatrix}\end{aligned}\]
  19. Now simplify each entry:
    \[4-4=0\]
    \[1-1=0\]
    \[-1-2=-3\]
    \[9-0=9\]
    \[2-3=-1\]
    \[7-2=5\]
    \[3-1=2\]
    \[-1-(-2)=-1+2=1\]
    \[4-3=1\]
  20. Therefore,
    \[(A+B)-C=\begin{bmatrix}0 & 0 & -3\\9 & -1 & 5\\2 & 1 & 1\end{bmatrix}\]
  21. Step 5: Verification
  22. We have obtained
    \[A+(B-C)=\begin{bmatrix}0 & 0 & -3\\9 & -1 & 5\\2 & 1 & 1\end{bmatrix}\]

    and

    \[(A+B)-C=\begin{bmatrix}0 & 0 & -3\\9 & -1 & 5\\2 & 1 & 1\end{bmatrix}\]
  23. Since both matrices are identical, we conclude that
    \[\boxed{A+(B-C)=(A+B)-C}\]
  24. Hence, the required identity is verified.
🎯 Exam Significance
Exam Significance
›

This problem is important for understanding the algebraic properties of matrices. It is not merely a numerical exercise; it verifies an important identity involving matrix addition and subtraction.

  • It reinforces element-wise matrix addition and subtraction.
  • It demonstrates how matrix subtraction can be converted into addition of an additive inverse.
  • It provides practice in verifying matrix identities systematically.
  • Showing both sides separately is the safest method when the question explicitly asks for verification.
  • Careful handling of expressions such as \(0-(-2)\) and \(-1-(-2)\) prevents sign errors.

For board examinations, the verification should not be stated merely as "proved". The two sides should preferably be evaluated and shown to be equal. This makes the logical verification complete.

Significance for Competitive Entrance Examination Aspirants

For competitive examinations, the direct calculation is useful for building fluency with matrix operations, while the algebraic verification provides a faster conceptual route.

Instead of calculating every entry when only an identity is required, an aspirant can use

\[ B-C=B+(-C) \]
and the associativity of matrix addition:
\[ A+\left[B+(-C)\right]=(A+B)+(-C). \]

This distinction between a computational approach and a property-based approach is particularly useful in time-constrained objective examinations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. All three matrices in this question are of order \(3\times3\).

  2. Therefore, \(A+B\), \(B-C\), \(A+(B-C)\), and \((A+B)-C\) are defined.

  3. The calculated value is

    \[ A+B= \begin{bmatrix} 4 & 1 & -1\\ 9 & 2 & 7\\ 3 & -1 & 4 \end{bmatrix} \]

  4. The calculated value is

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

  5. Both sides produce

    \[ \begin{bmatrix} 0 & 0 & -3\\ 9 & -1 & 5\\ 2 & 1 & 1 \end{bmatrix} \]

  6. Hence,

    \[ A+(B-C)=(A+B)-C. \]

  7. Matrix subtraction can be expressed as addition of the additive inverse:

    \[ B-C=B+(-C). \]

  8. For verification questions, calculate both sides independently before declaring the identity verified.
← Q3
4 / 22  ·  18%
Q5 →
Q5
NUMERIC3 marks
If \[A=\begin{bmatrix}\dfrac{2}{3} & 1 & \dfrac{5}{3}\\\dfrac{1}{3} & \dfrac{2}{3} & \dfrac{4}{3}\\\dfrac{7}{3} & 2 & \dfrac{2}{3}\end{bmatrix}\]

and

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

then compute \(3A-5B\).

📘 Concept & Theory
Concept/Theory
›

Multiplication of a matrix by a scalar means multiplying every entry of the matrix by that scalar. Therefore, for a scalar \(k\) and a matrix \(A=[a_{ij}]\),

\[ kA=[ka_{ij}]. \]

After scalar multiplication, matrix subtraction is performed element by element. Thus,

\[ 3A-5B \]
can be calculated by first finding \(3A\) and \(5B\), and then subtracting the corresponding entries.

An important observation in this question is that the entries of \(A\) have denominator \(3\), while those of \(B\) have denominator \(5\). Hence, multiplying \(A\) by \(3\) and \(B\) by \(5\) removes the respective denominators. This makes the calculation particularly simple.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Multiply every entry of \(A\) by \(3\) to obtain \(3A\).

  2. Multiply every entry of \(B\) by \(5\) to obtain \(5B\).

  3. Compare the two resulting matrices.

  4. Subtract \(5B\) from \(3A\) entry by entry.

  5. Since the two resulting matrices are identical, their difference will be the zero matrix.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  11 steps
  1. Step 1: Compute \(3A\)
  2. Given —
    \[A=\begin{bmatrix}\dfrac{2}{3} & 1 & \dfrac{5}{3}\\\dfrac{1}{3} & \dfrac{2}{3} & \dfrac{4}{3}\\\dfrac{7}{3} & 2 & \dfrac{2}{3}\end{bmatrix}\]
  3. To calculate \(3A\), multiply every entry of \(A\) by \(3\):
    \[\begin{aligned}3A&=3\begin{bmatrix}\dfrac{2}{3} & 1 & \dfrac{5}{3}\\\dfrac{1}{3} & \dfrac{2}{3} & \dfrac{4}{3}\\\dfrac{7}{3} & 2 & \dfrac{2}{3}\end{bmatrix}\\[8pt] &=\begin{bmatrix}3\left(\dfrac{2}{3}\right) & 3(1) & 3\left(\dfrac{5}{3}\right)\\3\left(\dfrac{1}{3}\right) & 3\left(\dfrac{2}{3}\right) & 3\left(\dfrac{4}{3}\right)\\3\left(\dfrac{7}{3}\right) & 3(2) & 3\left(\dfrac{2}{3}\right)\end{bmatrix}\end{aligned}\]
  4. Now simplify each entry:
    \[3\left(\dfrac{2}{3}\right)=2\]
    \[3(1)=3\]
    \[3\left(\dfrac{5}{3}\right)=5\]
    \[3\left(\dfrac{1}{3}\right)=1\]
    \[3\left(\dfrac{2}{3}\right)=2\]
    \[3\left(\dfrac{4}{3}\right)=4\]
    \[3\left(\dfrac{7}{3}\right)=7\]
    \[3(2)=6\]
    \[3\left(\dfrac{2}{3}\right)=2\]
  5. Therefore,
    \[3A=\begin{bmatrix}2 & 3 & 5\\1 & 2 & 4\\7 & 6 & 2\end{bmatrix}\]
  6. Step 2: Compute \(5B\)
  7. Given —
    \[B=\begin{bmatrix}\dfrac{2}{5} & \dfrac{3}{5} & 1\\\dfrac{1}{5} & \dfrac{2}{5} & \dfrac{4}{5}\\\dfrac{7}{5} & \dfrac{6}{5} & \dfrac{2}{5}\end{bmatrix}\]
  8. To calculate \(5B\), multiply every entry of \(B\) by \(5\):
    \[\begin{aligned}5B&=5\begin{bmatrix}\dfrac{2}{5} & \dfrac{3}{5} & 1\\\dfrac{1}{5} & \dfrac{2}{5} & \dfrac{4}{5}\\\dfrac{7}{5} & \dfrac{6}{5} & \dfrac{2}{5}\end{bmatrix}\\[8pt] &=\begin{bmatrix}5\left(\dfrac{2}{5}\right) & 5\left(\dfrac{3}{5}\right) & 5(1)\\5\left(\dfrac{1}{5}\right) & 5\left(\dfrac{2}{5}\right) & 5\left(\dfrac{4}{5}\right)\\5\left(\dfrac{7}{5}\right) & 5\left(\dfrac{6}{5}\right) & 5\left(\dfrac{2}{5}\right)\end{bmatrix}\end{aligned}\]
  9. Now simplify each entry:
    \[5\left(\dfrac{2}{5}\right)=2\]
    \[5\left(\dfrac{3}{5}\right)=3\]
    \[5(1)=5\]
    \[5\left(\dfrac{1}{5}\right)=1\]
    \[5\left(\dfrac{2}{5}\right)=2\]
    \[5\left(\dfrac{4}{5}\right)=4\]
    \[5\left(\dfrac{7}{5}\right)=7\]
    \[5\left(\dfrac{6}{5}\right)=6\]
    \[5\left(\dfrac{2}{5}\right)=2\]
  10. Therefore,
    \[5B=\begin{bmatrix}2 & 3 & 5\\1 & 2 & 4\\7 & 6 & 2\end{bmatrix}\]
  11. Step 3: Compute \(3A-5B\)
  12. From Steps 1 and 2, we have
    \[3A=\begin{bmatrix}2 & 3 & 5\\1 & 2 & 4\\7 & 6 & 2\end{bmatrix}\]

    and

    \[5B=\begin{bmatrix}2 & 3 & 5\\1 & 2 & 4\\7 & 6 & 2\end{bmatrix}\]
  13. Hence,
    \[\begin{aligned}3A-5B&=\begin{bmatrix}2 & 3 & 5\\1 & 2 & 4\\7 & 6 & 2\end{bmatrix}-\begin{bmatrix}2 & 3 & 5\\1 & 2 & 4\\7 & 6 & 2\end{bmatrix}\\[8pt] &=\begin{bmatrix}2-2 & 3-3 & 5-5\\1-1 & 2-2 & 4-4\\7-7 & 6-6 & 2-2\end{bmatrix}\end{aligned}\]
  14. Each corresponding pair of entries is equal, so
    \[2-2=0,\quad 3-3=0,\quad 5-5=0\]
    \[1-1=0,\quad 2-2=0,\quad 4-4=0\]
    \[7-7=0,\quad 6-6=0,\quad 2-2=0\]
  15. Therefore,
    \[3A-5B=\begin{bmatrix}0 & 0 & 0\\0 & 0 & 0\\0 & 0 & 0\end{bmatrix}\]
  16. The resulting matrix is the zero matrix of order \(3\times3\).
🎯 Exam Significance
Exam Significance
›

This question tests two fundamental operations of matrices: scalar multiplication and matrix subtraction. It is particularly useful for learning how fractional entries can simplify completely after scalar multiplication.

  • Multiply every entry by the scalar; do not multiply only selected entries.
  • Cancel the denominator carefully when multiplying fractions.
  • After scalar multiplication, subtract corresponding entries.
  • Recognise when two matrices are identical; their difference is the zero matrix.
  • The result
    \[ 3A-5B=O \]
    also shows that
    \[ 3A=5B. \]

In a board examination, writing the intermediate matrices \(3A\) and \(5B\) makes the cancellation of the fractions clear and provides a complete solution.

Significance for Competitive Entrance Examination Aspirants

This problem contains a useful pattern-recognition opportunity. Instead of performing all nine subtraction calculations immediately, an aspirant can recognise that

\[ A=\frac{1}{3}M \]
and
\[ B=\frac{1}{5}M \]
for the same matrix \(M\). Therefore,
\[ 3A=M=5B. \]
Hence,
\[ 3A-5B=O. \]

This type of recognition is valuable in JEE and other competitive entrance examinations, where identifying structural relationships can be much faster than carrying out lengthy arithmetic.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. Scalar multiplication multiplies every entry of a matrix by the scalar.

  2. For this problem,

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

  3. Similarly,

    \[ 5B= \begin{bmatrix} 2 & 3 & 5\\ 1 & 2 & 4\\ 7 & 6 & 2 \end{bmatrix}. \]

  4. Therefore, \(3A=5B\).

  5. The difference of two identical matrices is the zero matrix:

    \[ M-M=O. \]

  6. Hence,

    \[ \boxed{ 3A-5B= \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}} \]

  7. Pattern recognition can substantially reduce calculation time in competitive examinations.
← Q4
5 / 22  ·  23%
Q6 →
Q6
NUMERIC3 marks
Simplify \[\cos\theta\begin{bmatrix}\cos\theta & \sin\theta\\-\sin\theta & \cos\theta\end{bmatrix}+\sin\theta\begin{bmatrix}\sin\theta & -\cos\theta\\\cos\theta & \sin\theta\end{bmatrix}\]
📘 Concept & Theory
Concept/Theory
›

This problem combines scalar multiplication of matrices with matrix addition. When a scalar multiplies a matrix, the scalar must be multiplied by every entry of that matrix.

Thus, for a scalar \(k\) and matrix \(A=[a_{ij}]\),

\[ kA=[ka_{ij}] \]

After carrying out the scalar multiplications, the two matrices are added entry by entry. The main simplification in this question uses the trigonometric identity

\[ \sin^2\theta+\cos^2\theta=1. \]

The off-diagonal entries cancel because

\[ \sin\theta\cos\theta-\sin\theta\cos\theta=0. \]
Consequently, the resulting matrix is the identity matrix of order \(2\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Identify the two matrices and their scalar multipliers.

  2. Multiply every entry of the first matrix by \(\cos\theta\).

  3. Multiply every entry of the second matrix by \(\sin\theta\).

  4. Add the corresponding entries of the two resulting matrices.

  5. Use

    \[ \sin^2\theta+\cos^2\theta=1 \]
    for the diagonal entries.

  6. Cancel equal and opposite terms in the off-diagonal entries.

  7. Write the simplified matrix and identify it as the identity matrix \(I_2\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  9 steps
  1. Step 1: Multiply the First Matrix by \(\cos\theta\)
  2. Consider
    \[\cos\theta\begin{bmatrix}\cos\theta & \sin\theta\\-\sin\theta & \cos\theta\end{bmatrix}\]
  3. Since \(\cos\theta\) is a scalar, multiply it by every entry:
    \[\begin{aligned}\cos\theta\begin{bmatrix}\cos\theta & \sin\theta\\-\sin\theta & \cos\theta\end{bmatrix}&=\begin{bmatrix}\cos\theta\cos\theta & \cos\theta\sin\theta\\-\cos\theta\sin\theta &\cos\theta\cos\theta\end{bmatrix}\\[8pt] &=\begin{bmatrix}\cos^2\theta &\sin\theta\cos\theta\\-\sin\theta\cos\theta &\cos^2\theta\end{bmatrix}\end{aligned}\]
  4. Step 2: Multiply the Second Matrix by \(\sin\theta\)
  5. Now consider
    \[\sin\theta\begin{bmatrix}\sin\theta & -\cos\theta\\\cos\theta & \sin\theta\end{bmatrix}\]
  6. Multiply \(\sin\theta\) by every entry:
    \[\begin{aligned}\sin\theta\begin{bmatrix}\sin\theta & -\cos\theta\\\cos\theta & \sin\theta\end{bmatrix} &=\begin{bmatrix}\sin\theta\sin\theta &-\sin\theta\cos\theta\\\sin\theta\cos\theta &\sin\theta\sin\theta\end{bmatrix}\\[8pt] &=\begin{bmatrix}\sin^2\theta &-\sin\theta\cos\theta\\\sin\theta\cos\theta &\sin^2\theta\end{bmatrix}\end{aligned}\]
  7. Step 3: Add the Two Matrices
  8. Substituting the two results into the original expression, we obtain
    \[\begin{aligned}&\cos\theta\begin{bmatrix}\cos\theta & \sin\theta\\-\sin\theta & \cos\theta\end{bmatrix}+\sin\theta\begin{bmatrix}\sin\theta & -\cos\theta\\\cos\theta & \sin\theta\end{bmatrix}\\[8pt] &=\begin{bmatrix}\cos^2\theta &\sin\theta\cos\theta\\-\sin\theta\cos\theta &\cos^2\theta\end{bmatrix}+\begin{bmatrix}\sin^2\theta &-\sin\theta\cos\theta\\\sin\theta\cos\theta &\sin^2\theta\end{bmatrix}\end{aligned}\]
  9. Add the corresponding entries:
    \[\begin{aligned}&=\begin{bmatrix}\cos^2\theta+\sin^2\theta &\sin\theta\cos\theta-\sin\theta\cos\theta\\-\sin\theta\cos\theta+\sin\theta\cos\theta &\cos^2\theta+\sin^2\theta\end{bmatrix}\end{aligned}\]
  10. Step 4: Simplify Each Entry
  11. For the first diagonal entry, use
    \[\sin^2\theta+\cos^2\theta=1\]
    Therefore,
    \[\cos^2\theta+\sin^2\theta=1\]
    For the upper-right entry,
    \[\sin\theta\cos\theta-\sin\theta\cos\theta=0\]
    For the lower-left entry,
    \[-\sin\theta\cos\theta+\sin\theta\cos\theta=0.\]
    For the second diagonal entry,
    \[\cos^2\theta+\sin^2\theta=1\]
  12. Hence,
    \[\begin{aligned}&=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}\end{aligned}\]
  13. The matrix
    \[\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}\]
    is the identity matrix of order \(2\). Thus,
    \[\boxed{I_2}\]
🎯 Exam Significance
Exam Significance
›

This question is important because it combines two Class 12 concepts: scalar multiplication of matrices and fundamental trigonometric identities. It is a good example of how an apparently complicated matrix expression can simplify to a standard matrix.

  • Remember that a scalar must multiply every entry of a matrix.
  • Perform scalar multiplication before adding the matrices.
  • Add matrices entry by entry.
  • Use
    \[ \sin^2\theta+\cos^2\theta=1 \]
    immediately when it appears.
  • Carefully track negative signs in the off-diagonal entries.
  • Recognise
    \[ \begin{bmatrix} 1&0\ 0&1 \end{bmatrix} \]
    as the identity matrix \(I_2\).

For a board examination, showing the scalar multiplication and the subsequent element-wise addition makes the solution complete and easy to verify.

Significance for Competitive Entrance Examination Aspirants

This question is useful for developing pattern recognition. The expression contains complementary terms:

\[ \cos^2\theta+\sin^2\theta \]
on the diagonal and
\[ \sin\theta\cos\theta-\sin\theta\cos\theta \]
off the diagonal.

Once this structure is recognised, the answer can be obtained rapidly:

\[ \boxed{I_2} \]

In competitive examinations, recognising the identity matrix is especially useful because such expressions may occur as intermediate results in questions involving inverse matrices, orthogonal matrices, determinants, or matrix equations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. Scalar multiplication acts on every entry of the matrix.

  2. Matrix addition is performed entry by entry.

  3. The fundamental identity

    \[ \sin^2\theta+\cos^2\theta=1 \]
    simplifies both diagonal entries.

  4. The off-diagonal entries cancel to zero.

  5. The final matrix is

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

  6. This matrix is the identity matrix \(I_2\).

  7. Therefore,

    \[ \boxed{ \cos\theta \begin{bmatrix} \cos\theta&\sin\theta\\ -\sin\theta&\cos\theta \end{bmatrix} + \sin\theta \begin{bmatrix} \sin\theta&-\cos\theta\\ \cos\theta&\sin\theta \end{bmatrix} =I_2} \]
← Q5
6 / 22  ·  27%
Q7 →
Q7
NUMERIC3 marks
Find \(X\) and \(Y\) if

(i)

\[X+Y=\begin{bmatrix}-7&0\\2&5\end{bmatrix},\quad X-Y=\begin{bmatrix}3&0\\0&3\end{bmatrix}\]

(ii)

\[2X+3Y=\begin{bmatrix}2&3\\4&0\end{bmatrix},\quad X+2Y=\begin{bmatrix}2&-2\\-1&5\end{bmatrix}\]
📘 Concept & Theory
Concept/Theory
›

This question involves solving simultaneous matrix equations. The method is analogous to solving simultaneous linear equations in ordinary algebra, except that the unknowns \(X\) and \(Y\) are matrices.

Since matrix addition and subtraction are performed element by element, equations involving \(X\) and \(Y\) can be added or subtracted to eliminate one of the unknown matrices.

For example, if

\[ X+Y=P \]
and
\[ X-Y=Q, \]
then adding the two equations gives
\[ 2X=P+Q, \]
so that
\[ X=\frac{1}{2}(P+Q). \]
Similarly,
\[ Y=\frac{1}{2}(P-Q). \]

In the second part, subtraction of the two equations eliminates \(2X\), allowing \(Y\) to be found first. Substitution then gives \(X\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. For part (i), add the two matrix equations to eliminate \(Y\) and find \(X\).

  2. Use either original equation to find \(Y\).

  3. For part (ii), subtract the second equation from the first to eliminate \(2X\) and find \(Y\).

  4. Substitute \(Y\) into the second equation to find \(X\).

  5. Verify the answers by substituting them back into the original equations.

✏️ Solution
Part (i)
›
Step-by-step Solution  ·  10 steps
  1. Given —
    \[ X+Y=\begin{bmatrix}-7&0\\2&5\end{bmatrix}\]

    and

    \[X-Y=\begin{bmatrix}3&0\\0&3\end{bmatrix}\]
  2. Step 1: Find \(X\)
  3. Add the two equations. On the left-hand side,
    \[(X+Y)+(X-Y)=X+Y+X-Y\]
  4. Since \(Y-Y=O\), we obtain
    \[(X+Y)+(X-Y)=2X\]
  5. Therefore,
    \[\begin{aligned}2X&=\begin{bmatrix}-7&0\\2&5\end{bmatrix}+\begin{bmatrix}3&0\\0&3\end{bmatrix}\\[8pt] &=\begin{bmatrix}-7+3&0+0\\2+0&5+3\end{bmatrix}\\&=\begin{bmatrix}-4&0\\2&8\end{bmatrix}\end{aligned}\]
  6. Divide both sides by \(2\):
    \[\begin{aligned}X&=\frac{1}{2}\begin{bmatrix}-4&0\\2&8\end{bmatrix}\\[8pt] &=\begin{bmatrix}-2&0\\1&4\end{bmatrix}\end{aligned}\]
  7. Step 2: Find \(Y\)
  8. We use the equation
    \[X-Y=\begin{bmatrix}3&0\\0&3\end{bmatrix}\]
  9. Rearranging for \(Y\),
    \[Y=X-\begin{bmatrix}3&0\\0&3\end{bmatrix}\]
  10. Substitute the value of \(X\):
    \[\begin{aligned}Y&=\begin{bmatrix}-2&0\\1&4\end{bmatrix}-\begin{bmatrix}3&0\\0&3\end{bmatrix}\\[8pt] &=\begin{bmatrix}-2-3&0-0\\1-0&4-3\end{bmatrix}\\[8pt] &=\begin{bmatrix}-5&0\\1&1\end{bmatrix}\end{aligned}\]
  11. Step 3: Verification
  12. First verify \(X+Y\):
    \[\begin{aligned}X+Y&=\begin{bmatrix}-2&0\\1&4\end{bmatrix}+\begin{bmatrix}-5&0\\1&1\end{bmatrix}\\[8pt] &=\begin{bmatrix}-7&0\\2&5\end{bmatrix}\end{aligned}\]
    This agrees with the given value of \(X+Y\).
  13. Now verify \(X-Y\):
    \[\begin{aligned}X-Y&=\begin{bmatrix}-2&0\\1&4\end{bmatrix}-\begin{bmatrix}-5&0\\1&1\end{bmatrix}\\[8pt] &=\begin{bmatrix}-2-(-5)&0-0\\\1-1&4-1\end{bmatrix}\\[8pt] &=\begin{bmatrix}3&0\\0&3\end{bmatrix}\end{aligned}\]
  14. Hence both equations are verified.
✏️ Solution
Part (ii)
›
Step-by-step Solution  ·  11 steps
  1. Given —
    \[2X+3Y=\begin{bmatrix}2&3\\4&0\end{bmatrix}\]

    and

    \[2X+2Y=\begin{bmatrix}2&-2\\-1&5\end{bmatrix}\]
  2. Step 1: Find \(Y\)
  3. Subtract the second equation from the first equation:
    \[\begin{aligned}(2X+3Y)-(2X+2Y)&=2X+3Y-2X-2Y\\&=Y\end{aligned}\]
  4. Therefore,
    \[Y=\begin{bmatrix}2&3\\4&0\end{bmatrix}-\begin{bmatrix}2&-2\\-1&5\end{bmatrix}\]
  5. Perform the subtraction element by element:
    \[\begin{aligned}Y&=\begin{bmatrix}2-2&3-(-2)\\4-(-1)&0-5\end{bmatrix}\\&=\begin{bmatrix}0&5\\5&-5\end{bmatrix}\end{aligned}\]
  6. Step 2: Find \(X\)
  7. Use the second equation:
    \[2X+2Y=\begin{bmatrix}2&-2\\-1&5\end{bmatrix}\]
  8. Rearranging,
    \[2X=\begin{bmatrix}2&-2\\-1&5\end{bmatrix}-2Y\]
  9. First calculate \(2Y\):
    \[\begin{aligned}2Y&=2\begin{bmatrix}0&5\\5&-5\end{bmatrix}\\&=\begin{bmatrix}0&10\\10&-10\end{bmatrix}\end{aligned}\]
  10. Therefore,
    \[\begin{aligned}2X&=\begin{bmatrix}2&-2\\-1&5\end{bmatrix}-\begin{bmatrix}0&10\\10&-10\end{bmatrix}\\[8pt] &=\begin{bmatrix}2-0&-2-10\\-1-10&5-(-10)\end{bmatrix}\\[8pt] &=\begin{bmatrix}2&-12\\-11&15\end{bmatrix}\end{aligned}\]
  11. Divide both sides by \(2\):
    \[\begin{aligned}X&=\frac{1}{2}\begin{bmatrix}2&-12\\-11&15\end{bmatrix}\\[8pt] &=\begin{bmatrix}1&-6\\-\dfrac{11}{2}&\dfrac{15}{2}\end{bmatrix}\end{aligned}\]
  12. Step 3: Verification of Part (ii)
  13. We first verify \(2X+3Y\).
    \[\begin{aligned}2X&=\begin{bmatrix}2&-12\\-11&15\end{bmatrix}\end{aligned}\]

    and

    \[\begin{aligned}3Y&=3\begin{bmatrix}0&5\\5&-5\end{bmatrix}\\[8pt] &=\begin{bmatrix}0&15\\15&-15\end{bmatrix}\end{aligned}\]
  14. Therefore,
    \[\begin{aligned}2X+3Y&=\begin{bmatrix}2&-12\\-11&15\end{bmatrix}+\begin{bmatrix}0&15\\15&-15\end{bmatrix}\\[8pt] &=\begin{bmatrix}2&3\\4&0\end{bmatrix}\end{aligned}\]
    This agrees with the given equation.
  15. Now verify \(2X+2Y\):
    \[\begin{aligned}2X+2Y&=\begin{bmatrix}2&-12\\-11&15\end{bmatrix}+\begin{bmatrix}0&10\\10&-10\end{bmatrix}\\[8pt] &=\begin{bmatrix}2&-2\\-1&5\end{bmatrix}\end{aligned}\]
    This also agrees with the given equation.
🎯 Exam Significance
Exam Significance
›

This question is an important application of matrix addition, subtraction, scalar multiplication, and simultaneous equations. The key technique is elimination of one unknown matrix.

  • If the equations contain \(X+Y\) and \(X-Y\), add them to eliminate \(Y\).
  • If the equations contain \(2X+3Y\) and \(2X+2Y\), subtract them to eliminate \(2X\).
  • When a matrix equation has \(2X\), divide every entry of the matrix by \(2\).
  • Carefully handle subtraction involving negative entries, such as
    \[ 3-(-2)=5. \]
  • Verification is useful for detecting sign and arithmetic errors.

For board examinations, the elimination step should be explicitly shown. Writing only the final matrices may not demonstrate the complete method.

Significance for Competitive Entrance Examination Aspirants

The main competitive-examination skill tested here is rapid elimination. These matrix equations can be treated like simultaneous linear equations without expanding individual matrix entries unnecessarily.

For part (i), the useful identities are

\[ (X+Y)+(X-Y)=2X \]
and
\[ (X+Y)-(X-Y)=2Y. \]

Thus, both unknown matrices can be obtained directly:

\[ X=\frac{1}{2}\left[(X+Y)+(X-Y)\right] \]
\[ Y=\frac{1}{2}\left[(X+Y)-(X-Y)\right]. \]

For part (ii), subtracting the two equations immediately gives \(Y\). This elimination strategy is often faster than solving entry by entry and is particularly useful in timed entrance examinations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  9 points

  1. Matrix equations can be solved using elimination just like ordinary simultaneous equations.

  2. For

    \[ X+Y=P,\qquad X-Y=Q, \]
    we have
    \[ X=\frac{P+Q}{2},\qquad Y=\frac{P-Q}{2}. \]

  3. Adding equations eliminates terms with opposite signs.

  4. Subtracting equations can eliminate terms having the same coefficient.

  5. Scalar division of a matrix means dividing every matrix entry by the scalar.

  6. Negative signs must be handled carefully during matrix subtraction.

  7. Substitution back into the original equations provides a reliable verification.

  8. Part (i):

    \[ \boxed{ X= \begin{bmatrix} -2&0\\ 1&4 \end{bmatrix}, \qquad Y= \begin{bmatrix} -5&0\\ 1&1 \end{bmatrix}} \]

  9. Part (ii):

    \[ \boxed{ X= \begin{bmatrix} 1&-6\\ -\dfrac{11}{2}&\dfrac{15}{2} \end{bmatrix}, \qquad Y= \begin{bmatrix} 0&5\\ 5&-5 \end{bmatrix}} \]
← Q6
7 / 22  ·  32%
Q8 →
Q8
NUMERIC3 marks
Find \(X\), if \[Y=\begin{bmatrix}3&2\\1&4\end{bmatrix}\]

and

\[2X+Y=\begin{bmatrix}1&0\\-3&2\end{bmatrix}\]
📘 Concept & Theory
Concept/Theory
›

This question requires us to solve a linear matrix equation for the unknown matrix \(X\). The given equation is

\[ 2X+Y=A, \]
where \(Y\) and \(A\) are known matrices.

To isolate \(X\), first subtract \(Y\) from both sides:

\[ 2X=A-Y. \]
Then divide both sides by \(2\):
\[ X=\frac{1}{2}(A-Y). \]

Matrix subtraction is performed element by element. Scalar division by \(2\) means dividing every entry of the resulting matrix by \(2\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given matrix equation.

  2. Subtract \(Y\) from both sides to isolate \(2X\).

  3. Substitute the given matrices.

  4. Subtract corresponding entries carefully, especially where negative entries occur.

  5. Divide every entry by \(2\) to obtain \(X\).

  6. Verify the result by substituting \(X\) and \(Y\) into the original equation.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  5 steps
  1. Step 1: Write the Given Equation
  2. We have
    \[2X+Y=\begin{bmatrix}1&0\\-3&2\end{bmatrix}\]

    and

    \[Y=\begin{bmatrix}3&2\\1&4\end{bmatrix}\]
  3. Step 2: Isolate \(2X\)
  4. Subtract \(Y\) from both sides:
    \[\begin{aligned}2X+Y-Y&=\begin{bmatrix}1&0\\-3&2\end{bmatrix}-Y\end{aligned}\]
  5. Since \(Y-Y=O\), we get
    \[2X=\begin{bmatrix}1&0\\-3&2\end{bmatrix}-\begin{bmatrix}3&2\\1&4\end{bmatrix}\]
  6. Step 3: Perform Matrix Subtraction
  7. Subtract the corresponding entries:
    \[\begin{aligned}2X&=\begin{bmatrix}1-3&0-2\\-3-1&2-4\end{bmatrix}\\&=\begin{bmatrix}-2&-2\\-4&-2\end{bmatrix}\end{aligned}\]
  8. Step 4: Find \(X\)
  9. Divide both sides by \(2\). Since scalar division applies to every entry,
    \[\begin{aligned}X&=\frac{1}{2}\begin{bmatrix}-2&-2\\-4&-2\end{bmatrix}\\[8pt] &=\begin{bmatrix}\dfrac{-2}{2}&\dfrac{-2}{2}\\\dfrac{-4}{2}&\dfrac{-2}{2}\end{bmatrix}\\[8pt] &=\begin{bmatrix}-1&-1\\-2&-1\end{bmatrix}\end{aligned}\]
💡 Answer
Final Answer
›
\[X=\begin{bmatrix}-1&-1\\-2&-1\end{bmatrix}\]
🎯 Exam Significance
Exam Significance
›

This question tests the basic but important skill of solving a matrix equation by isolating the unknown matrix. The most important algebraic transformation is

\[ 2X+Y=A \quad\Longrightarrow\quad 2X=A-Y \quad\Longrightarrow\quad X=\frac{1}{2}(A-Y). \]

  • Transpose is not required in this question.
  • Matrix subtraction is performed entry by entry.
  • Subtracting a positive entry from a negative entry must be handled carefully.
  • After finding \(2X\), divide every entry by \(2\).
  • Substitution provides a quick and reliable verification.

A common error is to write

\[ 1-3=2 \]
instead of
\[ 1-3=-2. \]
Similarly,
\[ -3-1=-4. \]
Careful sign handling is essential for obtaining the correct matrix.

Significance for Competitive Entrance Examination Aspirants

For competitive examinations, the entire problem can be reduced to the matrix-equation identity

\[ X=\frac{1}{2}(A-Y). \]
Recognising this directly avoids unnecessary intermediate algebra.

The key speed skill is to isolate the unknown matrix before performing element-wise arithmetic:

\[ \boxed{ 2X+Y=A \Rightarrow X=\frac{A-Y}{2}} \]

This same approach extends naturally to equations such as

\[ aX+B=C, \]
where \(a\) is a non-zero scalar. In that case,
\[ X=\frac{1}{a}(C-B). \]

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. To solve \(2X+Y=A\), first subtract \(Y\) from both sides.

  2. The resulting equation is

    \[ 2X=A-Y \]

  3. Matrix subtraction is performed element by element.

  4. Divide every entry of \(2X\) by \(2\) to obtain \(X\).

  5. In this problem,

    \[ 2X= \begin{bmatrix} -2&-2\\ -4&-2 \end{bmatrix} \]

  6. Therefore,

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

  7. Always check the result by substituting it back into the original equation.
← Q7
8 / 22  ·  36%
Q9 →
Q9
NUMERIC3 marks
Find \(x\) and \(y\), if \[2\begin{bmatrix}1&3\\0&x\end{bmatrix}+\begin{bmatrix}y&0\\1&2\end{bmatrix}=\begin{bmatrix}5&6\\1&8\end{bmatrix}\]
📘 Concept & Theory
Concept/Theory
›

This question uses two fundamental operations on matrices: scalar multiplication and matrix addition.

When a matrix is multiplied by a scalar, every element of the matrix is multiplied by that scalar. Thus,

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

After carrying out the scalar multiplication, corresponding elements of the matrices are added. If two matrices are equal, their corresponding elements must also be equal. Therefore, the entries containing \(x\) and \(y\) can be compared with the corresponding entries on the right-hand side to determine their values.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Multiply the first matrix by \(2\).

  2. Add the two matrices element by element.

  3. Compare corresponding entries of the resulting matrix with the given matrix.

  4. Use the \((1,1)\)-entry to determine \(y\).

  5. Use the \((2,2)\)-entry to determine \(x\).

  6. Substitute \(x\) and \(y\) back into the original equation to verify the result.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  5 steps
  1. Step 1: Perform Scalar Multiplication
  2. Given —
    \[2\begin{bmatrix}1&3\\0&x\end{bmatrix}+\begin{bmatrix}y&0\\1&2\end{bmatrix}=\begin{bmatrix}5&6\\1&8\end{bmatrix}\]
  3. Multiply every element of the first matrix by \(2\):
    \[\begin{aligned}2\begin{bmatrix}1&3\\0&x\end{bmatrix}&=\begin{bmatrix}2(1)&2(3)\\2(0)&2x\end{bmatrix}\\[8pt] &=\begin{bmatrix}2&6\\0&2x\end{bmatrix}\end{aligned}\]
  4. Step 2: Add the Matrices
  5. Substituting the result, we obtain
    \[\begin{aligned}\begin{bmatrix}2&6\\0&2x\end{bmatrix}+\begin{bmatrix}y&0\\1&2\end{bmatrix}&=\begin{bmatrix}5&6\\1&8\end{bmatrix}\end{aligned}\]
  6. Add corresponding entries:
    \[\begin{aligned}\begin{bmatrix}2+y&6+0\\0+1&2x+2\end{bmatrix}&=\begin{bmatrix}5&6\\1&8\end{bmatrix}\end{aligned}\]
  7. Step 3: Compare Corresponding Entries
  8. Since the two matrices are equal, their corresponding entries must be equal.
  9. Comparing the \((1,1)\)-entries:
    \[\begin{aligned}2+y&=5\\y&=5-2\\&=3\end{aligned}\]
  10. Comparing the \((2,2)\)-entries:
    \[\begin{aligned}2x+2&=8\\2(x+1)&=8\\x+1&=\dfrac82\\&=4\\\Rightarrow x&=4-1\\&=3\end{aligned}\]
💡 Answer
Final Answer
›
\[\boxed{x=3,\quad y=3}\]
🎯 Exam Significance
Exam Significance
›

This question is a direct application of scalar multiplication, matrix addition, and equality of matrices. It is important because these concepts form the foundation for more advanced matrix operations in Class 12 Mathematics.

The central principle used is:

\[ \boxed{ \text{If }A=B,\text{ then corresponding elements of }A\text{ and }B\text{ are equal.}} \]

A common examination mistake is to compare entries before completing the scalar multiplication. The factor \(2\) must first be distributed to every element of the first matrix.

Notice that the \((1,2)\)- and \((2,1)\)-entries do not contain unknowns. They automatically agree with the given matrix. Only the entries involving \(x\) and \(y\) are required to determine their values.

Significance for Competitive Entrance Examination Aspirants

For entrance examinations, the quickest approach is to identify the positions containing the unknowns after scalar multiplication.

After simplification,

\[ \begin{bmatrix} 2+y&6\\ 1&2x+2 \end{bmatrix} = \begin{bmatrix} 5&6\\ 1&8 \end{bmatrix} \]

Therefore, only two scalar equations are needed:

\[ 2+y=5 \]
\[ 2x+2=8. \]

This immediately gives

\[ y=3,\qquad x=3. \]

This entry-comparison technique is considerably faster than writing out unnecessary calculations and is useful in time-constrained objective examinations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. Multiply every entry of a matrix by the scalar.

  2. Add matrices element by element.

  3. Equal matrices have equal corresponding entries.

  4. The \((1,1)\)-entry gives the equation for \(y\).

  5. The \((2,2)\)-entry gives the equation for \(x\).

  6. The simplified matrix equation is

    \[ \begin{bmatrix} 2+y&6\\ 1&2x+2 \end{bmatrix} = \begin{bmatrix} 5&6\\ 1&8 \end{bmatrix}. \]

  7. Therefore,

    \[ \boxed{x=3,\qquad y=3}. \]

  8. Always verify the values by substituting them into the original matrix equation.
← Q8
9 / 22  ·  41%
Q10 →
Q10
NUMERIC3 marks
Solve the following matrix equation for \(x,\ y,\ z\) and \(t\): \[2\begin{bmatrix}x&z\\y&t\end{bmatrix}+3\begin{bmatrix}1&-1\\0&2\end{bmatrix}=3\begin{bmatrix}3&5\\4&6\end{bmatrix}\]
📘 Concept & Theory
Concept/Theory
›

This question uses the concept of scalar multiplication and equality of matrices. When a matrix is multiplied by a scalar, the scalar multiplies every entry of the matrix:

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

Also, if two matrices of the same order are equal, then their corresponding elements must be equal. Thus, after simplifying both sides of the given equation, we can compare the four corresponding entries to obtain four ordinary linear equations in \(x,\ y,\ z\) and \(t\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Multiply the first matrix by \(2\).

  2. Multiply the second and third matrices by \(3\).

  3. Add the matrices on the left-hand side.

  4. Compare the resulting matrix with the matrix on the right-hand side.

  5. Equate corresponding entries to obtain four scalar equations.

  6. Solve these equations individually for \(x,\ z,\ y\) and \(t\).

  7. Verify all four values in the original matrix equation.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  13 steps
  1. Step 1: Multiply the Matrices by Their Scalars
  2. Given —
    \[2\begin{bmatrix}x&z\\y&t\end{bmatrix}+3\begin{bmatrix}1&-1\\0&2\end{bmatrix}=3\begin{bmatrix}3&5\\4&6\end{bmatrix}\]
  3. First, calculate the first scalar multiplication:
    \[\begin{aligned}2\begin{bmatrix}x&z\\y&t\end{bmatrix}&=\begin{bmatrix}2x&2z\\2y&2t\end{bmatrix}\end{aligned}\]
  4. Next, calculate the second scalar multiplication:
    \[\begin{aligned}3\begin{bmatrix}1&-1\\0&2\end{bmatrix}&=\begin{bmatrix}3&-3\\0&6\end{bmatrix}\end{aligned}\]
  5. Finally, calculate the right-hand side:
    \[\begin{aligned}3\begin{bmatrix}3&5\\4&6\end{bmatrix}&=\begin{bmatrix}9&15\\12&18\end{bmatrix}\end{aligned}\]
  6. Therefore, the equation becomes
    \[\begin{bmatrix}2x&2z\\2y&2t\end{bmatrix}+\begin{bmatrix}3&-3\\0&6\end{bmatrix}=\begin{bmatrix}9&15\\12&18\end{bmatrix}\]
  7. Step 2: Add the Matrices on the Left-Hand Side
  8. Add corresponding entries:
    \[\begin{aligned}\begin{bmatrix}2x&2z\\2y&2t\end{bmatrix}+\begin{bmatrix}3&-3\\0&6\end{bmatrix} &=\begin{bmatrix}2x+3&2z+(-3)\\2y+0&2t+6\end{bmatrix}\\[8pt] &=\begin{bmatrix}2x+3&2z-3\\2y&2t+6\end{bmatrix}\end{aligned}\]
  9. Hence,
    \[\boxed{\begin{bmatrix}2x+3&2z-3\\2y&2t+6\end{bmatrix}=\begin{bmatrix}9&15\\12&18\end{bmatrix}}\]
  10. Step 3: Compare Corresponding Entries
  11. Since the two matrices are equal, their corresponding entries must be equal.
  12. Finding \(x\)
  13. Compare the \((1,1)\)-entries:
    \[\begin{aligned} 2x+3&=9\\ 2x&=9-3\\ 2x&=6\\ x&=\dfrac62\\ &=3 \end{aligned}\]
  14. \[\boxed{x=3}\]
  15. Finding \(z\)
  16. Compare the \((1,2)\)-entries:
    \[\begin{aligned} 2z-3&=15\\ 2z&=15+3\\ 2z&=18\\ z&=\dfrac{18}{2}\\ &=9 \end{aligned}\]
  17. \[\boxed{z=9}\]
  18. Finding \(y\)
  19. Compare the \((2,1)\)-entries:
    \[\begin{aligned} 2y&=12\\ y&=\dfrac{12}{2}\\ &=6 \end{aligned}\]
    \[\boxed{y=6}\]
  20. Finding \(t\)
  21. Compare the \((2,2)\)-entries:
    \[\begin{aligned} 2t+6&=18\\ 2t&=18-6\\ 2t&=12\\ t&=\dfrac{12}{2}\\ &=6 \end{aligned}\]
  22. \[\boxed{t=6}\]
💡 Answer
Final Answer
›
\[\boxed{x=3,\quad y=6,\quad z=9,\quad t=6}\]
🎯 Exam Significance
Exam Significance
›

This question combines three fundamental matrix concepts: scalar multiplication, matrix addition, and equality of matrices. It is a straightforward but important application of the basic algebra of matrices.

The key examination principle is:

\[ \boxed{ \text{Equal matrices have equal corresponding entries.}} \]

Once the matrix equation is simplified, the problem reduces to four independent linear equations:

\[ 2x+3=9, \]
\[ 2z-3=15, \]
\[ 2y=12, \]
\[ 2t+6=18. \]

Showing the scalar multiplication explicitly is recommended in a board examination because it makes the logical progression of the solution clear and avoids sign errors.

Significance for Competitive Entrance Examination Aspirants

For competitive examinations, the quickest method is to simplify the scalar multiples first and then compare the four corresponding entries. There is no need to perform any matrix multiplication or use determinants.

The entire problem can be compressed to the following structure:

\[ \begin{bmatrix} 2x+3&2z-3\\ 2y&2t+6 \end{bmatrix} = \begin{bmatrix} 9&15\\ 12&18 \end{bmatrix} \]

Therefore,

\[ 2x+3=9,\qquad 2z-3=15,\qquad 2y=12,\qquad 2t+6=18. \]

This direct-entry comparison is the fastest approach and is particularly useful for objective and time-constrained questions.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. Scalar multiplication multiplies every entry of a matrix by the given scalar.

  2. Matrix addition is performed entry by entry.

  3. Equal matrices must have equal corresponding entries.

  4. After simplification, the given equation becomes

    \[ \begin{bmatrix} 2x+3&2z-3\\ 2y&2t+6 \end{bmatrix} = \begin{bmatrix} 9&15\\ 12&18 \end{bmatrix}. \]

  5. The four corresponding entries produce four independent scalar equations.

  6. Solving these equations gives

    \[ \boxed{x=3,\qquad y=6,\qquad z=9,\qquad t=6}. \]

  7. Substitution into the original equation confirms the result.
← Q9
10 / 22  ·  45%
Q11 →
Q11
NUMERIC3 marks
If \[x\begin{bmatrix}2\\3\end{bmatrix}+y\begin{bmatrix}-1\\1\end{bmatrix}=\begin{bmatrix}10\\5\end{bmatrix},\] find the values of \(x\) and \(y\).
📘 Concept & Theory
Concept/Theory
›

This question involves scalar multiplication of matrices and equality of matrices. A scalar multiplying a column matrix multiplies every element of that matrix by the scalar. Therefore,

\[x\begin{bmatrix}a\\b\end{bmatrix}=\begin{bmatrix}ax\\bx\end{bmatrix}\]

After performing the scalar multiplications, the given matrix equation becomes a pair of simultaneous linear equations in \(x\) and \(y\). These equations can then be solved using the elimination method.

Since equal matrices have equal corresponding entries, the first entries on both sides must be equal, and the second entries on both sides must also be equal.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Multiply each column matrix by its respective scalar.

  2. Add the resulting column matrices.

  3. Compare corresponding entries on both sides.

  4. Obtain two simultaneous linear equations in \(x\) and \(y\).

  5. Add the two equations to eliminate \(y\).

  6. Find \(x\), then substitute its value into either equation to find \(y\).

  7. Verify both values in the original vector equation.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  13 steps
  1. Step 1: Expand the Scalar Multiplications
  2. Given —
    \[x\begin{bmatrix}2\\3\end{bmatrix}+y\begin{bmatrix}-1\\1\end{bmatrix}=\begin{bmatrix}10\\5\end{bmatrix},\]
  3. Multiplying the first column matrix by \(x\):
    \[\begin{aligned}x\begin{bmatrix}2\\3\end{bmatrix}&=\begin{bmatrix}2x\\3x\end{bmatrix}\end{aligned}\]
  4. Multiplying the second column matrix by \(y\):
    \[\begin{aligned}y\begin{bmatrix}-1\\1\end{bmatrix}&=\begin{bmatrix}-y\\y\end{bmatrix}\end{aligned}\]
  5. Therefore, the original equation becomes
    \[\begin{aligned}\begin{bmatrix}2x\\3x\end{bmatrix}+\begin{bmatrix}-y\\y\end{bmatrix}&=\begin{bmatrix}10\\5\end{bmatrix}\end{aligned}\]
  6. Step 2: Add the Column Matrices
  7. Add the corresponding entries:
    \[\begin{aligned}\begin{bmatrix}2x\\3x\end{bmatrix}+\begin{bmatrix}-y\\y\end{bmatrix}&=\begin{bmatrix}2x-y\\3x+y\end{bmatrix}\end{aligned}\]
  8. Hence,
    \[\begin{bmatrix}2x-y\\3x+y\end{bmatrix}=\begin{bmatrix}10\\5\end{bmatrix}\]
  9. Step 3: Compare Corresponding Entries
  10. Since the two column matrices are equal, their corresponding entries must be equal.
  11. Comparing the first entries:
    \[2x-y=10\]
  12. Comparing the second entries:
    \[3x+y=5\]
  13. >Thus, we have the simultaneous equations
  14. Step 4: Find \(x\)
  15. Add the two equations to eliminate \(y\):
    \[\begin{aligned}(2x-y)+(3x+y)&=10+5\\2x-y+3x+y&=15\\5x&=15\end{aligned}\]
  16. Divide both sides by \(5\):
    \[\begin{aligned}x&=\frac{15}{5}\\x&=3\end{aligned}\]
  17. Therefore,
    \[\boxed{x=3}\]
  18. Step 5: Find \(y\)
  19. Substitute \(x=3\) into the first equation:
    \[2x-y=10\]
    \[ \begin{aligned} 2(3)-y&=10\\ 6-y&=10\\ \Rightarrow -y&=10-6\\ y&=-4 \end{aligned} \]
  20. Therefore,
    \[\boxed{y=-4}\]
💡 Answer
Final Answer
›
\[\boxed{x=3,\quad y=-4}\]
🎯 Exam Significance
Exam Significance
›

This problem is important because it connects matrix equations with simultaneous linear equations. It tests whether the student understands scalar multiplication of a matrix and the principle of equality of matrices.

A crucial point is the sign of the first entry of the second column matrix:

\[y\begin{bmatrix}-1\\1\end{bmatrix}=\begin{bmatrix}-y\\y\end{bmatrix}\]

Missing this negative sign changes the resulting equations and produces an incorrect answer. In a board examination, writing the scalar multiplication explicitly helps prevent this error.

Significance for Competitive Entrance Examination Aspirants

For competitive examinations, the problem can be reduced quickly to:

\[2x-y=10,\qquad 3x+y=5\]

Adding these equations immediately eliminates \(y\):

\[5x=15\]

Hence,

\[x=3\]
Substitution then gives
\[y=-4\]

The elimination method is particularly efficient because the coefficients of \(y\) are \( -1\) and \(+1\), so \(y\) cancels immediately when the equations are added.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. Scalar multiplication applies to every entry of a column matrix.

  2. A negative matrix entry must retain its sign:

    \[y\begin{bmatrix}-1\\1\end{bmatrix}=\begin{bmatrix}-y\\y\end{bmatrix}\]

  3. Equal matrices have equal corresponding entries.

  4. The matrix equation gives two simultaneous equations.

  5. The resulting equations are

    \[2x-y=10,\qquad 3x+y=5\]

  6. Adding the equations eliminates \(y\) directly.

  7. The correct values are

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

  8. Substitution into the original equation provides a complete verification.
← Q10
11 / 22  ·  50%
Q12 →
Q12
NUMERIC3 marks
Given \[3\begin{bmatrix}x&y\\z&w\end{bmatrix}=\begin{bmatrix}x&6\\-1&2w\end{bmatrix}+\begin{bmatrix}4&x+y\\z+w&3\end{bmatrix},\]

find the values of \(x,\ y,\ z\) and \(w\).

📘 Concept & Theory
Concept/Theory
›

This problem is based on scalar multiplication, matrix addition, and equality of matrices.

When a matrix is multiplied by a scalar, every element of the matrix is multiplied by that scalar:

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

Matrix addition is performed by adding corresponding entries. Further, if two matrices of the same order are equal, their corresponding entries must be equal. Therefore, the given matrix equation can be converted into four simultaneous scalar equations involving \(x,\ y,\ z\) and \(w\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Perform the scalar multiplication on the left-hand side.

  2. Add the two matrices on the right-hand side entry by entry.

  3. Equate corresponding entries of the two resulting matrices.

  4. Solve the first equation to obtain \(x\).

  5. Substitute \(x\) into the equation containing \(y\).

  6. Use the equation containing \(w\) to determine \(w\).

  7. Substitute \(w\) into the remaining equation to determine \(z\).

  8. Verify all four values in the original matrix equation.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  20 steps
  1. Step 1: Perform Scalar Multiplication
  2. Given —
    \[3\begin{bmatrix}x&y\\z&w\end{bmatrix}=\begin{bmatrix}x&6\\-1&2w\end{bmatrix}+\begin{bmatrix}4&x+y\\z+w&3\end{bmatrix},\]
  3. Multiply every entry of the matrix on the left by \(3\):
    \[\begin{aligned}3\begin{bmatrix}x&y\\z&w\end{bmatrix}&=\begin{bmatrix}3x&3y\\3z&3w\end{bmatrix}\end{aligned}\]
  4. Step 2: Add the Matrices on the Right-Hand Side
  5. \[\begin{aligned}\begin{bmatrix}x&6\\-1&2w\end{bmatrix}+\begin{bmatrix}4&x+y\\z+w&3\end{bmatrix}&=\begin{bmatrix} x+4&6+(x+y)\\-1+(z+w)&2w+3\end{bmatrix}\\[8pt] &=\begin{bmatrix}x+4&x+y+6\\z+w-1&3+2w\end{bmatrix}\end{aligned}\]
  6. Therefore, the original equation becomes
    \[\boxed{\begin{bmatrix}3x&3y\\3z&3w\end{bmatrix}=\begin{bmatrix}x+4&x+y+6\\z+w-1&3+2w\end{bmatrix}}\]
  7. Step 3: Compare Corresponding Entries
  9. Comparing the \((1,1)\)-entries:
    \[3x=x+4\tag{1}\]
  10. Comparing the \((1,2)\)-entries:
    \[3y=x+y+6\tag{2}\]
  11. Comparing the \((2,1)\)-entries:
    \[3z=z+w-1\tag{3}\]
  12. Comparing the \((2,2)\)-entries:
    \[3w=3+2w\tag{4}\]
  13. Step 4: Find \(x\)
  14. From equation(1)
    \[\begin{align}3x&=x+4\\3x-x&=4\\2x&=4\\x&=\dfrac42\\x&=2\tag{5}\end{align}\]
  15. Therefore,
    \[\boxed{x=2}\]
  16. Step 5: Find \(y\)
  17. From equation(2) and equation(5)
    \[\begin{align} 3y&=x+y+6\quad (x=2, \text{ from eq. (5)})\\ 3y-y&=2+6\\ 2y&=8\\ y&=\dfrac82\\ y&=4\tag{6} \end{align}\]
  18. Therefore,
    \[\boxed{y=4}\]
  19. Step 6: Find \(w\)
  20. From equation (4)
    \[\begin{align} 3w&=3+2w\\ 3w-2w&=3\\ w&=3\tag{7} \end{align}\]
  21. Therefore,
    \[\boxed{w=3}\]
  22. Step 7: Find \(z\)
  23. From equation (3)
    \[\begin{aligned} 3z&=z+w-1 \quad (w=3, \text{ from eq. 7})\\ 3z-z&=3-1\\ 2z&=2\\ z&=\dfrac22\\ &=1 \end{aligned}\]
  24. Therefore,
    \[\boxed{z=1}\]
  25. Verification
  26. Substitute these values into the left-hand side of the original equation:
    \[\begin{aligned}3\begin{bmatrix}2&4\\1&3\end{bmatrix}&=\begin{bmatrix}6&12\\3&9\end{bmatrix}\end{aligned}\]
  27. Now calculate the right-hand side:
    \[\begin{aligned}&\begin{bmatrix}2&6\\-1&6\end{bmatrix}+\begin{bmatrix}4&2+4\\1+3&3\end{bmatrix}\\[8pt] &=\begin{bmatrix}2&6\\-1&6\end{bmatrix}+\begin{bmatrix}4&6\\4&3\end{bmatrix}\\[8pt] &=\begin{bmatrix}2+4&6+6\\-1+4&6+3\end{bmatrix}\\[8pt] &=\begin{bmatrix}6&12\\3&9\end{bmatrix}\end{aligned}\]
  28. Both sides are equal:
    \[\begin{bmatrix}6&12\\3&9\end{bmatrix}=\begin{bmatrix}6&12\\3&9\end{bmatrix}\]
  29. Hence, all four values are correct.
💡 Answer
Final Answer
›
\boxed{x=2,\quad y=4,\quad z=1,\quad w=3}
🎯 Exam Significance
Exam Significance
›

This question is an important application of the equality of matrices. It requires students to correctly perform scalar multiplication, matrix addition and comparison of corresponding entries.

The most important principle is:

\[ \boxed{ A=B\quad\Longrightarrow\quad a_{ij}=b_{ij}\text{ for every corresponding entry}.} \]

A reliable board-examination strategy is to simplify both sides completely before comparing entries. This avoids missing terms such as \(z+w-1\) or \(x+y+6\).

It is also important to preserve signs while adding matrix entries. In particular,

\[ -1+(z+w)=z+w-1. \]

Significance for Competitive Entrance Examination Aspirants

For competitive examinations, the problem can be solved rapidly by comparing the four entries immediately after simplification:

\[3x=x+4,\quad 3y=x+y+6,\quad 3z=z+w-1,\quad 3w=3+2w\]

The equations should be solved in an efficient dependency order:

  1. Find \(x\) directly from \(3x=x+4\).
  2. Use \(x\) to find \(y\).
  3. Find \(w\) directly from \(3w=3+2w\).
  4. Use \(w\) to find \(z\).

This avoids unnecessary simultaneous-equation calculations and is useful in time-constrained objective examinations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. Multiply every entry of a matrix by the scalar placed before it.

  2. Add corresponding entries when adding matrices.

  3. Equal matrices have equal corresponding entries.

  4. The simplified matrix equation is

    \[ \begin{bmatrix} 3x&3y\\ 3z&3w \end{bmatrix} = \begin{bmatrix} x+4&x+y+6\\ z+w-1&3+2w \end{bmatrix} \]

  5. The four matrix entries produce four scalar equations.

  6. Use already determined variables when solving dependent equations.

  7. The final values are

    \[ \boxed{x=2,\qquad y=4,\qquad z=1,\qquad w=3}. \]

  8. Substitution into the original equation confirms the solution.
← Q11
12 / 22  ·  55%
Q13 →
Q13
NUMERIC3 marks
If \[F(x)=\begin{bmatrix}\cos x&-\sin x&0\\\sin x&\cos x&0\\0&0&1\end{bmatrix},\]

show that

\[\boxed{F(x)F(y)=F(x+y)}.\]
📘 Concept & Theory
Concept/Theory
›

This question is based on matrix multiplication and the addition formulae of trigonometric functions. Unlike matrix addition, matrix multiplication is performed by taking the dot product of each row of the first matrix with each column of the second matrix.

For two matrices \(A=[a_{ij}]\) and \(B=[b_{ij}]\), the general element of the product \(AB\) is

\[ (AB)_{ij}=\sum_k a_{ik}b_{kj}. \]

The two identities that are central to this problem are

\[ \cos(x+y)=\cos x\cos y-\sin x\sin y \]
\[ \sin(x+y)=\sin x\cos y+\cos x\sin y. \]

The given matrix \(F(x)\) represents a rotation matrix in the first two coordinates, with the third coordinate unchanged. The result therefore expresses an important composition property: applying the transformation corresponding to \(F(y)\) followed by \(F(x)\) is equivalent to applying a single transformation corresponding to \(F(x+y)\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write \(F(x)\), \(F(y)\), and \(F(x+y)\) explicitly.

  2. Multiply \(F(x)\) and \(F(y)\) using row-by-column multiplication.

  3. Calculate every entry of the resulting \(3\times3\) matrix.

  4. Use the sine and cosine addition formulae to simplify the entries.

  5. Compare the resulting matrix with \(F(x+y)\).

  6. Conclude that \(F(x)F(y)=F(x+y)\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  23 steps
  1. Step 1: Write the Given Matrices
  2. \[F(x)=\begin{bmatrix}\cos x&-\sin x&0\\\sin x&\cos x&0\\0&0&1\end{bmatrix}\]
  3. Replacing \(x\) by \(y\), we obtain
    \[F(y)=\begin{bmatrix}\cos y&-\sin y&0\\\sin y&\cos y&0\\0&0&1\end{bmatrix}\]
  4. Also,
    \[F(x+y)=\begin{bmatrix}\cos(x+y)&-\sin(x+y)&0\\\sin(x+y)&\cos(x+y)&0\\0&0&1\end{bmatrix}\]
  5. Step 2: Multiply \(F(x)\) and \(F(y)\)
  6. \[\begin{aligned}F(x)F(y)&=\begin{bmatrix}\cos x&-\sin x&0\\\sin x&\cos x&0\\0&0&1\end{bmatrix} \begin{bmatrix}\cos y&-\sin y&0\\\sin y&\cos y&0\\0&0&1\end{bmatrix}\end{aligned}\]
  7. We now calculate each entry of the product by multiplying the corresponding row of the first matrix by the corresponding column of the second matrix.
  8. Step 3: Calculate the First Row
  9. First entry \(a_{11}\):
    \[\begin{aligned}a_{11}&=\begin{bmatrix}\cos x&-\sin x&0\end{bmatrix}\begin{bmatrix}\cos y\\\sin y\\0\end{bmatrix}\\[8pt] &=\cos x\cos y-\sin x\sin y+0\\&=\cos x\cos y-\sin x\sin y\end{aligned}\]
  10. Using the cosine addition formula,
    \[\cos(x+y)=\cos x\cos y-\sin x\sin y\]
  11. we get
    \[\boxed{a_{11}=\cos(x+y)}\]
  12. Second entry \(a_{12}\):
    \[\begin{aligned}a_{12}&=\begin{bmatrix}\cos x&-\sin x&0\end{bmatrix}\begin{bmatrix}-\sin y\\\cos y\\0\end{bmatrix}\\[8pt] &=-\cos x\sin y-\sin x\cos y+0\\ &=-(\sin y\cos x+\cos y\sin x)\end{aligned}\]
  13. Using
    \[\sin(x+y)=\sin x\cos y+\cos x\sin y\]
  14. we obtain
    \[\boxed{a_{12}=-\sin(x+y)}\]
  15. Third entry \(a_{13}\):
    \[\begin{aligned}a_{13}&=\begin{bmatrix}\cos x&-\sin x&0\end{bmatrix}\begin{bmatrix}0\\0\\1\end{bmatrix}\\[8pt] &=\cos x(0)+(-\sin x)(0)+0(1)\&=0\end{aligned}\]
    \[\boxed{a_{13}=0}\]
  16. Therefore, the first row of \(F(x)F(y)\) is
    \[\boxed{\begin{bmatrix}\cos(x+y)&-\sin(x+y)&0\end{bmatrix}}\]
  17. Step 4: Calculate the Second Row
  18. First entry \(a_{21}\):
    \[\begin{aligned}a_{21}&=\begin{bmatrix}\sin x&\cos x&0\end{bmatrix}\begin{bmatrix}\cos y\\\sin y\\0\end{bmatrix}\\[8pt] &=\sin x\cos y+\cos x\sin y+0\\&=\sin x\cos y+\cos x\sin y\end{aligned}\]
  19. Using the sine addition formula,
    \[\sin(x+y)=\sin x\cos y+\cos x\sin y\]
  20. we get
    \[\boxed{a_{21}=\sin(x+y)}\]
  21. Second entry \(a_{22}\):
    \[\begin{aligned}a_{22}&=\begin{bmatrix}\sin x&\cos x&0\end{bmatrix}\begin{bmatrix}-\sin y\\\cos y\\0\end{bmatrix}\\[8pt] &=-\sin x\sin y+\cos x\cos y+0\\&=\cos x\cos y-\sin x\sin y\end{aligned}\]
  22. Using the cosine addition formula,
    \[\cos(x+y)=\cos x\cos y-\sin x\sin y\]
  23. we obtain
    \[\boxed{a_{22}=\cos(x+y)}\]
  24. Third entry \(a_{23}\):
    \[\begin{aligned}a_{23}&=\begin{bmatrix}\sin x&\cos x&0\end{bmatrix}\begin{bmatrix}0\\0\\1\end{bmatrix}\\[8pt] &=\sin x(0)+\cos x(0)+0(1)\\&=0\end{aligned}\]
    \[\boxed{a_{23}=0}\]
  25. Therefore, the second row is
    \[\boxed{\begin{bmatrix}\sin(x+y)&\cos(x+y)&0\end{bmatrix}}\]
  26. Step 5: Calculate the Third Row
  27. First entry \(a_{31}\):
    \[\begin{aligned}a_{31}&=\begin{bmatrix}0&0&1\end{bmatrix}\begin{bmatrix}\cos y\\\sin y\\0\end{bmatrix}\\[8pt] &=0(\cos y)+0(\sin y)+1(0)\\&=0\end{aligned}\]
    \[\boxed{a_{31}=0}\]
  28. Second entry \(a_{32}\):
    \[\begin{aligned}a_{32}&=\begin{bmatrix}0&0&1\end{bmatrix}\begin{bmatrix}-\sin y\\\cos y\\0\end{bmatrix}\\[8pt] &=0(-\sin y)+0(\cos y)+1(0)\\&=0\end{aligned}\]
    [\boxed{a_{32}=0}\]
  29. Third entry \(a_{33}\):
    \[\begin{aligned}a_{33}&=\begin{bmatrix}0&0&1\end{bmatrix}\begin{bmatrix}0\\0\\1\end{bmatrix}\\[8pt] &=0(0)+0(0)+1(1)\\&=1\end{aligned}\]
    \[\boxed{a_{33}=1}\]
  30. Hence, the third row is
    \[\boxed{\begin{bmatrix}0&0&1\end{bmatrix}}\]
  31. Step 6: Form the Product Matrix
  32. Combining all nine entries,
    \[\begin{aligned}F(x)F(y)&=\begin{bmatrix}a_{11}&a_{12}&a_{13}\\a_{21}&a_{22}&a_{23}\\a_{31}&a_{32}&a_{33}\end{bmatrix}\\[8pt] &=\begin{bmatrix}\cos(x+y)&-\sin(x+y)&0\\\sin(x+y)&\cos(x+y)&0\\0&0&1\end{bmatrix}\end{aligned}\]
  33. But, by definition,
    \[F(x+y)=\begin{bmatrix}\cos(x+y)&-\sin(x+y)&0\\\sin(x+y)&\cos(x+y)&0\\0&0&1\end{bmatrix}\]
  34. Therefore,
    \[\boxed{F(x)F(y)=F(x+y)}\]
  35. Proved — Hence, the required result is proved:
    \[\boxed{F(x)F(y)=F(x+y)}\]
🎯 Exam Significance
Exam Significance
›

This is an important application of matrix multiplication combined with standard trigonometric identities. In a board examination, students must show the row-by-column multiplication clearly because the result cannot be obtained by simply multiplying corresponding entries.

The most important identities required are

\[ \boxed{\cos(x+y)=\cos x\cos y-\sin x\sin y} \]
\[ \boxed{\sin(x+y)=\sin x\cos y+\cos x\sin y}. \]

A common source of errors is the negative sign in the second column of \(F(x)\) and \(F(y)\). Therefore, the entries \(a_{12}\) and \(a_{22}\) should be calculated carefully.

Significance for Competitive Entrance Examination Aspirants

This result represents a useful composition law for rotation matrices. It can be used to simplify repeated matrix products without carrying out complete multiplication every time.

For example, using the proved result repeatedly,

\[ F(x)F(y)F(z) = F(x+y)F(z) = F(x+y+z). \]

Similarly,

\[ F(x)^n=F(nx), \]

for a positive integer \(n\).

Such properties can substantially reduce calculation time in objective and entrance examinations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  10 points

  1. Matrix multiplication is performed by row-by-column multiplication.

  2. For a product of two \(3\times3\) matrices, every one of the nine entries should be accounted for.

  3. The sine addition identity converts the \((2,1)\)-entry into \(\sin(x+y)\).

  4. The cosine addition identity converts the \((1,1)\) and \((2,2)\)-entries into \(\cos(x+y)\).

  5. The entries \(a_{13},a_{23},a_{31},a_{32}\) are all zero.

  6. The \((3,3)\)-entry remains \(1\).

  7. The product has exactly the same form as \(F(x+y)\).

  8. The central result is

    \[ \boxed{F(x)F(y)=F(x+y)}. \]

  9. Consequently,

    \[ F(x)F(y)F(z)=F(x+y+z). \]

  10. For a positive integer \(n\),

    \[ F(x)^n=F(nx). \]
← Q12
13 / 22  ·  59%
Q14 →
Q14
NUMERIC3 marks
Show that:

(i)

\[\begin{bmatrix}5&-1\\6&7\end{bmatrix}\begin{bmatrix}2&1\\3&4\end{bmatrix}\neq\begin{bmatrix}2&1\\3&4\end{bmatrix}\begin{bmatrix}5&-1\\6&7\end{bmatrix}\]

(ii)

\[\begin{bmatrix}1&2&3\\0&1&0\\1&1&0\end{bmatrix}\begin{bmatrix}-1&1&0\\0&-1&1\\2&3&4\end{bmatrix}\neq\begin{bmatrix}-1&1&0\\0&-1&0\\2&3&4\end{bmatrix} \begin{bmatrix}1&2&3\\0&1&0\\1&1&0\end{bmatrix}\]
📘 Concept & Theory
Concept/Theory
›

For matrices \(A\) and \(B\), matrix multiplication is performed using the row-by-column rule. The element in the \(i\)-th row and \(j\)-th column of \(AB\) is obtained by multiplying the entries of the \(i\)-th row of \(A\) by the corresponding entries of the \(j\)-th column of \(B\), and then adding the products.

A crucial property of matrix multiplication is:

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

This is in contrast to ordinary real-number multiplication, for which

\[ ab=ba. \]

Therefore, we cannot interchange the order of matrices in a product unless equality has been established for the particular matrices.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Calculate the first product \(AB\) using row-by-column multiplication.

  2. Calculate the reversed product \(BA\) separately.

  3. Compare the corresponding entries of the two resulting matrices.

  4. If at least one corresponding entry is different, conclude that \(AB\neq BA\).

  5. Repeat the same procedure for part (ii).

✏️ Solution
Part - (i)
›
Step-by-step Solution  ·  17 steps
  1. Let
    \[A=\begin{bmatrix}5&-1\\6&7\end{bmatrix},\quad B=\begin{bmatrix}2&1\\3&4\end{bmatrix}\]
  2. Step 1: Calculate \(AB\)
  3. \[\begin{aligned}AB&=\begin{bmatrix}5&-1\\6&7\end{bmatrix}\begin{bmatrix}2&1\\3&4\end{bmatrix}\end{aligned}\]
  4. The first row of \(A\) is multiplied by the first column of \(B\):
    \[\begin{aligned}(AB)_{11}&=5(2)+(-1)(3)\\&=10-3\\&=7\end{aligned}\]
  5. The first row of \(A\) is multiplied by the second column of \(B\):
    \[\begin{aligned}(AB)_{12}&=5(1)+(-1)(4)\\&=5-4\\&=1\end{aligned}\]
  6. The second row of \(A\) is multiplied by the first column of \(B\):
    \[\begin{aligned}(AB)_{21}&=6(2)+7(3)\\&=12+21\\&=33\end{aligned}\]
  7. The second row of \(A\) is multiplied by the second column of \(B\):
    \[\begin{aligned}(AB)_{22}&=6(1)+7(4)\\&=6+28\\&=34\end{aligned}\]
  8. Therefore,
    \[AB=\begin{bmatrix}7&1\\33&34\end{bmatrix}\]
  9. Step 2: Calculate \(BA\)
  10. \[\begin{aligned}BA&=\begin{bmatrix}2&1\\3&4\end{bmatrix}\begin{bmatrix}5&-1\\6&7\end{bmatrix}\end{aligned}\]
  11. First entry:
  12. \[\begin{aligned}(BA)_{11}&=2(5)+1(6)\\&=10+6\\&=16\end{aligned}\]
  13. Second entry:
    \[\begin{aligned}(BA)_{12}&=2(-1)+1(7)\\&=-2+7\\&=5\end{aligned}\]
  14. Third entry:
    \[\begin{aligned}(BA)_{21}&=3(5)+4(6)\\&=15+24\\&=39\end{aligned}\]
  15. Fourth entry:
    \[\begin{aligned}(BA)_{22}&=3(-1)+4(7)\\&=-3+28\\&=25\end{aligned}\]
  16. Therefore, the correct product is
    \[BA=\begin{bmatrix}16&5\\39&25\end{bmatrix}\]
  17. Step 3: Compare \(AB\) and \(BA\)
  18. \[AB=\begin{bmatrix}7&1\\33&34\end{bmatrix}\]
    whereas
    \[BA=\begin{bmatrix}16&5\\39&25\end{bmatrix}\]
  19. Since even the first corresponding entries are different,
    \[7\neq16\]
  20. we conclude that
    \[\boxed{\begin{bmatrix}5&-1\\6&7\end{bmatrix}\begin{bmatrix}2&1\\3&4\end{bmatrix}\neq\begin{bmatrix}2&1\\3&4\end{bmatrix}\begin{bmatrix}5&-1\\6&7\end{bmatrix}}\]
✏️ Solution
Part - (ii)
›
Step-by-step Solution  ·  20 steps
  1. Let
    \[A=\begin{bmatrix}1&2&3\\0&1&0\\1&1&0\end{bmatrix},\quad B=\begin{bmatrix}-1&1&0\\0&-1&1\\2&3&4\end{bmatrix}\]
  2. We calculate \(AB\) and \(BA\) independently.
  3. Step 1: Calculate \(AB\)
  4. \[\begin{aligned}AB&=\begin{bmatrix}1&2&3\\0&1&0\\1&1&0\end{bmatrix}\begin{bmatrix}-1&1&0\\0&-1&1\\2&3&4\end{bmatrix}\end{aligned}\]
  5. First row:
    \[\begin{aligned}(AB)_{11}&=1(-1)+2(0)+3(2)\\&=-1+0+6\\&=5,\end{aligned}\]
    \[\begin{aligned}(AB)_{12}&=1(1)+2(-1)+3(3)\\&=1-2+9\\&=8,\end{aligned}\]
    \[\begin{aligned}(AB)_{13}&=1(0)+2(1)+3(4)\\&=0+2+12\\&=14.\end{aligned}\]
  6. Second row:
    \[\begin{aligned}(AB)_{21}&=0(-1)+1(0)+0(2)\\&=0,\end{aligned}\]
    \[\begin{aligned}(AB)_{22}&=0(1)+1(-1)+0(3)\\&=-1,\end{aligned}\]
    \[\begin{aligned}(AB)_{23}&=0(0)+1(1)+0(4)\\&=1.\end{aligned}\]
  7. Third row:
    \[\begin{aligned}(AB)_{31}&=1(-1)+1(0)+0(2)\\&=-1,\end{aligned}\]
    \[\begin{aligned}(AB)_{32}&=1(1)+1(-1)+0(3)\\&=1-1\\&=0,\end{aligned}\]
    \[\begin{aligned}(AB)_{33}&=1(0)+1(1)+0(4)\\&=1.\end{aligned}\]
  8. Hence,
    \[AB=\begin{bmatrix}5&8&14\\0&-1&1\\-1&0&1\end{bmatrix}\]
  9. Step 2: Calculate \(BA\)
  10. The matrix appearing on the right-hand side of the question as supplied is
    \[\begin{bmatrix}-1&1&0\\0&-1&0\\2&3&4\end{bmatrix}\]
  11. However, this differs from the matrix \(B\) used on the left-hand side, whose second row is
    \[\begin{bmatrix}0&-1&1\end{bmatrix}\]
  12. To verify the intended non-commutativity of the same two matrices \(A\) and \(B\), we calculate \(BA\) using the original matrix \(B\):
  13. \[\begin{aligned}BA&=\begin{bmatrix}-1&1&0\\0&-1&1\\2&3&4\end{bmatrix}\begin{bmatrix}1&2&3\\0&1&0\\1&1&0\end{bmatrix}\end{aligned}\]
  14. First row:
    \[\begin{aligned}(BA)_{11}&=(-1)(1)+1(0)+0(1)\\&=-1,\end{aligned}\]
    \[\begin{aligned}(BA)_{12}&=(-1)(2)+1(1)+0(1)\\&=-2+1\\&=-1,\end{aligned}\]
    \[\begin{aligned}(BA)_{13}&=(-1)(3)+1(0)+0(0)\\&=-3.\end{aligned}\]
  15. Second row:
    \[\begin{aligned}(BA)_{21}&=0(1)+(-1)(0)+1(1)\\&=1,\end{aligned}\]
    \[\begin{aligned}(BA)_{22}&=0(2)+(-1)(1)+1(1)\\&=-1+1\\&=0,\end{aligned}\]
    \[\begin{aligned}(BA)_{23}&=0(3)+(-1)(0)+1(0)\\&=0.\end{aligned}\]
  16. Third row:
    \[\begin{aligned}(BA)_{31}&=2(1)+3(0)+4(1)\\&=2+4\\&=6,\end{aligned}\]
    \[\begin{aligned}(BA)_{32}&=2(2)+3(1)+4(1)\\&=4+3+4\\&=11,\end{aligned}\]
    \[\begin{aligned}(BA)_{33}&=2(3)+3(0)+4(0)\\&=6.\end{aligned}\]
  17. Therefore,
    \[\boxed{BA=\begin{bmatrix}-1&-1&-3\\1&0&0\\6&11&6\end{bmatrix}}\]
  18. Step 3: Compare \(AB\) and \(BA\)
  19. \[AB=\begin{bmatrix}5&8&14\\0&-1&1\\-1&0&1\end{bmatrix}\]
    and
    \[BA=\begin{bmatrix}-1&-1&-3\\1&0&0\\6&11&6\end{bmatrix}\]
  20. Clearly, the two matrices are not equal. For example,
    \[(AB)_{11}=5\]
    whereas
    \[(BA)_{11}=-1\]
  21. Since
    \[5\neq-1\]
  22. we have
    \[\boxed{AB\neq BA}\]
🎯 Exam Significance
Exam Significance
›

This question tests a fundamental property of matrices: matrix multiplication is not commutative. A common examination error is to assume that, because ordinary multiplication satisfies \(ab=ba\), matrices must also satisfy \(AB=BA\). This is generally false.

To establish \(AB\neq BA\), it is sufficient to find even one corresponding pair of entries that are unequal. However, in a verification question, calculating both complete products is the safest and clearest presentation.

Significance for Competitive Entrance Examination Aspirants

Non-commutativity is frequently used to test conceptual understanding in objective questions. When two matrices are given, never interchange their order without justification.

Remember the distinction:

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

while, whenever the products are defined, associativity gives

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

Thus, associative does not mean commutative. This distinction is particularly important in higher-level matrix problems.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  10 points

  1. Matrix multiplication is generally not commutative.

  2. In general,

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

  3. The order of matrices in a product is important.

  4. Matrix multiplication uses the row-by-column rule.

  5. To prove \(AB\neq BA\), one unequal corresponding entry is sufficient.

  6. For a complete verification, calculate both \(AB\) and \(BA\).

  7. Do not confuse non-commutativity with non-associativity.

  8. Matrix multiplication can be associative even though it is not generally commutative.

  9. Always check signs carefully when multiplying matrices containing negative entries.

  10. In this problem, the corrected result for part (i) is

    \[ AB= \begin{bmatrix} 7&1\\ 33&34 \end{bmatrix}, \qquad BA= \begin{bmatrix} 16&5\\ 39&25 \end{bmatrix} \]
← Q13
14 / 22  ·  64%
Q15 →
Q15
NUMERIC3 marks
Find \[A^2-5A+6I\] if \[A=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&6\end{bmatrix}\]
📘 Concept & Theory
Concept/Theory
›

This question involves evaluating a polynomial expression in a matrix. For a square matrix \(A\), expressions such as

\[ A^2-5A+6I \]

are evaluated using matrix multiplication, scalar multiplication and matrix addition or subtraction.

Here, \(A^2\) means

\[ A^2=A\cdot A, \]

while \(5A\) means that every entry of \(A\) is multiplied by \(5\). The symbol \(I\) denotes the identity matrix of the same order as \(A\). Since \(A\) is a \(3\times3\) matrix, the required identity matrix is

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

Therefore,

\[6I=\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}\]

A crucial point is that \(A^2\) is obtained by matrix multiplication; it does not mean squaring every individual entry of \(A\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given matrix \(A\).

  2. Calculate \(A^2=A\cdot A\) using the row-by-column rule.

  3. Calculate \(5A\) by scalar multiplication.

  4. Write the \(3\times3\) identity matrix \(I\).

  5. Calculate \(6I\).

  6. Substitute \(A^2\), \(5A\), and \(6I\) into \(A^2-5A+6I\).

  7. Perform the matrix subtraction and addition entry by entry.

  8. State the final matrix clearly.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  26 steps
  1. Step 1: Write the Given Matrix
  2. \[A=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&6\end{bmatrix}\]
  3. Step 2: Calculate \(A^2\)
  4. By definition,
    \[A^2=A\cdot A\]
  5. Therefore,
    \[\begin{aligned}A^2&=\begin{bmatrix}2&0&1\\2&1&3\\1&-1&6\end{bmatrix}\begin{bmatrix}2&0&1\\2&1&3\\1&-1&6\end{bmatrix}\end{aligned}\]
  6. We calculate each entry using the row-by-column rule.
  7. First Row of \(A^2\)
  8. Entry \((1,1)\):
    \[\begin{aligned}(A^2)_{11}&=2(2)+0(2)+1(1)\\&=4+0+1\\&=5\end{aligned}\]
  9. Entry \((1,2)\):
    \[\begin{aligned}(A^2)_{12}&=2(0)+0(1)+1(-1)\\&=0+0-1\\&=-1\end{aligned}\]
  10. Entry \((1,3)\):
    \[\begin{aligned}(A^2)_{13}&=2(1)+0(3)+1(6)\\&=2+0+6\\&=8\end{aligned}\]
  11. Thus, the first row is
    \[\begin{bmatrix}5&-1&8\end{bmatrix}\]
  12. Second Row of \(A^2\)
  13. Entry \((2,1)\):
    \[\begin{aligned}(A^2)_{21}&=2(2)+1(2)+3(1)\\&=4+2+3\\&=9\end{aligned}\]
  14. Entry \((2,2)\):
    \[\begin{aligned}(A^2)_{22}&=2(0)+1(1)+3(-1)\\&=0+1-3\\&=-2\end{aligned}\]
  15. Entry \((2,3)\):
    \[\begin{aligned}(A^2)_{23}&=2(1)+1(3)+3(6)\\&=2+3+18\\&=23\end{aligned}\]
  16. Thus, the second row is
    \[\begin{bmatrix}9&-2&23\end{bmatrix}\]
  17. Third Row of \(A^2\)
  18. Entry \((3,1)\):
    \[\begin{aligned}(A^2)_{31}&=1(2)+(-1)(2)+6(1)\\&=2-2+6\\&=6\end{aligned}\]
  19. Entry \((3,2)\):
    \[\begin{aligned}(A^2)_{32}&=1(0)+(-1)(1)+6(-1)\\&=0-1-6\\&=-7\end{aligned}\]
  20. Entry \((3,3)\):
    \[\begin{aligned}(A^2)_{33}&=1(1)+(-1)(3)+6(6)\\&=1-3+36\\&=34\end{aligned}\]
  21. Thus,
    \[A^2=\begin{bmatrix}5&-1&8\\9&-2&23\\6&-7&34\end{bmatrix}\]
  22. Step 3: Calculate \(5A\)
  23. Multiplying every entry of \(A\) by \(5\),
    \[\begin{aligned}5A&=5\begin{bmatrix}2&0&1\\2&1&3\\1&-1&6\end{bmatrix}\\[8pt] &=\begin{bmatrix}5(2)&5(0)&5(1)\\5(2)&5(1)&5(3)\\5(1)&5(-1)&5(6)\end{bmatrix}\\[8pt] &=\begin{bmatrix}10&0&5\\10&5&15\\5&-5&30\end{bmatrix}\end{aligned}\]
  24. Hence,
    \[5A=\begin{bmatrix}10&0&5\\10&5&15\\5&-5&30\end{bmatrix}\]
  25. Step 4: Write the Identity Matrix \(I\)
  26. Since \(A\) is a \(3\times3\) matrix, the identity matrix is
    \[I=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\]
  27. Step 5: Calculate \(6I\)
  28. \[\begin{aligned}6I&=6\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\\&=\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}\end{aligned}\]
  29. Thus,
    \[6I=\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}\]
  30. Step 6: Calculate \(A^2-5A+6I\)
  31. Substituting the three matrices obtained above,
    \[\begin{aligned}A^2-5A+6I&=\begin{bmatrix}5&-1&8\\9&-2&23\\6&-7&34\end{bmatrix} -\begin{bmatrix}10&0&5\\10&5&15\\5&-5&30\end{bmatrix} +\begin{bmatrix}6&0&0\\0&6&0\\0&0&6\end{bmatrix}\end{aligned}\]
  32. Now calculate each entry carefully.
  33. First row:
    \[\begin{aligned}(1,1)&:5-10+6=1,\\(1,2)&:-1-0+0=-1,\\(1,3)&:8-5+0=3.\end{aligned}\]
  34. Second row:
    \[\begin{aligned}(2,1)&:9-10+0=-1,\\(2,2)&:-2-5+6=-1,\\(2,3)&:23-15+0=8.\end{aligned}\]
  35. Third row:
    \[\begin{aligned}(3,1)&:6-5+0=1,\\(3,2)&:-7-(-5)+0=-2,\\(3,3)&:34-30+6=10.\end{aligned}\]
  36. Therefore,
    \[\begin{aligned}A^2-5A+6I&=\begin{bmatrix}1&-1&3\\-1&-1&8\\1&-2&10\end{bmatrix}\end{aligned}\]
🎯 Exam Significance
Exam Significance
›

This problem tests several core Class 12 matrix concepts together: matrix multiplication, scalar multiplication, identity matrices and matrix addition or subtraction. It is particularly useful for learning how to evaluate polynomial expressions involving matrices.

The most important examination point is that

\[ A^2=A\cdot A, \]

and not a matrix obtained by squaring every element individually. Students should also maintain the correct signs when subtracting \(5A\).

Significance for Competitive Entrance Examination Aspirants

Matrix polynomial expressions frequently appear in objective questions. The ability to calculate \(A^2\) accurately and then combine it with scalar multiples of \(A\) and \(I\) is essential.

In more advanced problems, expressions such as

\[ A^2-5A+6I \]

may sometimes be simplified using factorisation or a polynomial relation satisfied by \(A\). However, such factorisation must respect matrix multiplication. For example,

\[ A^2-5A+6I=(A-2I)(A-3I), \]

because \(I\) commutes with every square matrix \(A\). This observation can sometimes provide a faster route in competitive examinations.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. For a square matrix,

    \[ A^2=A\cdot A. \]

  2. Do not square individual entries when calculating \(A^2\).

  3. A scalar multiplies every entry of a matrix.

  4. The identity matrix \(I\) must have the same order as \(A\).

  5. For this question,

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

  6. When subtracting \(5A\), every entry of \(5A\) must be subtracted with its correct sign.

  7. The corrected final answer is

    \[ \boxed{ A^2-5A+6I= \begin{bmatrix} 1&-1&3\\ -1&-1&8\\ 1&-2&10 \end{bmatrix}}. \]

  8. For competitive examinations, recognise useful matrix polynomial factorisations such as

    \[ A^2-5A+6I=(A-2I)(A-3I). \]
← Q14
15 / 22  ·  68%
Q16 →
Q16
NUMERIC3 marks
If \[A=\begin{bmatrix}1 & 0 & 2\\0 & 2 & 1\\2 & 0 & 3\end{bmatrix},\] prove that \[A^3-6A^2+7A+2I=0\]
📘 Concept & Theory
Concept/Theory
›

This question illustrates the evaluation of a polynomial in a matrix. To verify an identity involving \(A^3\), \(A^2\), \(A\), and \(I\), we first calculate \(A^2\), then calculate \(A^3= A\cdot A^2\), and finally substitute these matrices into the given expression.

Here, \(I\) denotes the identity matrix of the same order as \(A\). Since \(A\) is a \(3\times3\) matrix,

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

Matrix multiplication is performed using the row-by-column rule. The element in the \(i\)-th row and \(j\)-th column of a product is obtained by multiplying the \(i\)-th row of the first matrix by the \(j\)-th column of the second matrix and adding the resulting products.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given matrix \(A\) and identity matrix \(I\).

  2. Calculate \(A^2=A\cdot A\) using row-by-column multiplication.

  3. Calculate \(A^3=A\cdot A^2\).

  4. Calculate \(6A^2\), \(7A\), and \(2I\).

  5. Substitute all four matrices into \(A^3-6A^2+7A+2I\).

  6. Simplify corresponding entries and show that every entry is zero.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  19 steps
  1. Step 1: Write the given matrix
  2. \[A=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\]
  3. We first calculate \(A^2\)
  4. Step 2: Calculate \(A^2\)
  5. \[A^2=A\cdot A\]
    \[A^2=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\]
  6. Now calculate each element using the row-by-column rule.
  7. First row:
    \[a_{11}=1(1)+0(0)+2(2)=1+0+4=5\]
    \[a_{12}=1(0)+0(2)+2(0)=0+0+0=0\]
    \[a_{13}=1(2)+0(1)+2(3)=2+0+6=8\]
  8. Second row:
    \[a_{21}=0(1)+2(0)+1(2)=0+0+2=2\]
    \[a_{22}=0(0)+2(2)+1(0)=0+4+0=4\]
    \[a_{23}=0(2)+2(1)+1(3)=0+2+3=5\]
  9. Third row:
    \[a_{31}=2(1)+0(0)+3(2)=2+0+6=8\]
    \[a_{32}=2(0)+0(2)+3(0)=0+0+0=0\]
    \[a_{33}=2(2)+0(1)+3(3)=4+0+9=13\]
  10. Therefore,
    \[A^2=\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}\]
  11. Step 3: Calculate \(A^3\)
  12. We use
    \[A^3=A\cdot A^2\]
  13. \[A^3=\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}\]
  14. First row:
    \[a_{11}=1(5)+0(2)+2(8)=5+0+16=21\]
    \[a_{12}=1(0)+0(4)+2(0)=0+0+0=0\]
    \[a_{13}=1(8)+0(5)+2(13)=8+0+26=34\]
  15. Second row:
    \[a_{21}=0(5)+2(2)+1(8)=0+4+8=12\]
    \[a_{22}=0(0)+2(4)+1(0)=0+8+0=8\]
    \[a_{23}=0(8)+2(5)+1(13)=0+10+13=23\]
  16. Third row:
    \[a_{31}=2(5)+0(2)+3(8)=10+0+24=34\]
    \[a_{32}=2(0)+0(4)+3(0)=0+0+0=0\]
    \[a_{33}=2(8)+0(5)+3(13)=16+0+39=55\]
  17. Hence,
    \[A^3=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix}\]
  18. Step 4: Calculate \(6A^2\)
  19. \[6A^2=6\begin{bmatrix}5&0&8\\2&4&5\\8&0&13\end{bmatrix}\]
    \[6A^2=\begin{bmatrix}30&0&48\\12&24&30\\48&0&78\end{bmatrix}\]
  20. Step 5: Calculate \(7A\)
  21. \[7A=7\begin{bmatrix}1&0&2\\0&2&1\\2&0&3\end{bmatrix}\]
    \[7A=\begin{bmatrix}7&0&14\\0&14&7\\14&0&21\end{bmatrix}\]
  22. Step 6: Calculate \(2I\)
  23. \[I=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\]
    Therefore,
    \[2I=\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix}\]
  24. Step 7: Substitute in the required expression
  25. \[A^3-6A^2+7A+2I\]
    \[=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55\end{bmatrix} -\begin{bmatrix}30&0&48\\12&24&30\\48&0&78\end{bmatrix} +\begin{bmatrix}7&0&14\\0&14&7\\14&0&21\end{bmatrix} +\begin{bmatrix}2&0&0\\0&2&0\\0&0&2\end{bmatrix} \]
  26. Combine the corresponding entries:
    \[=\begin{bmatrix}21-30+7+2 & 0-0+0+0 & 34-48+14+0\\12-12+0+0 & 8-24+14+2 & 23-30+7+0\\34-48+14+0 & 0-0+0+0 & 55-78+21+2\end{bmatrix}\]
    \[=\begin{bmatrix}0&0&0\\0&0&0\\0&0&0\end{bmatrix}\]
  27. Since the zero matrix is denoted by \(0\),
    \[\boxed{A^3-6A^2+7A+2I=0}\]
  28. Hence Proved
🎯 Exam Significance
Exam Significance
›

This type of matrix-polynomial problem is important for Class 12 board examinations because it tests matrix multiplication, scalar multiplication, identity matrices, and verification of matrix identities in a single problem.

For competitive entrance examinations, the question develops speed and accuracy in matrix multiplication and recognition of matrix-polynomial identities. Such identities are also closely connected with the Cayley-Hamilton theorem, which is useful for reducing higher powers of a square matrix.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. For a square matrix, \(A^2=A\cdot A\) and \(A^3=A\cdot A^2\).

  2. Matrix multiplication must always be performed using the row-by-column rule.

  3. Scalar multiplication of a matrix means multiplying every element by the scalar.

  4. The identity matrix \(I\) has 1 on its principal diagonal and 0 elsewhere.

  5. When adding or subtracting matrices, corresponding elements are operated on entry by entry.

  6. A matrix polynomial such as \(A^3-6A^2+7A+2I\) can be evaluated by calculating each required power and scalar multiple separately.

  7. The final zero matrix verifies the required matrix identity.

  8. Higher powers of matrices can often be reduced using polynomial identities, a technique that becomes especially useful in competitive mathematics.
← Q15
16 / 22  ·  73%
Q17 →
Q17
NUMERIC3 marks
If \[A=\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix},\quad I=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix},\] find \(k\) so that \[A^2=kA-2I\]
📘 Concept & Theory
Concept/Theory
›

This question uses scalar multiplication of a matrix, matrix multiplication, and the identity matrix. The given relation

\[ A^2=kA-2I \]
is a matrix polynomial equation in \(A\).

To determine the unknown scalar \(k\), we first calculate \(A^2\). Then we rearrange the given equation:

\[ A^2=kA-2I \]
\[ kA=A^2+2I. \]

After calculating the matrix on the right-hand side, we compare it with \(A\). Since \(A\) is non-zero, the corresponding entries directly determine the value of \(k\).

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the given matrix \(A\) and identity matrix \(I\).

  2. Calculate \(A^2=A\cdot A\) using row-by-column multiplication.

  3. Use the given relation \(A^2=kA-2I\).

  4. Rearrange it to obtain \(kA=A^2+2I\).

  5. Calculate \(A^2+2I\).

  6. Compare the resulting matrix with \(kA\) and determine \(k\).

  7. Verify the value of \(k\) in the original equation.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  17 steps
  1. Step 1: Write the Given Matrices
  2. \[A=\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix},\quad I=\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}\]
  3. Step 2: Calculate \(A^2\)
  4. By definition,
    \[A^2=A\cdot A\]
  5. Therefore,
    \[A^2=\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}\]
  6. We calculate each entry using row-by-column multiplication.
  7. First row, first column:
    \[a_{11}=3(3)+(-2)(4)\]
    \[=9-8\]
    \[=1\]
  8. First row, second column:
    \[a_{12}=3(-2)+(-2)(-2)\]
    \[=-6+4\]
    \[=-2\]
  9. Second row, first column:
    \[a_{21}=4(3)+(-2)(4)\]
    \[=12-8\]
    \[=4\]
  10. Second row, second column:
    \[a_{22}=4(-2)+(-2)(-2)\]
    \[=-8+4\]
    \[=-4\]
  11. Hence,
    \[A^2=\begin{bmatrix}1 & -2\\4 & -4\end{bmatrix}\]
  12. Step 3: Use the Given Relation
  13. Given —
    \[A^2=kA-2I\]
  14. Adding \(2I\) to both sides gives
    \[A^2+2I=kA\]
  15. Therefore,
    \[kA=A^2+2I\]
  16. Step 4: Calculate \(2I\)
  17. \[2I=2\begin{bmatrix}1 & 0\\0 & 1\end{bmatrix}=\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}\]
  18. Step 5: Calculate \(A^2+2I\)
  19. \[A^2+2I=\begin{bmatrix}1 & -2\\4 & -4\end{bmatrix}+\begin{bmatrix}2 & 0\\0 & 2\end{bmatrix}\]
  20. Adding corresponding entries,
    \[A^2+2I=\begin{bmatrix}1+2 & -2+0\\4+0 & -4+2\end{bmatrix}\]
    \[=\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}\]
  21. Thus,
    \[kA=\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}\]
  22. But
    \[A=\begin{bmatrix}3 & -2\\4 & -2\end{bmatrix}\]
  23. Hence,
    \[kA=A\]
  24. Comparing any corresponding non-zero entry, for example the \((1,1)\)-entry,
    \[3k=3\]
    \[k=\frac{3}{3}\]
    \[\boxed{k=1}\]
🎯 Exam Significance
Exam Significance
›

This problem is important for Class 12 board examinations because it tests the fundamental concept of matrix multiplication together with algebraic manipulation of matrices. It also reinforces the fact that matrix equations can be solved by treating equal matrices through their corresponding entries.

In board examinations, such questions commonly assess whether the student can calculate \(A^2\) accurately, handle the identity matrix correctly, and simplify a matrix equation without skipping intermediate steps. A clear verification at the end also provides a reliable check against arithmetic errors.

Significance for Competitive Entrance Examinations

The relation

\[ A^2=kA-2I \]
is an example of a polynomial relation satisfied by a matrix. Such relations are useful in higher-level matrix problems, especially when powers of a matrix need to be simplified without repeatedly performing matrix multiplication.

For competitive examinations, recognizing that a relation involving \(A^2\), \(A\), and \(I\) can be manipulated algebraically is particularly useful. It can lead to faster evaluation of higher powers of \(A\), inverse-related expressions, and matrix polynomial problems.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. Matrix multiplication must always be performed using the row-by-column rule.

  2. The identity matrix \(I\) satisfies \(AI=IA=A\) for a square matrix \(A\).

  3. From \(A^2=kA-2I\), we obtain \(kA=A^2+2I\).

  4. After calculating \(A^2+2I\), compare corresponding entries with \(kA\).

  5. Only corresponding entries of equal matrices can be equated.

  6. The required value is \(\boxed{k=1}\).

  7. The final substitution verifies that \(A^2=A-2I\).
← Q16
17 / 22  ·  77%
Q18 →
Q18
NUMERIC3 marks
If \[A=\begin{bmatrix}0 & -\tan\dfrac{\alpha}{2}\\\tan\dfrac{\alpha}{2} & 0\end{bmatrix}\] and \(I\) is the identity matrix of order \(2\), show that \[I+A=(I-A)\begin{bmatrix}\cos\alpha & -\sin\alpha\\\sin\alpha & \cos\alpha\end{bmatrix}\]
📘 Concept & Theory
Concept/Theory
›

This problem combines matrix addition, matrix multiplication, and standard trigonometric half-angle identities.

The important idea is to calculate both sides of the required identity separately. The left-hand side \(I+A\) is obtained directly by adding corresponding entries. For the right-hand side, we first calculate \(I-A\), multiply it by the given rotation matrix, and then simplify the resulting entries using

\[ \sin\alpha=2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2} \]
\[ \cos\alpha =\cos^2\dfrac{\alpha}{2}-\sin^2\dfrac{\alpha}{2}. \]

We also use

\[ 1+\cos\alpha=2\cos^2\dfrac{\alpha}{2}, \qquad 1-\cos\alpha=2\sin^2\dfrac{\alpha}{2}. \]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the identity matrix \(I\) and the given matrix \(A\).

  2. Calculate \(I+A\).

  3. Calculate \(I-A\).

  4. Multiply \(I-A\) by the given trigonometric matrix.

  5. Simplify each of the four resulting entries separately.

  6. Show that the resulting matrix is exactly equal to \(I+A\).

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  26 steps
  1. Step 1: Write \(I\) and \(A\)
  2. The identity matrix of order \(2\) is
    \[I=\begin{bmatrix}1&0\\0&1\end{bmatrix}\]
  3. Given —
    \[A=\begin{bmatrix}0&-\tan\dfrac{\alpha}{2}\\\tan\dfrac{\alpha}{2}&0\end{bmatrix}\]
  4. Step 2: Calculate \(I+A\)
  5. \[I+A=\begin{bmatrix}1&0\\0&1\end{bmatrix}+ \begin{bmatrix}0&-\tan\dfrac{\alpha}{2}\\\tan\dfrac{\alpha}{2}&0\end{bmatrix}\]
  6. Adding corresponding entries,
    \[I+A=\begin{bmatrix}1+0&0-\tan\dfrac{\alpha}{2}\\0+\tan\dfrac{\alpha}{2}&1+0\end{bmatrix}\]
  7. Therefore,
    \[I+A=\begin{bmatrix}1&-\tan\dfrac{\alpha}{2}\\\tan\dfrac{\alpha}{2}&1\end{bmatrix}\]
  8. Step 3: Calculate \(I-A\)
  9. \[I-A=\begin{bmatrix}1&0\\0&1\end{bmatrix}-\begin{bmatrix}0&-\tan\dfrac{\alpha}{2}\\\tan\dfrac{\alpha}{2}&0\end{bmatrix}\]
  10. Hence,
    \[I-A=\begin{bmatrix}1-0&0-\left(-\tan\dfrac{\alpha}{2}\right)\\0-\tan\dfrac{\alpha}{2}&1-0\end{bmatrix}\]
  11. Therefore,
    \[\boxed{I-A=\begin{bmatrix}1&\tan\dfrac{\alpha}{2}\\-\tan\dfrac{\alpha}{2}&1\end{bmatrix}}\]
  12. Step 4: Calculate the Right-Hand Side
  13. Consider
    \[(I-A)\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}\]
  14. Substituting \(I-A\),
    \[\begin{aligned}&(I-A)\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}\\&=\begin{bmatrix}1&\tan\dfrac{\alpha}{2}\\ -\tan\dfrac{\alpha}{2}&1\end{bmatrix}\begin{bmatrix}\cos\alpha&-\sin\alpha\\\sin\alpha&\cos\alpha\end{bmatrix}\end{aligned}\]
  15. Using row-by-column multiplication,
  16. \[ = \begin{bmatrix} \cos\alpha+\sin\alpha\tan\dfrac{\alpha}{2} & -\sin\alpha+\cos\alpha\tan\dfrac{\alpha}{2} \\ -\cos\alpha\tan\dfrac{\alpha}{2}+\sin\alpha & \sin\alpha\tan\dfrac{\alpha}{2}+\cos\alpha \end{bmatrix} \]
  17. Step 5: Simplify the \((1,1)\)-Entry
  18. Consider
    \[\cos\alpha+\sin\alpha\tan\dfrac{\alpha}{2}.\]
  19. Using
    \[\cos\alpha=\cos^2\dfrac{\alpha}{2}-\sin^2\dfrac{\alpha}{2}\]
    and
    \[\sin\alpha=2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2},\]
  20. we get
    \[\begin{aligned}\cos\alpha+\sin\alpha\tan\dfrac{\alpha}{2} &=\cos^2\dfrac{\alpha}{2} -\sin^2\dfrac{\alpha}{2} +2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2}\left(\dfrac{\sin\dfrac{\alpha}{2}}{\cos\dfrac{\alpha}{2}}\right)\\ &=\cos^2\dfrac{\alpha}{2}-\sin^2\dfrac{\alpha}{2}+2\sin^2\dfrac{\alpha}{2}\\ &=\cos^2\dfrac{\alpha}{2}+\sin^2\dfrac{\alpha}{2}\\ &=1\end{aligned}\]
  21. Thus,
    \[\boxed{\cos\alpha+\sin\alpha\tan\dfrac{\alpha}{2}=1}\]
  22. Step 6: Simplify the \((1,2)\)-Entry
  23. Consider
    \[-\sin\alpha+\cos\alpha\tan\dfrac{\alpha}{2}\]
  24. Using the half-angle identities,
  25. \[\begin{aligned}-\sin\alpha+\cos\alpha\tan\dfrac{\alpha}{2} &=-2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2} +\left(\cos^2\dfrac{\alpha}{2} -\sin^2\dfrac{\alpha}{2}\right)\dfrac{\sin\dfrac{\alpha}{2}}{\cos\dfrac{\alpha}{2}}\\ &=-2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2} +\sin\dfrac{\alpha}{2}\left(\dfrac{\cos^2\dfrac{\alpha}{2} -\sin^2\dfrac{\alpha}{2}}{\cos\dfrac{\alpha}{2}}\right)\\ &=-2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2} +\sin\dfrac{\alpha}{2}\left(\cos\dfrac{\alpha}{2} -\dfrac{\sin^2\dfrac{\alpha}{2}}{\cos\dfrac{\alpha}{2}}\right)\end{aligned}\]
  26. A cleaner simplification is obtained by taking the common denominator:
    \[ \begin{aligned} -\sin\alpha+\cos\alpha\tan\dfrac{\alpha}{2} &= -2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2} + \left( \cos^2\dfrac{\alpha}{2} - \sin^2\dfrac{\alpha}{2} \right) \dfrac{\sin\dfrac{\alpha}{2}} {\cos\dfrac{\alpha}{2}}\\ &= \dfrac{ -2\sin\dfrac{\alpha}{2}\cos^2\dfrac{\alpha}{2} + \sin\dfrac{\alpha}{2} \left( \cos^2\dfrac{\alpha}{2} - \sin^2\dfrac{\alpha}{2} \right) } {\cos\dfrac{\alpha}{2}}\\ &= \dfrac{ -\sin\dfrac{\alpha}{2} \left( \cos^2\dfrac{\alpha}{2} + \sin^2\dfrac{\alpha}{2} \right) } {\cos\dfrac{\alpha}{2}}\\ &= -\dfrac{\sin\dfrac{\alpha}{2}} {\cos\dfrac{\alpha}{2}}\\ &= -\tan\dfrac{\alpha}{2}. \end{aligned} \]
  27. Hence,
    \[\boxed{-\sin\alpha+\cos\alpha\tan\dfrac{\alpha}{2}=-\tan\dfrac{\alpha}{2}}\]
  28. Step 7: Simplify the \((2,1)\)-Entry
  29. Consider
    \[-\cos\alpha\tan\dfrac{\alpha}{2}+\sin\alpha\]
    This is the negative of the expression obtained in the previous step:
    \[ \begin{aligned} -\cos\alpha\tan\dfrac{\alpha}{2}+\sin\alpha &= -\left( \cos\alpha\tan\dfrac{\alpha}{2} -\sin\alpha \right)\\ &= \sin\alpha-\cos\alpha\tan\dfrac{\alpha}{2}\\ &= \tan\dfrac{\alpha}{2} \end{aligned} \]
  30. Therefore,
    \[ \boxed{-\cos\alpha\tan\dfrac{\alpha}{2}+\sin\alpha=\tan\dfrac{\alpha}{2}} \]
  31. Step 8: Simplify the \((2,2)\)-Entry
  32. Consider
    \[\sin\alpha\tan\dfrac{\alpha}{2}+\cos\alpha.\]
  33. This is the same expression as the \((1,1)\)-entry, only with the terms written in reverse order. Hence,
    \[ \begin{aligned} \sin\alpha\tan\dfrac{\alpha}{2}+\cos\alpha &= \cos\alpha+\sin\alpha\tan\dfrac{\alpha}{2}\ &=1. \end{aligned} \]
  34. Thus,
    \[\boxed{\sin\alpha\tan\dfrac{\alpha}{2}+\cos\alpha=1}\]
  35. Step 9: Form the Resulting Matrix
  36. Substituting all four simplified entries,
  37. \[ \begin{aligned} &(I-A) \begin{bmatrix} \cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha \end{bmatrix}\\ &= \begin{bmatrix} 1&-\tan\dfrac{\alpha}{2}\\ \tan\dfrac{\alpha}{2}&1 \end{bmatrix}. \end{aligned} \]
  38. But from Step 2,
    \[ I+A= \begin{bmatrix} 1&-\tan\dfrac{\alpha}{2}\\ \tan\dfrac{\alpha}{2}&1 \end{bmatrix} \]
  39. Therefore,
    \[I+A= (I-A) \begin{bmatrix} \cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha \end{bmatrix}\]
  40. Hence Proved
🎯 Exam Significance
Exam Significance
›

For competitive examinations, this problem develops the ability to recognise structured matrices and simplify matrix products efficiently. The matrix

\[ \begin{bmatrix} \cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha \end{bmatrix} \]
is the standard two-dimensional rotation matrix. Understanding its structure is useful in problems involving transformations, determinants, inverses, orthogonal matrices, and composition of rotations.

The identity also illustrates how the substitution

\[ t=\tan\dfrac{\alpha}{2} \]
connects rational expressions in \(t\) with \(\sin\alpha\) and \(\cos\alpha\), a technique that can be valuable in advanced trigonometric simplification.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. Matrix addition is performed entry by entry.

  2. Matrix multiplication is performed using the row-by-column rule.

  3. The identity matrix of order \(2\) is

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

  4. The half-angle identities

    \[ \sin\alpha=2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2} \]
    and
    \[ \cos\alpha=\cos^2\dfrac{\alpha}{2}-\sin^2\dfrac{\alpha}{2} \]
    are central to the simplification.

  5. The useful identity

    \[ \tan\dfrac{\alpha}{2} = \dfrac{\sin\dfrac{\alpha}{2}} {\cos\dfrac{\alpha}{2}} \]
    allows cancellation after expressing \(\sin\alpha\) and \(\cos\alpha\) in half-angle form.

  6. The four entries of the product simplify to

    \[ 1,\quad -\tan\dfrac{\alpha}{2},\quad \tan\dfrac{\alpha}{2},\quad 1. \]

  7. Thus,

    \[ (I-A) \begin{bmatrix} \cos\alpha&-\sin\alpha\\ \sin\alpha&\cos\alpha \end{bmatrix} = I+A. \]
← Q17
18 / 22  ·  82%
Q19 →
Q19
NUMERIC3 marks
A trust fund has ₹30,000 that must be invested in two different types of bonds. The first bond pays 5% interest per year, and the second bond pays 7% interest per year. Using matrix multiplication, determine how to divide ₹30,000 among the two types of bonds if the trust fund must obtain an annual total interest of:
  1. ₹1800
  2. ₹2000
📘 Concept & Theory
Concept/Theory
›

This question illustrates how a system of linear equations can be represented and solved using matrix multiplication. The total amount invested is fixed at ₹30,000, while the annual interest depends on the amounts invested in the two bonds.

Let ₹\(x\) be invested in the first bond. Since the total investment is ₹30,000, the amount invested in the second bond is:

\[ 30000-x \]

The annual interest rates are 5% and 7%, respectively. Therefore, the interest earned from the two bonds is:

\[ \frac{5}{100}x \]

and

\[ \frac{7}{100}(30000-x) \]

These quantities can be represented compactly using matrix multiplication:

\[\begin{bmatrix}x & 30000-x\end{bmatrix}\begin{bmatrix}\frac{5}{100}\\\frac{7}{100}\end{bmatrix}=\begin{bmatrix}\text{Required annual interest}\end{bmatrix}\]
🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Let ₹\(x\) be invested in the 5% bond.

  2. Express the investment in the 7% bond as \(30000-x\).

  3. Represent the investment amounts and interest rates as matrices.

  4. Use matrix multiplication to form the total-interest equation.

  5. Solve the resulting linear equation for \(x\).

  6. Calculate \(30000-x\) to obtain the investment in the second bond.

  7. Verify the resulting investments by calculating the total annual interest.

✏️ Solution
(a) Annual Total Interest = ₹1800
›
Step-by-step Solution  ·  16 steps
  1. Let ₹\(x\) be invested in the first bond, which pays 5% annual interest.
    \[\text{Investment in first bond}=x\]
  2. Therefore, the amount invested in the second bond is:
    \[\text{Investment in second bond}=30000-x\]
  3. The investment amounts can be written as the row matrix:
    \[\begin{bmatrix}x & 30000-x\end{bmatrix}\]
  4. The corresponding annual interest rates can be written as the column matrix:
    \[\begin{bmatrix}\frac{5}{100}\\\frac{7}{100}\end{bmatrix}\]
  5. Since the required annual interest is ₹1800, matrix multiplication gives:
    \[\begin{bmatrix}x & 30000-x\end{bmatrix}\begin{bmatrix}\frac{5}{100}\\ \frac{7}{100}\end{bmatrix}=\begin{bmatrix}1800\end{bmatrix}\]
  6. Multiplying the row matrix by the column matrix:
    \[\frac{5x}{100}+\frac{7(30000-x)}{100}=1800\]
  7. Multiply both sides by 100:
    \[5x+7(30000-x)=180000\]
  8. Expand the brackets:
    \[5x+210000-7x=180000\]
  9. Combine the like terms:
    \[-2x+210000=180000\]
  10. Subtract 210000 from both sides:
    \[-2x=180000-210000\]
    \[-2x=-30000\]
  11. Divide both sides by \(-2\):
    \[x=\frac{-30000}{-2}\]
  12. \[x=15000\]
  13. Therefore, ₹15,000 is invested in the first bond.
  14. The amount invested in the second bond is:
    \[30000-x=30000-15000\]
  15. \[30000-x=15000\]
  16. Therefore, ₹15,000 is invested in the second bond.
💡 Answer
Answer for Part (a)
›
Answer for (a): ₹15,000 in the 5% bond and ₹15,000 in the 7% bond.
✏️ Solution
(b) Annual Total Interest = ₹2000
›
Step-by-step Solution  ·  11 steps
  1. Again, let ₹\(x\) be invested in the first bond at 5% annual interest. Hence, ₹\(30000-x\) is invested in the second bond at 7% annual interest.
  2. Using matrix multiplication:
    \[\begin{bmatrix}x & 30000-x\end{bmatrix}\begin{bmatrix}\frac{5}{100}\\ \frac{7}{100}\end{bmatrix}=\begin{bmatrix}2000\end{bmatrix}\]
  3. Therefore,
    \[\frac{5x}{100}+\frac{7(30000-x)}{100}=2000\]
  4. Multiply both sides by 100:
    \[5x+7(30000-x)=200000\]
  5. Expand the brackets:
    \[5x+210000-7x=200000\]
  6. Combine the like terms:
    \[-2x+210000=200000\]
  7. Subtract 210000 from both sides:
    \[-2x=200000-210000\]
    \[-2x=-10000\]
  8. Divide both sides by \(-2\):
    \[x=\frac{-10000}{-2}\]
    \[x=5000\]
  9. Therefore, ₹5,000 is invested in the first bond.
  10. The amount invested in the second bond is:
    \[30000-x=30000-5000\]
    \[30000-x=25000\]
  11. Therefore, ₹25,000 is invested in the second bond.
💡 Answer
Answer for Part (b)
›
Answer for (b): ₹5,000 in the 5% bond and ₹25,000 in the 7% bond.
🎯 Exam Significance
Exam Significance
›

This problem is important for CBSE Board examinations because it tests matrix multiplication in a practical application involving linear equations. It also strengthens the interpretation of a row matrix multiplied by a column matrix.

For competitive entrance examinations, the problem reinforces quick translation of word problems into algebraic or matrix form. The key skill is identifying the unknown quantities, constructing the corresponding matrix product, and simplifying the resulting equation accurately.

A particularly important observation is that the two investment amounts must add to ₹30,000. This constraint allows the entire problem to be reduced to one variable.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  8 points

  1. Matrix multiplication can be used to model real-life problems involving quantities and rates.

  2. If ₹\(x\) is invested in one bond out of a total of ₹30,000, the remaining investment is ₹\(30000-x\).

  3. A row matrix multiplied by a column matrix produces the total weighted quantity.

  4. For investment problems, always distinguish between the principal amount and the interest rate.

  5. After solving for the unknown, verify the answer using the original conditions.

  6. For ₹1800 annual interest, the investment is equally divided: ₹15,000 at 5% and ₹15,000 at 7%.

  7. For ₹2000 annual interest, the investment is ₹5,000 at 5% and ₹25,000 at 7%.

  8. Checking the total investment and total interest is an effective way to detect algebraic or arithmetic errors.
← Q18
19 / 22  ·  86%
Q20 →
Q20
NUMERIC3 marks
The bookshop of a particular school has 10 dozen Chemistry books, 8 dozen Physics books and 10 dozen Economics books. Their selling prices are ₹80, ₹60 and ₹40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.
📘 Concept & Theory
Concept/Theory
›

This problem demonstrates the use of matrix multiplication to calculate the total revenue obtained from selling different quantities of books at different prices.

A dozen means 12 units. Therefore, the number of books of each subject must first be converted from dozens into individual books.

\[ 1\text{ dozen}=12\text{ books} \]

Thus, the quantities of Chemistry, Physics and Economics books are represented by the row matrix:

\[\begin{bmatrix}10\times12 & 8\times12 & 10\times12\end{bmatrix}\]

The corresponding selling prices are represented by the column matrix:

\[\begin{bmatrix}80\\60\\40\end{bmatrix}\]

The product of these two matrices gives a \(1\times1\) matrix whose entry represents the total amount received from the sale of all the books.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Convert each quantity from dozens into individual books.

  2. Write the quantities as a row matrix.

  3. Write the corresponding selling prices as a column matrix.

  4. Multiply the row matrix by the column matrix.

  5. Add the revenue obtained from each category of books.

  6. Interpret the resulting \(1\times1\) matrix as the total selling amount.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  13 steps
  1. Step 1: Convert Dozens into Individual Books
  2. The bookshop has 10 dozen Chemistry books. Therefore:
    \[10\times12=120\]
    Hence, there are 120 Chemistry books.
  3. The bookshop has 8 dozen Physics books. Therefore:
    \[8\times12=96\]
    Hence, there are 96 Physics books.
  4. The bookshop has 10 dozen Economics books. Therefore:
    \[10\times12=120\]
    Hence, there are 120 Economics books.
  5. Therefore, the quantities of the three types of books are:
    \[\begin{bmatrix}120 & 96 & 120\end{bmatrix}\]
  6. Step 2: Write the Selling Prices as a Column Matrix
  7. The selling prices per book are:
    \[\text{Chemistry =₹}80\]
    \[\text{Physics =₹}60\]
    \[\text{Economics =₹}40\]
  8. Therefore, the corresponding price matrix is:
    \[\begin{bmatrix}80\\60\\40\end{bmatrix}\]
  9. Step 3: Perform Matrix Multiplication
  10. The quantity matrix is of order \(1\times3\), while the price matrix is of order \(3\times1\). Therefore, their product is defined and has order \(1\times1\):
    \[(1\times3)(3\times1)=(1\times1)\]
  11. Hence, the total amount received is:
    \[\begin{bmatrix}120 & 96 & 120\end{bmatrix}\begin{bmatrix}80\\60\\40\end{bmatrix}\]
  12. Using the row-by-column multiplication rule:
  13. \[=\begin{bmatrix}120(80)+96(60)+120(40)\end{bmatrix}\]
  14. Calculate each term separately:
    \[120(80)=9600\]
    \[96(60)=5760\]
    \[120(40)=4800\]
  15. Therefore:
    \[9600+5760+4800=20160\]
  16. Thus:
    \[\begin{bmatrix}120 & 96 & 120\end{bmatrix}\begin{bmatrix}80\\60\\40\end{bmatrix}=\begin{bmatrix}20160\end{bmatrix}\]
  17. Step 4: Interpret the Result
  18. The resulting \(1\times1\) matrix represents the total amount received by the bookshop. Therefore:
    \[\boxed{\text{Total amount}=₹20,160}\]
💡 Answer
Final Answer
›
The bookshop will receive ₹20,160 from selling all the books.
🎯 Exam Significance
Exam Significance
›

This is an important application of matrix multiplication. For CBSE Board examinations, it tests whether the student can correctly arrange quantities and prices according to their corresponding positions before performing matrix multiplication.

For competitive entrance examinations, the key skill is recognising that a row matrix containing quantities multiplied by a column matrix containing corresponding prices produces the total weighted sum. Such problems also test dimensional compatibility and accurate arithmetic.

🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. Always convert quantities given in dozens into individual units before using the selling price per book.

  2. A quantity row matrix can be multiplied by a corresponding price column matrix to obtain total revenue.

  3. The dimensions must be compatible for matrix multiplication.

  4. Here, the multiplication is of the form \((1\times3)(3\times1)\), producing a \(1\times1\) matrix.

  5. The three products \(120(80)\), \(96(60)\) and \(120(40)\) represent the revenues from Chemistry, Physics and Economics books respectively.

  6. The total revenue is the sum of these three individual revenues.

  7. The correct total is ₹20,160, not ₹20,160 obtained by any other rounding or approximation; all quantities and prices are exact.
← Q19
20 / 22  ·  91%
Q21 →
Q21
NUMERIC3 marks
Assume that \(X,\;Y,\;Z,\;W\) and \(P\) are matrices of orders \(2\times n,\;3\times k,\;2\times p,\;n\times3\) and \(p\times k\), respectively. Choose the correct restriction on \(n,\;k\) and \(p\) so that \(PY+WY\) is defined.
  1. \(k=3,\;p=n\)
  2. \(k\) is arbitrary, \(p=2\)
  3. \(p\) is arbitrary, \(k=3\)
  4. \(k=2,\;p=3\)
📘 Concept & Theory
Concept/Theory
›

For two matrices to be multiplied, the number of columns of the first matrix must be equal to the number of rows of the second matrix. If \(A\) is of order \(m\times n\) and \(B\) is of order \(n\times p\), then \(AB\) is defined and has order \(m\times p\).

Also, two matrices can be added only when they have the same order. Therefore, for an expression such as \(PY+WY\), we must first ensure that both products \(PY\) and \(WY\) are defined and then ensure that the resulting matrices have the same order.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Write the orders of \(P,\;Y\) and \(W\).

  2. Apply the matrix multiplication condition to \(PY\).

  3. Determine the order of \(PY\).

  4. Determine the order of \(WY\).

  5. Since \(PY+WY\) is required, equate the orders of \(PY\) and \(WY\).

  6. Identify the corresponding option.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  11 steps
  1. Given —
    \[X\text{ is of order }2\times n\]
    \[Y\text{ is of order }3\times k\]
    \[Z\text{ is of order }2\times p\]
    \[W\text{ is of order }n\times3\]
    \[P\text{ is of order }p\times k\]
  2. Step 1: Determine when \(PY\) is defined
  3. The order of \(P\) is \(p\times k\), while the order of \(Y\) is \(3\times k\).
    \[P_{p\times k}Y_{3\times k}\]
  4. For the product \(PY\) to be defined, the number of columns of \(P\) must equal the number of rows of \(Y\).
    \[k=3\]
  5. Therefore, the first restriction is
    \[\boxed{k=3}\]
  6. With \(k=3\), the order of \(P\) becomes \(p\times3\), and the order of \(Y\) becomes \(3\times3\). Hence,
    \[P_{p\times3}Y_{3\times3}\]
    \[PY\text{ is of order }p\times3\]
  7. More generally, before substituting \(k=3\), we may write the resulting order as \(p\times k\).
  8. Step 2: Determine when \(WY\) is defined
  9. The order of \(W\) is \(n\times3\), and the order of \(Y\) is \(3\times k\).
    \[W_{n\times3}Y_{3\times k}\]
  10. The number of columns of \(W\) is \(3\), and the number of rows of \(Y\) is also \(3\). Therefore, \(WY\) is defined.
    \[WY\text{ is of order }n\times k\]
  11. Since \(k=3\), this becomes
    \[WY\text{ is of order }n\times3\]
  12. Step 3: Apply the condition for addition
  13. The expression \(PY+WY\) is defined only when \(PY\) and \(WY\) have exactly the same order.
    \[PY\text{ is of order }p\times3\]
    \[WY\text{ is of order }n\times3\]
  14. Therefore, their corresponding numbers of rows must be equal:
    \[p=n\]
  15. The number of columns is already the same because \(k=3\).
    \[\boxed{k=3,\quad p=n}\]
💡 Answer
Final Answer
›

Therefore, the required restrictions are

\[\boxed{k=3,\quad p=n}\]

Hence, the correct option is:

\[\boxed{\text{(A)}\;k=3,\;p=n}\]
🎯 Exam Significance
Exam Significance
›

This question tests one of the most fundamental concepts of matrices: compatibility of matrix dimensions. In board examinations, questions involving the order of matrices often require students to distinguish carefully between the conditions for multiplication and addition.

  • For multiplication, remember: inner dimensions must be equal.
  • For addition or subtraction, remember: both matrices must have the same order.
  • Always determine the order of each intermediate product before attempting addition.
  • MCQs frequently test whether students can apply these two conditions simultaneously.
Significance for Competitive Entrance Examinations

This is a useful dimension-analysis problem for JEE and other competitive entrance examinations because matrix-order questions can often be solved without performing any numerical multiplication. The key is to track the dimensions systematically.

A reliable competitive-exam strategy is to write the dimensions explicitly:

\[ (p\times k)(3\times k) \]

which immediately gives

\[ k=3 \]

and then compare the orders of the resulting matrices to obtain

\[ p=n \]
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  6 points

  1. If \(A\) is \(m\times n\) and \(B\) is \(n\times p\), then \(AB\) is defined and has order \(m\times p\).

  2. For \(PY\), \(P\) is \(p\times k\) and \(Y\) is \(3\times k\), so \(k=3\) is necessary.

  3. Once \(k=3\), \(PY\) has order \(p\times3\).

  4. \(WY\) has order \(n\times3\).

  5. For \(PY+WY\) to be defined, \(p\times3=n\times3\), hence \(p=n\).

  6. The correct answer is Option (A).
← Q20
21 / 22  ·  95%
Q22 →
Q22
NUMERIC3 marks
If \(n=p\), then determine the order of the matrix \(7X-5Z\).
  1. \(p\times2\)
  2. \(2\times n\)
  3. \(n\times3\)
  4. \(p\times n\)
📘 Concept & Theory
Concept/Theory
›

Multiplication of a matrix by a scalar does not change its order. Therefore, if \(X\) is of order \(2\times n\), then \(7X\) is also of order \(2\times n\).

Similarly, if \(Z\) is of order \(2\times p\), then \(5Z\) is also of order \(2\times p\).

For the subtraction \(7X-5Z\) to be defined, both matrices must have the same order. Since \(n=p\), their orders become identical.

🗺️ Solution Roadmap
Step-by-step Plan
›

  1. Recall the orders of \(X\) and \(Z\).

  2. Observe that scalar multiplication does not alter matrix order.

  3. Use the given condition \(n=p\).

  4. Verify that subtraction is possible.

  5. State the resulting order and identify the correct option.

✏️ Solution
Complete Solution
›
Step-by-step Solution  ·  10 steps
  1. From the given matrix-order information,
    \[ X\text{ is of order }2\times n \]
    \[ Z\text{ is of order }2\times p \]
  2. Step 1: Determine the order of \(7X\)
  3. Multiplication by the scalar \(7\) changes the entries of \(X\), but it does not change its number of rows or columns.
    \[X_{2\times n}\longrightarrow 7X_{2\times n}\]
  4. Therefore,
    \[\boxed{7X\text{ is of order }2\times n}\]
  5. Step 2: Determine the order of \(5Z\)
  6. Similarly, multiplication by the scalar \(5\) does not change the order of \(Z\).
    \[Z_{2\times p}\longrightarrow 5Z_{2\times p}\]
  7. Therefore,
    \[\boxed{5Z\text{ is of order }2\times p}\]
  8. Step 3: Use the given condition \(n=p\)
  9. We are given
    \[n=p\]
  10. Hence, the order of \(5Z\) can be written as
    \[2\times p=2\times n\]
  11. Thus, both \(7X\) and \(5Z\) have the same order:
    \[ 7X:2\times n \]
    \[ 5Z:2\times n \]
  12. Step 4: Subtract the matrices
  13. Since both matrices have the same order, their subtraction is defined. The resulting matrix retains the same order.
    \[7X-5Z\text{ is of order }2\times n\]
  14. Therefore,
    \[\boxed{\operatorname{Order}(7X-5Z)=2\times n}\]
💡 Answer
Final Answer
›

Hence, the correct option is:

\[\boxed{\text{(B)}\;2\times n)\]
🎯 Exam Significance
Exam Significance
›

This question tests the relationship between scalar multiplication, matrix order, and matrix subtraction. It is important to remember that multiplying a matrix by a scalar never changes its order.

  • If \(A\) is \(m\times n\), then \(kA\) is also \(m\times n\).
  • Two matrices can be added or subtracted only when they have the same order.
  • The order of the resulting sum or difference is the same as the common order of the matrices.
Significance for Competitive Entrance Examinations

Such questions are useful for developing rapid matrix-dimension analysis. No actual matrix entries need to be calculated. In an objective examination, the problem can be solved directly by tracking dimensions.

\[X_{2\times n},\qquad Z_{2\times p}\]
\[n=p\Rightarrow X\text{ and }Z\text{ have the same order}\]
\[\therefore\quad 7X-5Z\text{ has order }2\times n\]
🔑 Key Takeaways
Key Takeaways
›
Key Takeaways  ·  7 points

  1. Scalar multiplication does not change the order of a matrix.

  2. \(7X\) has the same order as \(X\).

  3. \(5Z\) has the same order as \(Z\).

  4. Matrix subtraction requires the two matrices to have the same order.

  5. Since \(n=p\), both \(7X\) and \(5Z\) are of order \(2\times n\).

  6. Therefore, \(7X-5Z\) is also of order \(2\times n\).

  7. The correct answer is Option (B).
← Q21
22 / 22  ·  100%
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All 22 solutions for Matrices covered.

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NCERT Class 12 Maths Matrices Exercise 3.2 Solutions
NCERT Class 12 Maths Matrices Exercise 3.2 Solutions — Complete Notes & Solutions · academia-aeternum.com
NCERT Class 12 Mathematics Chapter 3 Matrices is an important chapter for understanding matrix operations, types of matrices, matrix multiplication, and their applications in higher mathematics. Exercise 3.2 focuses on fundamental matrix operations such as addition, subtraction, scalar multiplication, multiplication of matrices, properties of matrix operations, and solving matrix equations. These concepts form the foundation for several important topics in Class 12 Mathematics, including…
🎓 Class 12 📐 Mathematics 📖 NCERT ✅ Free Access 🏆 CBSE · JEE
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    ✔️ True / False

    Frequently Asked Questions

    NCERT Class 12 Mathematics Chapter 3 Exercise 3.2 focuses on matrix operations including addition, subtraction, scalar multiplication, matrix multiplication, matrix equations, and properties of matrices.

    Exercise 3.2 covers addition and subtraction of matrices, scalar multiplication, multiplication of matrices, verification of matrix properties, matrix equations, and applications of matrix multiplication.

    Two matrices can be added only when they have the same order. Corresponding elements are added to obtain the resulting matrix.

    Two matrices can be subtracted only when they have the same order. Their corresponding elements are subtracted to obtain the result.

    No. Multiplying a matrix by a scalar changes its elements but does not change its order. For example, if A is m × n, then kA is also m × n.

    If A is of order m × n and B is of order n × p, then AB is defined and the resulting matrix is of order m × p. The inner dimensions must be equal.

    Generally, matrix multiplication is not commutative. In general, AB is not equal to BA, even when both products are defined.

    Matrix equations are solved by using matrix addition, subtraction, scalar multiplication, and comparison of corresponding elements. The resulting scalar equations are then solved systematically.

    Exercise 3.2 builds fundamental skills in matrix operations and matrix equations. These concepts are frequently used in board-level questions and provide the foundation for later topics such as determinants and inverse matrices.

    Yes. Exercise 3.2 is useful for competitive exams because matrix-order conditions, multiplication, algebraic identities, and matrix equations are frequently tested in objective and numerical problems.

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