NCERT Solutions for Exercise 3.2 Class 12 Maths Chapter 3 - Matrices

NCERT Solutions for Exercise 3.2 Class 12 Maths Chapter 3 - Matrices

Updated on 23 Apr 2025, 12:51 PM IST

In this exercise, you will learn about the operations on matrices: addition of matrices, multiplication of a matrix by a scalar, subtraction, and multiplication of matrices. Class 12 maths chapter 3 exercise 3.2 of NCERT consists of questions related to properties of multiplication of matrices, like associative, distributive, and existence of multiplicative identity, which are included in this exercise. There are 22 questions given in Exercise 3.2 Class 12 Maths Solutions. These NCERT solutions are created to develop an easy approach to solving the questions. You can take help from these Class 12 Maths Chapter 3 exercise 3.2 solutions to learn the proper approach for solving the questions. It will also learn the concept in a better manner.

Class 12 Maths Chapter 3 Exercise 3.2 Solutions: Download PDF

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Matrices Exercise: 3.2

Question 1(i) Let $A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$, $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$, $C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

Find each of the following:

A + B

Answer:

$A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$

(i) A + B

The addition of matrix can be done as follows

$A+B = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $+ \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$

$A+B = \begin{bmatrix} 2+1 &4+3 \\ 3+(-2) & 2+5 \end{bmatrix}$

$A+B = \begin{bmatrix} 3 &7 \\ 1 & 7 \end{bmatrix}$

Question 1(ii) Let $A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$, $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$, $C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

Find each of the following:

A - B

Answer:

$A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$

(ii) A - B

$A-B = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $- \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$

$A-B = \begin{bmatrix} 2-1 &4-3 \\ 3-(-2) & 2-5 \end{bmatrix}$

$A-B = \begin{bmatrix} 1 &1 \\ 5 & -3 \end{bmatrix}$

Question 1(iii) Let $A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$, $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$, $C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

Find each of the following:

3A - C

Answer:

$A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

(iii) 3A - C

First multiply each element of A with 3 and then subtract C

$3A -C = 3\begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $- \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

$3A -C = \begin{bmatrix} 6 &12 \\ 9 & 6 \end{bmatrix}$ $- \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

$3A -C = \begin{bmatrix} 6-(-2) &12-5 \\ 9-3 & 6-4 \end{bmatrix}$

$3A -C = \begin{bmatrix} 8 &7 \\ 6 & 2 \end{bmatrix}$

Question 1(iv)Let $A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$, $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$, $C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

Find each of the following:

AB

Answer:

$A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$

(iv) AB

$AB = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ $\times \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$

$AB = \begin{bmatrix} 2\times 1+4\times -2 & \, \, \, 2\times 3+4\times 5 \\ 3\times 1+2\times -2 & \, \, \, 3\times 3+2 \times 5 \end{bmatrix}$

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

Question 1(v) Let $A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$, $B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$, $C = \begin{bmatrix} -2 &5 \\ 3 & 4 \end{bmatrix}$

Find each of the following:

BA

Answer:

The multiplication is performed as follows

$A = \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$ ,$B = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$

$BA = \begin{bmatrix} 1 &3 \\ -2 & 5 \end{bmatrix}$ $\times \begin{bmatrix} 2 &4 \\ 3 & 2 \end{bmatrix}$

$BA = \begin{bmatrix} 1\times 2+3\times 3 &1\times 4+3\times 2 \\ -2\times 2+5\times 3& -2\times 4+2\times 5 \end{bmatrix}$

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

Question 2(i). Compute the following:

$\begin{bmatrix} a &b \\ -b& a \end{bmatrix} + \begin{bmatrix} a &b \\ b& a \end{bmatrix}$

Answer:

(i) $\begin{bmatrix} a &b \\ -b& a \end{bmatrix} + \begin{bmatrix} a &b \\ b& a \end{bmatrix}$

$= \begin{bmatrix} a+a &b+b \\ -b+b & a+a \end{bmatrix}$

$= \begin{bmatrix} 2a &2b \\ 0 & 2a \end{bmatrix}$

Question 2(ii). Compute the following:

$\begin{bmatrix} a^2 + b^2& b^2+c^2\\ a^2 + c^2& a^2 + b^2 \end{bmatrix} + \begin{bmatrix} 2ab &2bc \\ -2ac & -2ab \end{bmatrix}$

Answer:

(ii) The addition operation can be performed as follows

$\begin{bmatrix} a^2 + b^2& b^2+c^2\\ a^2 + c^2& a^2 + b^2 \end{bmatrix} + \begin{bmatrix} 2ab &2bc \\ -2ac & -2ab \end{bmatrix}$

$=\begin{bmatrix} a^2 + b^2+2ab& b^2+c^2+2bc\\ a^2 + c^2-2ac& a^2 + b^2-2ab \end{bmatrix}$

$=\begin{bmatrix} (a+b)^2 & (b+c)^2\\ (a-c)^2 & (a-b)^2 \end{bmatrix}$

Question 2(iii). Compute the following:

$\begin{bmatrix} -1 & 4 & -6\\ 8 & 5 & 16\\ 2 & 8 & 5 \end{bmatrix} + \begin{bmatrix} 12 & 7 & 6\\ 8 & 0 &5 \\ 3 & 2 & 4 \end{bmatrix}$

Answer:

(iii) The addition of given three by three matrix is performed as follows

$\begin{bmatrix} -1 & 4 & -6\\ 8 & 5 & 16\\ 2 & 8 & 5 \end{bmatrix} + \begin{bmatrix} 12 & 7 & 6\\ 8 & 0 &5 \\ 3 & 2 & 4 \end{bmatrix}$

$=\begin{bmatrix} -1+12 & 4+7 & -6+6\\ 8+8 & 5+0 & 16+5\\ 2+3 & 8+2 & 5+4 \end{bmatrix}$

$=\begin{bmatrix} 11 & 11 & 0\\ 16 & 5 & 21\\ 5 & 10 & 9 \end{bmatrix}$

Question 2(iv). Compute the following:

$\begin{bmatrix} \cos^2 x &\sin^2 x\\ \sin^2 x & \cos^2x \end{bmatrix} + \begin{bmatrix} \sin^2 x &\cos^2 x\\ \cos^2 x & \sin^2x \end{bmatrix}$

Answer:

(iv) the addition is done as follows

$\begin{bmatrix} \cos^2 x &\sin^2 x\\ \sin^2 x & \cos^2x \end{bmatrix} + \begin{bmatrix} \sin^2 x &\cos^2 x\\ \cos^2 x & \sin^2x \end{bmatrix}$

$=\begin{bmatrix} \cos^2+ \sin^2 x &\sin^2 x+\cos^2 x\\ \sin^2 x+\cos^2 x & \cos^2x+ \sin^2 x \end{bmatrix}$ since $sin^2x+cos^2x=1$

$=\begin{bmatrix} 1 &1\\ 1 & 1 \end{bmatrix}$

Question 3(i). Compute the indicated products.

$\begin{bmatrix} a &b \\ -b &a \end{bmatrix} \begin{bmatrix} a & -b \\ b &a \end{bmatrix}$

Answer:

(i) The multiplication is performed as follows

$\begin{bmatrix} a &b \\ -b &a \end{bmatrix} \begin{bmatrix} a & -b \\ b &a \end{bmatrix}$

$=\begin{bmatrix} a &b \\ -b &a \end{bmatrix} \times \begin{bmatrix} a & -b \\ b &a \end{bmatrix}$

$=\begin{bmatrix} a\times a+b\times b &a\times -b+b\times a \\ -b\times a+a\times b &-b\times -b+a\times a \end{bmatrix}$

$=\begin{bmatrix} a^{2}+b^{2} & 0 \\ 0 & b^{2}+a^{2} \end{bmatrix}$

Question 3(ii). Compute the indicated products.

$\begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}\begin{bmatrix} 2 &3 & 4 \end{bmatrix}$

Answer:

(ii) the multiplication can be performed as follows

$\begin{bmatrix} 1\\ 2\\ 3 \end{bmatrix}\begin{bmatrix} 2 &3 & 4 \end{bmatrix}$

$=\begin{bmatrix} 1\times 2 &1\times 3&1\times 4\\ 2\times 2&2\times 3&2\times 4\\3\times 2&3\times 3&3\times 4 \end{bmatrix}$

$=\begin{bmatrix} 2 &3& 4\\ 4&6&8\\6&9&12 \end{bmatrix}$

Question 3(iii). Compute the indicated products.

$\begin{bmatrix} 1 & -2\\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 &2 &3\\ 2 & 3 & 1 \end{bmatrix}$

Answer:

(iii) The multiplication can be performed as follows

$\begin{bmatrix} 1 & -2\\ 2 & 3 \end{bmatrix}\begin{bmatrix} 1 &2 &3\\ 2 & 3 & 1 \end{bmatrix}$

$=\begin{bmatrix} 1\times 1+(-2)\times 2 & 1\times 2+(-2)\times 3&1\times 3+(-2)\times 1\\ 2\times 1+3\times 2 & 2\times 2+3\times 3&2\times 3+3\times 1 \end{bmatrix}$

Question 3(iv). Compute the indicated products.

$\begin{bmatrix} 2 & 3 & 4\\ 3 & 4 & 5\\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & -3 & 5\\ 0& 2 & 4\\ 3 & 0 & 5 \end{bmatrix}$

Answer:

(iv) The multiplication is performed as follows

$\begin{bmatrix} 2 & 3 & 4\\ 3 & 4 & 5\\ 4 & 5 & 6 \end{bmatrix} \begin{bmatrix} 1 & -3 & 5\\ 0& 2 & 4\\ 3 & 0 & 5 \end{bmatrix}$

$=\begin{bmatrix} 2 & 3 & 4\\ 3 & 4 & 5\\ 4 & 5 & 6 \end{bmatrix}\times \begin{bmatrix} 1 & -3 & 5\\ 0& 2 & 4\\ 3 & 0 & 5 \end{bmatrix}$

$=\begin{bmatrix} 2\times 1+3\times 0+4\times 3 \, \, & 2\times (-3)+3\times 2+4\times 0 \, \, & 2\times 5+3\times 4+4\times 5 \\ 3\times 1+4\times 0+5\times 3 \, \, & 3\times (-3)+4\times 2+5\times 0 & 3\times 5+4\times 4+5\times 5 \\ 4\times 1+5\times 0+6\times 3 \, \, & 4\times (-3)+5\times 2+6\times 0\, \, & 4\times 5+5\times 4+6\times 5 \end{bmatrix}$


$= \begin{bmatrix} 14 & 0 & 42\\ 18 & -1 & 56\\ 22 & -2 & 70 \end{bmatrix}$

Question 3(v). Compute the indicated products.

$\begin{bmatrix} 2 &1 \\ 3 & 2\\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1\\ -1 &2 & 1 \end{bmatrix}$

Answer:

(v) The product can be computed as follows

$\begin{bmatrix} 2 &1 \\ 3 & 2\\ -1 & 1 \end{bmatrix}\begin{bmatrix} 1 & 0 & 1\\ -1 &2 & 1 \end{bmatrix}$

$=\begin{bmatrix} 2 &1 \\ 3 & 2\\ -1 & 1 \end{bmatrix}\times \begin{bmatrix} 1 & 0 & 1\\ -1 &2 & 1 \end{bmatrix}$

$=\begin{bmatrix} 2\times 1+1\times (-1) &2\times 0+1\times (2) & 2\times 1+1\times (1) \\ 3\times 1+2\times (-1) & 3\times 0+2\times (2) &3\times 1+2\times (1) \\ (-1)\times 1+1\times (-1) & (-1)\times 0+1\times (2) & (-1)\times 1+1\times (1) \end{bmatrix}$

$=\begin{bmatrix} 1 &2&3 \\ 1 & 4&5\\ -2 & 2&0 \end{bmatrix}$

Question 3(vi). Compute the indicated products.

$\begin{bmatrix} 3 & -1 & 3\\ -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3\\ 1 & 0\\ 3 & 1 \end{bmatrix}$

Answer:

(vi) The given product can be computed as follows

$\begin{bmatrix} 3 & -1 & 3\\ -1 & 0 & 2 \end{bmatrix}\begin{bmatrix} 2 & -3\\ 1 & 0\\ 3 & 1 \end{bmatrix}$

$=\begin{bmatrix} 3 & -1 & 3\\ -1 & 0 & 2 \end{bmatrix}\times \begin{bmatrix} 2 & -3\\ 1 & 0\\ 3 & 1 \end{bmatrix}$

$=\begin{bmatrix} 3 \times 2+(-1)\times 1+3\times 3\, \, \, & 3 \times (-3)+(-1)\times 0+3\times 1 \\ (-1) \times 2+ 0 \times 1+2\times 3 \, \, \, & (-1) \times -3+0\times 0+2\times 1 \end{bmatrix}$

$=\begin{bmatrix} 14 & -6 \\ 4 & 5 \end{bmatrix}$

Question 4. If $A = \begin{bmatrix} 1 &2 &-3 \\ 5 &0 &2 \\ 1 & -1 &1 \end{bmatrix}$, $B = \begin{bmatrix} 3 &-1 &2 \\ 4 &2 &5 \\ 2 & 0 &3 \end{bmatrix}$ and $C = \begin{bmatrix} 4 &1 &2 \\ 0 &3 &2 \\ 1 & -2 &3 \end{bmatrix}$, then compute (A+B) and (B-C). Also verify that A + (B - C) = (A + B) - C

Answer:

$A = \begin{bmatrix} 1 &2 &-3 \\ 5 &0 &2 \\ 1 & -1 &1 \end{bmatrix}$, $B = \begin{bmatrix} 3 &-1 &2 \\ 4 &2 &5 \\ 2 & 0 &3 \end{bmatrix}$ and $C = \begin{bmatrix} 4 &1 &2 \\ 0 &3 &2 \\ 1 & -2 &3 \end{bmatrix}$

$A+B = \begin{bmatrix} 1 &2 &-3 \\ 5 &0 &2 \\ 1 & -1 &1 \end{bmatrix}$ $+ \begin{bmatrix} 3 &-1 &2 \\ 4 &2 &5 \\ 2 & 0 &3 \end{bmatrix}$

$A+B = \begin{bmatrix} 1+3 &2+(-1) &-3+2 \\ 5+4 &0+2 &2+5 \\ 1+2 & -1+0 &1+3 \end{bmatrix}$

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

$B-C = \begin{bmatrix} 3 &-1 &2 \\ 4 &2 &5 \\ 2 & 0 &3 \end{bmatrix}$ $-\begin{bmatrix} 4 &1 &2 \\ 0 &3 &2 \\ 1 & -2 &3 \end{bmatrix}$

$B-C = \begin{bmatrix} 3-4 &-1-1 &2-2 \\ 4-0 &2-3 &5-2 \\ 2-1 & 0-(-2) &3-3 \end{bmatrix}$

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

Now, to prove A + (B - C) = (A + B) - C

$L.H.S\, \, :\, A+(B-C)$

$A+(B-C)=\begin{bmatrix} 1 &2 &-3 \\ 5 &0 &2 \\ 1 & -1 &1 \end{bmatrix}$ $+ \begin{bmatrix} -1 &-2 &0 \\ 4 &-1 &3 \\ 1 & 2 &0 \end{bmatrix}$ (Puting value of $B-C$ from above)

$A+(B-C)=\begin{bmatrix} 1-1 &2-2 &-3+0 \\ 5+4 &0+(-1) &2+3 \\ 1+1 & -1+2 &1+0 \end{bmatrix}$

$A+(B-C)=\begin{bmatrix} 0 &0 &-3 \\ 9 &-1 &5 \\ 2 & 1 &1 \end{bmatrix}$

$R.H.S\, \, :\, (A+B)-C$

$(A+B)-C = \begin{bmatrix} 4 &1 &-1 \\ 9 &2 &7 \\ 3 & -1 &4 \end{bmatrix}$ $- \begin{bmatrix} 4 &1 &2 \\ 0 &3 &2 \\ 1 & -2 &3 \end{bmatrix}$

$(A+B)-C = \begin{bmatrix} 4-4 &1-1 &-1-2 \\ 9-0 &2-3 &7-2 \\ 3-1 & -1-(-2) &4-3 \end{bmatrix}$

$(A+B)-C = \begin{bmatrix} 0 &0 &-3 \\ 9 &-1 &5 \\ 2 & 1 &1 \end{bmatrix}$

Hence, we can see L.H.S = R.H.S = $\begin{bmatrix} 0 &0 &-3 \\ 9 &-1 &5 \\ 2 & 1 &1 \end{bmatrix}$

Question 5. If $A = \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3}\\ \frac{1}{3} & \frac{2}{3} &\frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix}$ and $B = \begin{bmatrix} \frac{2}{5} & \frac{3}{5}&1\\ \frac{1}{5} & \frac{2}{5} &\frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix}$, then compute 3A - 5B

Answer:

$A = \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3}\\ \frac{1}{3} & \frac{2}{3} &\frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix}$ and $B = \begin{bmatrix} \frac{2}{5} & \frac{3}{5}&1\\ \frac{1}{5} & \frac{2}{5} &\frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix}$

$3A-5B = 3\times \begin{bmatrix} \frac{2}{3} & 1 & \frac{5}{3}\\ \frac{1}{3} & \frac{2}{3} &\frac{4}{3} \\ \frac{7}{3} & 2 & \frac{2}{3} \end{bmatrix}$ $-5\times \begin{bmatrix} \frac{2}{5} & \frac{3}{5}&1\\ \frac{1}{5} & \frac{2}{5} &\frac{4}{5} \\ \frac{7}{5} & \frac{6}{5} & \frac{2}{5} \end{bmatrix}$

$3A-5B = \begin{bmatrix} 2 & 3 & 5\\ 1 & 2 &4 \\ 7 & 6 & 2 \end{bmatrix}$ $- \begin{bmatrix} 2 & 3 & 5\\ 1 & 2 &4 \\ 7 & 6 & 2 \end{bmatrix}$

$3A-5B = \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0 \end{bmatrix}$

$3A-5B = 0$

Question 6. Simplify $\cos\theta\begin{bmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{bmatrix} + \sin\theta\begin{bmatrix} \sin\theta & -\cos\theta\\ \cos\theta & \sin\theta \end{bmatrix}$.

Answer:

The simplification is explained in the following step

$\cos\theta\begin{bmatrix} \cos\theta & \sin\theta\\ -\sin\theta & \cos\theta \end{bmatrix} + \sin\theta\begin{bmatrix} \sin\theta & -\cos\theta\\ \cos\theta & \sin\theta \end{bmatrix}$

$= \begin{bmatrix} \cos^{2}\theta & \sin\theta \cos\theta \\ -\sin\theta \cos\theta & \cos^{2}\theta \end{bmatrix} +\begin{bmatrix} \sin^{2}\theta & - \sin\theta \cos\theta\\ \sin\theta\cos\theta & \sin^{2}\theta \end{bmatrix}$

$= \begin{bmatrix} \cos^{2}\theta+\sin^{2}\theta & \sin\theta \cos\theta - \sin\theta \cos\theta \\ -\sin\theta \cos\theta + \sin\theta \cos\theta & \cos^{2}\theta + \sin^{2}\theta\end{bmatrix}$

$= \begin{bmatrix} 1&0 \\ 0 & 1\end{bmatrix} =I$

the final answer is an identity matrix of order 2

Question 7(i). Find X and Y, if

$X + Y = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix}$ and $X - Y = \begin{bmatrix} 3 &0 \\ 0 &3 \end{bmatrix}$

Answer:

(i) The given matrices are

$X + Y = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix}$ and $X - Y = \begin{bmatrix} 3 &0 \\ 0 &3 \end{bmatrix}$

$X + Y = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix}.............................1$

$X - Y = \begin{bmatrix} 3 &0 \\ 0 &3 \end{bmatrix}.............................2$

Adding equation 1 and 2, we get

$2 X = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix}$ $+ \begin{bmatrix} 3 &0 \\ 0 &3 \end{bmatrix}$

$2 X = \begin{bmatrix} 7+3 &0+0 \\ 2+0 &5+3 \end{bmatrix}$

$2 X = \begin{bmatrix} 10 &0 \\ 2 &8 \end{bmatrix}$

$X = \begin{bmatrix} 5 &0 \\ 1 &4 \end{bmatrix}$

Putting the value of X in equation 1, we get

$\begin{bmatrix} 5 &0 \\ 1 &4 \end{bmatrix}$ $+Y = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix}$

$Y = \begin{bmatrix} 7 &0 \\ 2 &5 \end{bmatrix} -$ $\begin{bmatrix} 5 &0 \\ 1 &4 \end{bmatrix}$

$Y = \begin{bmatrix} 7-5 &0-0 \\ 2-1 &5-4 \end{bmatrix}$

$Y = \begin{bmatrix} 2 &0 \\ 1 &1 \end{bmatrix}$

Question 7(ii). Find X and Y, if

$2X + 3Y = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}$ and $3X + 2Y = \begin{bmatrix} 2 &-2 \\ -1 & 5 \end{bmatrix}$

Answer:

(ii) $2X + 3Y = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}$ and $3X + 2Y = \begin{bmatrix} 2 &-2 \\ -1 & 5 \end{bmatrix}$

$2X + 3Y = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}..........................1$

$3X + 2Y = \begin{bmatrix} 2 &-2 \\ -1 & 5 \end{bmatrix}......................2$

Multiply equation 1 by 3 and equation 2 by 2 and subtract them,

$3(2X + 3Y)-2(3X+2Y) = 3 \times \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}$ $- \, \, \, 2\times \begin{bmatrix} 2 &-2 \\ -1 & 5 \end{bmatrix}$

$6X + 9Y-6X-4Y= \begin{bmatrix} 6 &9 \\ 12 & 0 \end{bmatrix}$ $- \begin{bmatrix} 4 &-4 \\ -2 & 10 \end{bmatrix}$

$9Y-4Y= \begin{bmatrix} 6-4 &9-(-4) \\ 12-(-2) & 0-10 \end{bmatrix}$

$5Y= \begin{bmatrix} 2 &13 \\ 14 & -10 \end{bmatrix}$

$Y= \begin{bmatrix} \frac{2}{5} &\frac{13}{5} \\ \frac{14}{5} & -2 \end{bmatrix}$

Putting value of Y in equation 1 , we get

$2X + 3Y = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}$

$2X + 3 \begin{bmatrix} \frac{2}{5} &\frac{13}{5} \\ \frac{14}{5} & -2 \end{bmatrix} = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}$

$2X + \begin{bmatrix} \frac{6}{5} &\frac{39}{5} \\ \frac{42}{5} & -6 \end{bmatrix} = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix}$

$2X = \begin{bmatrix} 2 &3 \\ 4 & 0 \end{bmatrix} - \begin{bmatrix} \frac{6}{5} &\frac{39}{5} \\ \frac{42}{5} & -6 \end{bmatrix}$

$2X = \begin{bmatrix} 2-\frac{6}{5} &3-\frac{39}{5} \\ 4-\frac{42}{5} & 0 -(-6)\end{bmatrix}$

$2X = \begin{bmatrix} \frac{4}{5} &-\frac{24}{5} \\ -\frac{22}{5} & 6\end{bmatrix}$

$X = \begin{bmatrix} \frac{2}{5} &-\frac{12}{5} \\ -\frac{11}{5} & 3\end{bmatrix}$

Question 8. Find X, if $Y = \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix}$ and $2X+ Y = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix}$

Answer:

$Y = \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix}$

$2X+ Y = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix}$

Substituting the value of Y in the above equation

$2X+ \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix} = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix}$

$2X = \begin{bmatrix} 1 &0 \\ -3 & 2 \end{bmatrix}- \begin{bmatrix} 3 &2 \\ 1 & 4 \end{bmatrix}$

$2X = \begin{bmatrix} 1-3 &0-2 \\ -3-1 & 2-4 \end{bmatrix}$

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

$X = \begin{bmatrix} -1 &-1 \\ -2 & -1 \end{bmatrix}$

Question 9. Find x and y, if $2\begin{bmatrix} 1 & 3\\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0\\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6\\ 1 & 8\end{bmatrix}$

Answer:

$2\begin{bmatrix} 1 & 3\\ 0 & x \end{bmatrix} + \begin{bmatrix} y & 0\\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6\\ 1 & 8\end{bmatrix}$

$\begin{bmatrix} 2 & 6\\ 0 & 2x \end{bmatrix} + \begin{bmatrix} y & 0\\ 1 & 2 \end{bmatrix} = \begin{bmatrix} 5 & 6\\ 1 & 8\end{bmatrix}$

$\begin{bmatrix} 2+y & 6+0\\ 0+1 & 2x+2 \end{bmatrix} = \begin{bmatrix} 5 & 6\\ 1 & 8\end{bmatrix}$

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

Now equating LHS and RHS we can write the following equations

$2+y=5$ $2x+2=8$

$y=5-2$ $2x=8-2$

$y=3$ $2x=6$

$x=3$

Question 10. Solve the equation for x, y, z and t, if $2\begin{bmatrix}x & z \\ y &t \end{bmatrix} + 3\begin{bmatrix} 1 & -1\\ 0 & 2 \end{bmatrix} = 3\begin{bmatrix} 3 & 5\\ 4 & 6 \end{bmatrix}$

Answer:

$2\begin{bmatrix}x & z \\ y &t \end{bmatrix} + 3\begin{bmatrix} 1 & -1\\ 0 & 2 \end{bmatrix} = 3\begin{bmatrix} 3 & 5\\ 4 & 6 \end{bmatrix}$

Multiplying with constant terms and rearranging we can rewrite the matrix as

$\begin{bmatrix}2x &2 z \\ 2y &2t \end{bmatrix} = \begin{bmatrix} 9 &15\\ 12 & 18 \end{bmatrix} - 3\begin{bmatrix} 1& -1\\ 0 & 2 \end{bmatrix}$

$\begin{bmatrix}2x &2 z \\ 2y &2t \end{bmatrix} = \begin{bmatrix} 9 &15\\ 12 & 18 \end{bmatrix} - \begin{bmatrix} 3& -3\\ 0 & 6 \end{bmatrix}$

$\begin{bmatrix}2x &2 z \\ 2y &2t \end{bmatrix} = \begin{bmatrix} 9-3 &15-(-3)\\ 12-0 & 18-6 \end{bmatrix}$

$\begin{bmatrix}2x &2 z \\ 2y &2t \end{bmatrix} = \begin{bmatrix} 6 &18\\ 12 & 12 \end{bmatrix}$

Dividing by 2 on both sides

$\begin{bmatrix}x & z \\ y &t \end{bmatrix} = \begin{bmatrix} 3 &9\\ 6 & 6 \end{bmatrix}$

$x=3,y=6,z=9\, \, and\, \, t=6$

Question 11. 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.

Answer:

$x\begin{bmatrix}2\\3 \end{bmatrix} + y\begin{bmatrix} -1\\1 \end{bmatrix} = \begin{bmatrix} 10\\5 \end{bmatrix}$

$\begin{bmatrix}2x\\3x \end{bmatrix} + \begin{bmatrix} -y\\y \end{bmatrix} = \begin{bmatrix} 10\\5 \end{bmatrix}$

Adding both the matrix in LHS and rewriting

$\begin{bmatrix}2x-y\\3x+y \end{bmatrix} = \begin{bmatrix} 10\\5 \end{bmatrix}$

$2x-y=10........................1$

$3x+y=5........................2$

Adding equation 1 and 2, we get

$5x=15$

$x=3$

Put the value of x in equation 2, we have

$3x+y=5$

$3\times 3+y=5$

$9+y=5$

$y=5-9$

$y=-4$

Question 12. 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.

Answer:

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

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

If two matrices are equal than corresponding elements are also equal.

Thus, we have

$3x=x+4$

$3x-x=4$

$2x=4$

$x=2$

$3y=6+x+y$

Put the value of x

$3y-y=6+2$

$2y=8$

$y=4$

$3w=2w+3$

$3w-2w=3$

$w=3$

$3z=-1+z+w$

$3z-z=-1+3$

$2z=2$

$z=1$

Hence, we have $x=2,y=4,z=1\, \, and\, \, w=3.$

Question 13. If $F(x) = \begin{bmatrix} \cos x & -\sin x& 0\\\sin x &\cos x & 0 \\ 0 &0&1\end{bmatrix}$, show that $F(x) F(y) = F(x + y)$.

Answer:

$F(x) = \begin{bmatrix} \cos x & -\sin x& 0\\\sin x &\cos x & 0 \\ 0 &0&1\end{bmatrix}$

To prove : $F(x) F(y) = F(x + y)$

$R.H.S : F(x + y)$

$F(x+y) = \begin{bmatrix} \cos (x+y) & -\sin (x+y)& 0\\\sin (x+y) &\cos (x+y) & 0 \\ 0 &0&1\end{bmatrix}$

$L.H.S : F(x) F(y)$

$F(x)F(y) = \begin{bmatrix} \cos x & -\sin x& 0\\\sin x &\cos x & 0 \\ 0 &0&1\end{bmatrix}\times \begin{bmatrix} \cos y & -\sin y& 0\\\sin y &\cos y & 0 \\ 0 &0&1\end{bmatrix}$

$F(x)F(y) = \begin{bmatrix} \cos x \cos y- \sin x\sin y+0 & -\cos x \sin y-\sin x\cos y+0& 0+0+0\\\ sin x\cos y+\cos x \sin y+0 & - \sin x\sin y+\cos x \cos y+0 &0+0+0 \\ 0+0+0 &0+0+0&0+0+1\end{bmatrix}$


$F(x) F(y)= \begin{bmatrix} \cos (x+y) & -\sin (x+y)& 0\\\sin (x+y) &\cos (x+y) & 0 \\ 0 &0&1\end{bmatrix}$

Hence, we have L.H.S. = R.H.S i.e. $F(x) F(y) = F(x + y)$.

Question 14(i). Show that

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

Answer:

To prove:

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

$L.H.S : \begin{bmatrix}5&-1\\6&7 \end{bmatrix}\begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix}$

$= \begin{bmatrix}5\times 2+(-1)\times 3 &5\times 1+(-1)\times 4\\6\times 2+7\times 3&6\times 1+7\times 4 \end{bmatrix}$

$= \begin{bmatrix}7 &1\\33&34 \end{bmatrix}$

$R.H.S : \begin{bmatrix} 2 & 1\\ 3 & 4 \end{bmatrix}\begin{bmatrix}5&-1\\6&7 \end{bmatrix}$

$= \begin{bmatrix} 2\times 5+1\times 6 & 2\times (-1)+1\times 7\\ 3\times 5+4\times 6 & 3\times (-1)+4\times 7 \end{bmatrix}$

$= \begin{bmatrix} 16 & 5\\ 39 & 25 \end{bmatrix}$

Hence, the right-hand side not equal to the left-hand side, that is

Question 14(ii). Show that

$\begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \neq \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix}$

Answer:

To prove the following multiplication of three by three matrices are not equal

$\begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \neq \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix}$

$L.H.S: \begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix}$

$= \begin{bmatrix}1\times(-1)+2\times 0+3\times 2 \, \, \, & 1\times(1)+2\times (-1)+3\times 3\, \, \, &1\times(0)+2\times 1+3\times 4\\0\times(-1)+1\times 0+0\times 2\, \, \, &0\times(1)+1\times (-1)+0\times 3\, \, \, &0\times(0)+1\times 1+0\times 4\\1\times(-1)+1\times 0+0\times 2\, \, \, &1\times(1)+1\times (-1)+0\times 3\, \, \, &1\times(0)+1\times 1+0\times 4 \end{bmatrix}$

$= \begin{bmatrix}5& 8&14\\0&-1&1\\-1&0&1\end{bmatrix}$


$R.H.S : \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix}$

$= \begin{bmatrix}-1\times(1)+1\times 0+0\times 1 \, \, \, & -1\times(2)+1\times (1)+0\times 1\, \, \, &-1\times(3)+1\times 0+0\times 0\\0\times(1)+-(1)\times 0+1\times 1\, \, \, &0\times(2)+(-1)\times (1)+1\times 1\, \, \, &0\times(3)+(-1)\times 0+1\times 0\\2\times(1)+3\times 0+4\times 1\, \, \, &2\times(2)+3\times (1)+4\times 1\, \, \, &2\times(3)+3\times 0+4\times 0 \end{bmatrix}$

$= \begin{bmatrix}-1& -1&-3\\1&0&0\\6&11&6\end{bmatrix}$

Hence, $L.H.S \neq R.H.S$ i.e. $\begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix} \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \neq \begin{bmatrix} -1 & 1 & 0\\ 0 & -1 & 1\\ 2 & 3 & 4 \end{bmatrix} \begin{bmatrix}1& 2&3\\0&1&0\\1&1&0 \end{bmatrix}$.

Question 15. Find$A^2 -5A + 6I$, if

$A = \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}$

Answer:

$A = \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}$

First, we will find ou the value of the square of matrix A

$A\times A = \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}\times \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 2\times 2+0\times 2+1\times 1 & 2\times 0+0\times 1+1\times -1 & 2\times 1+0\times 3+1\times 0\\ 2\times 2+1\times 2+3\times 1& 2\times 0+1\times 1+3\times -1 &2\times 1+1\times 3+3\times 0 \\ 1\times 2+(-1)\times 2+0\times 1 & 1\times 0+(-1)\times 1+0\times -1 & 1\times 1+(-1)\times 3+0\times 0 \end{bmatrix}$

$A^{2} = \begin{bmatrix} 5 & -1 & 2\\ 9 & -2 &5 \\ 0 & -1 & -2 \end{bmatrix}$

$I= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$

$\therefore$ $A^2 -5A + 6I$

$= \begin{bmatrix} 5 & -1 & 2\\ 9 & -2 &5 \\ 0 & -1 & -2 \end{bmatrix}$ $-5 \begin{bmatrix} 2 & 0 & 1\\ 2 & 1 &3 \\ 1 & -1 & 0 \end{bmatrix}$$+6 \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$

$= \begin{bmatrix} 5 & -1 & 2\\ 9 & -2 &5 \\ 0 & -1 & -2 \end{bmatrix}$$- \begin{bmatrix} 10 & 0 & 5\\ 10 & 5 &15 \\ 5 & -5 & 0 \end{bmatrix}$$+\begin{bmatrix} 6 & 0 & 0\\ 0 & 6 &0 \\ 0 & 0 & 6 \end{bmatrix}$

$= \begin{bmatrix} 5-10+6 & -1-0+0 & 2-5+0\\ 9-10+0 & -2-5+6 &5-15+0 \\ 0-5+0 & -1-(-5)+0 & -2-0+6 \end{bmatrix}$

$= \begin{bmatrix} 1 & -1 & -3\\ -1 & -1 &-10 \\ -5 & 4 & 4 \end{bmatrix}$

Question 16. If $A = \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$ prove that $A^3 - 6A^2 + 7A + 2I = 0$.

Answer:

$A = \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$

First, find the square of matrix A and then multiply it with A to get the cube of matrix A

$A\times A = \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$$\times \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$

$A^{2} = \begin{bmatrix}1+0+4&0+0+0&2+0+6\\0+0+2&0+4+0&0+2+3\\2+0+6&0+0+0&4+0+9 \end{bmatrix}$

$A^{2} = \begin{bmatrix}5&0&8\\2&4&5\\8&0&13 \end{bmatrix}$

$A^{3}=A^{2}\times A$

$A^{2}\times A = \begin{bmatrix}5&0&8\\2&4&5\\8&0&13 \end{bmatrix}$ $\times \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$

$A^{3} = \begin{bmatrix}5+0+16&0+0+0&10+0+24\\2+0+10&0+8+0&4+4+15\\8+0+26&0+0+0&16+0+39 \end{bmatrix}$

$A^{3} = \begin{bmatrix}21&0&34\\12&8&23\\34&0&55 \end{bmatrix}$

$I= \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$

$\therefore$ $A^3 - 6A^2 + 7A + 2I = 0$

L.H.S :

$\begin{bmatrix}21&0&34\\12&8&23\\34&0&55 \end{bmatrix}$$- 6\begin{bmatrix}5&0&8\\2&4&5\\8&0&13 \end{bmatrix}$$+7 \begin{bmatrix}1&0&2\\0&2&1\\2&0&3 \end{bmatrix}$$+2 \begin{bmatrix} 1 & 0 & 0\\ 0 & 1 &0 \\ 0 & 0 & 1 \end{bmatrix}$

$=\begin{bmatrix}21&0&34\\12&8&23\\34&0&55 \end{bmatrix}$ $- \begin{bmatrix}30&0&48\\12&24&30\\48&0&78 \end{bmatrix}$ $+ \begin{bmatrix}7&0&14\\0&14&7\\14&0&21 \end{bmatrix}$ $+ \begin{bmatrix} 2 & 0 & 0\\ 0 & 2 &0 \\ 0 & 0 & 2 \end{bmatrix}$

$=\begin{bmatrix}21-30+7+2&0-0+0+0&34-48+14+0\\12-12+0+0&8-24+14+2&23-30+7+0\\34-48+14+0&0-0+0+0&55-78+21+2 \end{bmatrix}$

$=\begin{bmatrix}30-30&0&48-48\\12-12&24-24&30-30\\48-48&0&78-78 \end{bmatrix}$

$= \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 &0 \\ 0 & 0 & 0 \end{bmatrix}=0$

Hence, L.H.S = R.H.S

i.e.$A^3 - 6A^2 + 7A + 2I = 0$.

Question 17. If $A = \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$ and $I = \begin{bmatrix}1 &0\\0&1 \end{bmatrix}$, find k so that $A^{2} = kA - 2I$.

Answer:

$A = \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$

$I = \begin{bmatrix}1 &0\\0&1 \end{bmatrix}$

$A \times A= \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$$\times \begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$

$A^{2} = \begin{bmatrix}9-8 &-6+4\\12-8&-8+4 \end{bmatrix}$

$A^{2} = \begin{bmatrix}1 &-2\\4&-4 \end{bmatrix}$

$A^{2} = kA - 2I$

$\begin{bmatrix}1 &-2\\4&-4 \end{bmatrix}=$$k\begin{bmatrix}3 &-2\\4&-2 \end{bmatrix} -$$2 \begin{bmatrix}1 &0\\0&1 \end{bmatrix}$

$\begin{bmatrix}1 &-2\\4&-4 \end{bmatrix}=$$k\begin{bmatrix}3 &-2\\4&-2 \end{bmatrix} -$$\begin{bmatrix}2 &0\\0&2 \end{bmatrix}$

$\begin{bmatrix}1 &-2\\4&-4 \end{bmatrix}+$ $\begin{bmatrix}2 &0\\0&2 \end{bmatrix}$ $=k\begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$

$\begin{bmatrix}1+2 &-2+0\\4+0&-4+2 \end{bmatrix}$$=\begin{bmatrix}3k&-2k\\4k&-2k \end{bmatrix}$

$\begin{bmatrix}3 &-2\\4&-2 \end{bmatrix}$ $=\begin{bmatrix}3k&-2k\\4k&-2k \end{bmatrix}$

We have,$3=3k$

$k=\frac{3}{3}=1$

Hence, the value of k is 1.

Question 18. If $A = \begin{bmatrix} 0&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}& 0\end{bmatrix}$ and I is the identity matrix of order 2, show that$I + A = (I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

Answer:

$A = \begin{bmatrix} 0&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}& 0\end{bmatrix}$

$I = \begin{bmatrix}1 &0\\0&1 \end{bmatrix}$

To prove : $I + A = (I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

L.H.S : $I+A$

$I+A = \begin{bmatrix}1 &0\\0&1 \end{bmatrix}$$+ \begin{bmatrix} 0&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}& 0\end{bmatrix}$

$I+A = \begin{bmatrix} 1+0&0-\tan\frac{\alpha}{2}\\0+\tan\frac{\alpha}{2}&1+ 0\end{bmatrix}$

$I+A = \begin{bmatrix} 1&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}&1\end{bmatrix}$

R.H.S : $(I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

$(I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$$= (\begin{bmatrix}1 &0\\0&1 \end{bmatrix}-$ $\begin{bmatrix} 0&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}& 0\end{bmatrix})$$\times \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

$(I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$ $=\begin{bmatrix} 1-0&0-(-\tan\frac{\alpha}{2})\\0-\tan\frac{\alpha}{2}&1- 0\end{bmatrix}$ $\times \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

$(I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$$=\begin{bmatrix} 1&\tan\frac{\alpha}{2}\\-\tan\frac{\alpha}{2}&1\end{bmatrix}$ $\times \begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$

$=\begin{bmatrix} \cos\alpha + \sin \alpha\tan\frac{\alpha}{2} &- \sin \alpha+ \cos \alpha \tan\frac{\alpha}{2}\\-\tan\frac{\alpha}{2} \cos\alpha + \sin \alpha &\tan\frac{\alpha}{2} \sin\alpha + \cos \alpha \end{bmatrix}$

$=\begin{bmatrix} 1-2\sin^{2}\frac{\alpha }{2} + 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2}\tan\frac{\alpha}{2} &- 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2}+ (2\cos^{2} \frac{\alpha }{2} -1)\tan\frac{\alpha}{2}\\-\tan\frac{\alpha}{2} (2\cos^{2} \frac{\alpha }{2} -1) + 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2} &\tan\frac{\alpha}{2} 2\sin\frac{\alpha } {2} \ cos \frac{\alpha }{2} + 1-2\sin^{2}\frac{\alpha }{2} \end{bmatrix}$

$=\begin{bmatrix} 1-2\sin^{2}\frac{\alpha }{2} + 2\sin^{2}\frac{\alpha }{2} &- 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2}+ 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2} -\tan\frac{\alpha}{2}\\-2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2}+\tan\frac{\alpha}{2} + 2\sin\frac{\alpha }{2} \ cos \frac{\alpha }{2} & 2\sin^{2}\frac{\alpha } {2} + 1-2\sin^{2}\frac{\alpha }{2} \end{bmatrix}$


$= \begin{bmatrix} 1&-\tan\frac{\alpha}{2}\\\tan\frac{\alpha}{2}&1\end{bmatrix}$

Hence, we can see L.H.S = R.H.S

i.e. $I + A = (I- A)\begin{bmatrix} \cos\alpha & -\sin\alpha\\ \sin\alpha & \cos\alpha \end{bmatrix}$.

Question 19(i). A trust fund has Rs. 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 Rs 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

Rs. 1800

Answer:

Let Rs. x be invested in the first bond.

Money invested in second bond = Rs (3000-x)

The first bond pays 5% interest per year and the second bond pays 7% interest per year.

To obtain an annual total interest of Rs. 1800, we have

$\begin{bmatrix}x &(30000-x) \end{bmatrix}$ $\begin{bmatrix} \frac{5}{100} \\ \frac{7}{100} \end{bmatrix}$ $=1800$ (simple interest for 1 year $=\frac{pricipal\times rate}{100}$ )

$\frac{5}{100}x+\frac{7}{100}(30000-x) = 1800$

$5x+210000-7x=180000$

$210000-180000=7x-5x$

$30000=2x$

$x=15000$

Thus, to obtain an annual total interest of Rs. 1800, the trust fund should invest Rs 15000 in the first bond and Rs 15000 in the second bond.

Question 19(ii). A trust fund has Rs. 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 Rs 30,000 among the two types of bonds. If the trust fund must obtain an annual total interest of:

Rs. 2000

Answer:

Let Rs. x be invested in the first bond.

Money invested in second bond = Rs (3000-x)

The first bond pays 5% interest per year and the second bond pays 7% interest per year.

To obtain an annual total interest of Rs. 1800, we have

$\begin{bmatrix} x & (30000 - x) \end{bmatrix} \begin{bmatrix} \frac{5}{100} \\ \frac{7}{100} \end{bmatrix} = 2000$

(Simple interest for 1 year = $\frac{\text{Principal} \times \text{Rate}}{100}$)

$\frac{5}{100}x + \frac{7}{100}(30000 - x) = 2000$

$\frac{5x + 210000 - 7x}{100} = 2000$

$\frac{210000 - 2x}{100} = 2000$

$210000 - 2x = 200000$

$210000 - 200000 = 2x$

$10000 = 2x$

$x = 5000$

Thus, to obtain an annual total interest of Rs. 2000, the trust fund should invest Rs 5000 in the first bond and Rs 25000 in the second bond.

Question 20. The bookshop of a particular school has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books. Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively. Find the total amount the bookshop will receive from selling all the books using matrix algebra.

Answer:

The bookshop has 10 dozen chemistry books, 8 dozen physics books, 10 dozen economics books.

Their selling prices are Rs 80, Rs 60 and Rs 40 each respectively.

The total amount the bookshop will receive from selling all the books:

$12$$\begin{bmatrix}10 &8&10 \end{bmatrix}$ $\begin{bmatrix}80\\60\\40 \end{bmatrix}$

$=12(10\times 80+8\times 60+10\times 40)$

$= 12(800+480+ 400)$

$= 12(1680)$

$=20160$

The total amount the bookshop will receive from selling all the books is 20160.

Question 21 Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k, respectively. Choose the correct answer in Exercises 21 and 22.

Q21. The restriction on n, k and p so that PY + WY will be defined are:
(A) $k = 3, p = n$

(B) k is arbitrary,$p = 2$

(C) p is arbitrary, $k = 3$

(D) $k = 2, p = 3$

Answer:

P and Y are of order \( p \times k \) and \( 3 \times k \) respectively.

Therefore, \( PY \) will be defined only if \( k = 3 \), i.e., the order of \( PY \) is \( p \times k \).

W and Y are of order \( n \times 3 \) and \( 3 \times k \) respectively.

Therefore, \( WY \) is defined because the number of columns of \( W \) is equal to the number of rows of \( Y \), which is 3, i.e., the order of \( WY \) is \( n \times k \).

Matrices \( PY \) and \( WY \) can only be added if they both have the same order, i.e., \( p \times k = n \times k \Rightarrow p = n \).

Therefore, \( k = 3 \), \( p = n \) are restrictions on \( n \), \( k \), and \( p \) so that \( PY + WY \) will be defined.

Option (A) is correct.

Question 22 Assume X, Y, Z, W and P are matrices of order 2 × n, 3 × k, 2 × p, n × 3 and p × k,
respectively. Choose the correct answer in Exercises 21 and 22. If n = p, then the order of the matrix $7X - 5Z$ is:
(A) p × 2
(B) 2 × n
(C) n × 3
(D) p × n

Answer:

$X$ has order $2 \times n$.

Therefore, $7X$ also has order $2 \times n$.

$Z$ has order $2 \times p$.

Therefore, $5Z$ also has order $2 \times p$.

Matrices $7X$ and $5Z$ can only be subtracted if they both have the same order, i.e., $2 \times n = 2 \times p$, and it is given that $p = n$.

We can say that both matrices have order $2 \times n$.

Therefore, the order of $7X - 5Z$ is $2 \times n$.

Option (B) is correct.

Also Read,

Topics covered in Chapter 3: Matrices: Exercise 3.2

  • Introduction
  • Topics covered
  • Addition of matrices: If $\mathrm{A}=\left[a_{i j}\right]$ and $\mathrm{B}=\left[b_{i j}\right]$ are two matrices of the same order, say $m \times n$. Then, the sum of the two matrices A and B is defined as a matrix $\mathrm{C}=\left[c_{i j}\right]_{m \times n}$, where $c_{i j}=a_{i j}+b_{i j}$, for all possible values of $i$ and $j$.
  • Multiplication of a matrix by a scalar: Multiplication of a matrix by a scalar as follows: if $\mathrm{A}=\left[a_{i j}\right]_{m \times n}$ is a matrix and $k$ is a scalar, then $k \mathrm{~A}$ is another matrix which is obtained by multiplying each element of A by the scalar $k$.
  • Difference of matrices: If $\mathrm{A}=\left[a_{i j}\right], \mathrm{B}=\left[b_{i j}\right]$ are two matrices of the same order, say $m \times n$, then difference $\mathrm{A}-\mathrm{B}$ is defined as a matrix $\mathrm{D}=\left[d_{i j}\right]$, where $d_{i j}=a_{i j}-b_{i j}$, for all value of $i$ and $j$. In other words, $\mathrm{D}=\mathrm{A}-\mathrm{B}=\mathrm{A}+(-1) \mathrm{B}$, that is, the sum of the matrix A and the matrix - B.
  • Properties of matrix addition
  1. Commutative Law: If $\mathrm{A}=\left[a_{i j}\right], \mathrm{B}=\left[b_{i j}\right]$ are matrices of the same order, say $m \times n$, then $\mathrm{A}+\mathrm{B}=\mathrm{B}+\mathrm{A}$.
  2. Associative Law: For any three matrices $\mathrm{A}=\left[a_{i j}\right], \mathrm{B}=\left[b_{i j}\right], \mathrm{C}=\left[c_{i j}\right]$ of the same order, say $m \times n,(A+B)+C=A+(B+C)$.
  3. Existence of additive identity: Let $\mathrm{A}=\left[a_{i j}\right]$ be an $m \times n$ matrix and O be an $m \times n$ zero matrix, then $\mathrm{A}+\mathrm{O}=\mathrm{O}+\mathrm{A}=\mathrm{A}$. In other words, O is the additive identity for matrix addition.
  4. The existence of additive inverse: Let $\mathrm{A}=\left[a_{i j}\right]_{m \times n}$ be any matrix, then we have another matrix as $-\mathrm{A}=\left[-a_{i j}\right]_{m \times n}$ such that $\mathrm{A}+(-\mathrm{A}) = (-\mathrm{A})+\mathrm{A} =\mathrm{O}$. So, - A is the additive inverse of A or the negative of A.
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  • Properties of scalar multiplication of a matrix
  1. $k(\mathrm{~A}+\mathrm{B})=k \mathrm{~A}+k \mathrm{~B}$
  2. $k(\mathrm{~A}+\mathrm{B})=k \mathrm{~A}+k \mathrm{~B}$
  3. $(k+l) \mathrm{A}=k \mathrm{~A}+l \mathrm{~A}$
  • Multiplication of matrices: If $\mathrm{A}=\left[a_{i j}\right]_{m \times n}, \mathrm{~B}=\left[b_{j k}\right]_{n \times p}$, then the $i^{\text {th }}$ row of A is $\left[a_{i 1} a_{i 2} \ldots a_{i n}\right]$ and the $k^{\text {th }}$ column of B is $\left[\begin{array}{c}b_{1 k} \\ b_{2 k} \\ \vdots \\ b_{n k}\end{array}\right]$, then $c_{i k}=a_{i 1} b_{1 k}+a_{i 2} b_{2 k}+a_{i 3} b_{3 k}+\ldots+a_{i n} b_{n k}=\sum_{j=1}^n a_{i j} b_{j k}$.

The matrix $\mathrm{C}=\left[c_{ik}\right]_{m \ m\times p}$ is the product of A and B.

  • Properties of Multiplication
  1. The associative law: For any three matrices A, B, and C. We have $(A B) C=A(B C)$ whenever both sides of the equality are defined.
  2. The distributive law: For three matrices $\mathrm{A}, \mathrm{B}$ and C.

(i) $\mathrm{A}(\mathrm{B}+\mathrm{C})=\mathrm{AB}+\mathrm{AC}$

(ii) $(\mathrm{A}+\mathrm{B}) \mathrm{C}=\mathrm{AC}+\mathrm{BC}$, whenever both sides of equality are defined.

  1. The existence of multiplicative identity: For every square matrix $A$, there exists an identity matrix of the same order such that $\mathrm{IA}=\mathrm{AI}=\mathrm{A}$.

Also, read,

Subject-wise NCERT Exemplar solutions

Students can check these NCERT exemplar links for additional practice.

Frequently Asked Questions (FAQs)

Q: What is the use of matrix ?
A:

Matrix is an important tool useful in science, statistics, research, representation of data, mechanics, optics, electromagnetism, quantum mechanics, and quantum electrodynamics, etc.

Q: What is zero matrix ?
A:

A matrix all of whose entries are zero is called a zero matrix.

Q: What is equal matrix ?
A:

Two matrices are equal matrices if the order and correspondence entities of both matrices are the same.

Q: Does scalar matrix is square matrix ?
A:

Yes, the scalar matrix is a square matrix.

Q: How do you identify a scalar matrix ?
A:

A scalar matrix is a square matrix whose all diagonal elements are equal and all other elements are zero.

Q: Can I get free NCERT solutions for Class 12 maths?
A:

Click here to get NCERT solutions for class 12 maths. Links for solutions to each chapters of NCERT syllabus Class 12 Mathematics is available here. All the exercise questions of NCERT textbook are covered. For more questions solve use NCERT exemplar.

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Questions related to CBSE Class 12th

On Question asked by student community

Have a question related to CBSE Class 12th ?

Hello,

The date of 12 exam is depends on which board you belongs to . You should check the exact date of your exam by visiting the official website of your respective board.

Hope this information is useful to you.

Hello,

Class 12 biology questions papers 2023-2025 are available on cbseacademic.nic.in , and other educational website. You can download PDFs of questions papers with solution for practice. For state boards, visit the official board site or trusted education portal.

Hope this information is useful to you.

Hello Pruthvi,

Taking a drop year to reappear for the Karnataka Common Entrance Test (KCET) is a well-defined process. As a repeater, you are fully eligible to take the exam again to improve your score and secure a better rank for admissions.

The main procedure involves submitting a new application for the KCET through the official Karnataka Examinations Authority (KEA) website when registrations open for the next academic session. You must pay the required application fee and complete all formalities just like any other candidate. A significant advantage for you is that you do not need to retake your 12th board exams. Your previously secured board marks in the qualifying subjects will be used again. Your new KCET rank will be calculated by combining these existing board marks with your new score from the KCET exam. Therefore, your entire focus during this year should be on preparing thoroughly for the KCET to achieve a higher score.

For more details about the KCET Exam preparation, CLICK HERE.

I hope this answer helps you. If you have more queries, feel free to share your questions with us, and we will be happy to assist you.

Thank you, and I wish you all the best in your bright future.

Yes, you can switch from Science in Karnataka State Board to Commerce in CBSE for 12th. You will need a Transfer Certificate from your current school and meet the CBSE school’s admission requirements. Since you haven’t studied Commerce subjects like Accountancy, Economics, and Business Studies, you may need to catch up before or during 12th. Not all CBSE schools accept direct admission to 12th from another board, so some may ask you to join Class 11 first. Make sure to check the school’s rules and plan your subject preparation.



Hello

For the 12th CBSE Hindi Medium board exam, important questions usually come from core chapters like “Madhushala”, “Jhansi ki Rani”, and “Bharat ki Khoj”.
Questions often include essay writing, letter writing, and comprehension passages. Grammar topics like Tenses, Voice Change, and Direct-Indirect Speech are frequently asked.
Students should practice poetry questions on themes and meanings. Important questions also cover summary writing and translation from Hindi to English or vice versa.
Previous years’ question papers help identify commonly asked questions.
Focus on writing practice to improve handwriting and presentation. Time management during exams is key to answering all questions effectively.