Exercise 3.2 (Revised) - Chapter 3 - Matrices - Ncert Solutions class 12 - Maths

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NCERT Solutions Class 12 Maths: Chapter 3 - Matrices | Comprehensive Guide

Ex 3.2 Question 1:

Let

Find each of the following

Answer:

(i)

(ii)

(iii)

(iv) Matrix A has 2 columns. This number is equal to the number of rows in matrix B. Therefore, AB is defined as:

(v) Matrix B has 2 columns. This number is equal to the number of rows in matrix A. Therefore, BA is defined as:

Ex 3.2 Question 2:
Compute the following:
(i) $\left[\begin{array}{rr}a & b \\ -b & a\end{array}\right]+\left[\begin{array}{ll}a & b \\ b & a\end{array}\right]_{\text {(ii) }}\left[\begin{array}{ll}a^2+b^2 & b^2+c^2 \\ a^2+c^2 & a^2+b^2\end{array}\right]+\left[\begin{array}{ll}2 a b & 2 b c \\ -2 a c & -2 a b\end{array}\right]$
(iii) $\left[\begin{array}{lll}-1 & 4 & -6 \\ 8 & 5 & 16 \\ 2 & 8 & 5\end{array}\right]+\left[\begin{array}{lll}12 & 7 & 6 \\ 8 & 0 & 5 \\ 3 & 2 & 4\end{array}\right]$
(v) $\left[\begin{array}{ll}\cos ^2 x & \sin ^2 x \\ \sin ^2 x & \cos ^2 x\end{array}\right]+\left[\begin{array}{cc}\sin ^2 x & \cos ^2 x \\ \cos ^2 x & \sin ^2 x\end{array}\right]$

Answer:
(i)
$
\begin{aligned}
& {\left[\begin{array}{rr}
a & b \\
-b & a
\end{array}\right]+\left[\begin{array}{ll}
a & b \\
b & a
\end{array}\right]=\left[\begin{array}{ll}
a+a & b+b \\
-b+b & a+a
\end{array}\right]=\left[\begin{array}{ll}
2 a & 2 b \\
0 & 2 a
\end{array}\right]} \\
& \text { (ii) }\left[\begin{array}{ll}
a^2+b^2 & b^2+c^2 \\
a^2+c^2 & a^2+b^2
\end{array}\right]+\left[\begin{array}{ll}
2 a b & 2 b c \\
-2 a c & -2 a b
\end{array}\right]
\end{aligned}
$

$\begin{aligned}
&\begin{aligned}
& =\left[\begin{array}{ll}
a^2+b^2+2 a b & b^2+c^2+2 b c \\
a^2+c^2-2 a c & a^2+b^2-2 a b
\end{array}\right] \\
& =\left[\begin{array}{ll}
(a+b)^2 & (b+c)^2 \\
(a-c)^2 & (a-b)^2
\end{array}\right]
\end{aligned}\\
&\text { (iii) }\left[\begin{array}{lll}
-1 & 4 & -6 \\
8 & 5 & 16 \\
2 & 8 & 5
\end{array}\right]+\left[\begin{array}{lll}
12 & 7 & 6 \\
8 & 0 & 5 \\
3 & 2 & 4
\end{array}\right]\\
&\begin{aligned}
& =\left[\begin{array}{ccc}
-1+12 & 4+7 & -6+6 \\
8+8 & 5+0 & 16+5 \\
2+3 & 8+2 & 5+4
\end{array}\right] \\
& =\left[\begin{array}{lll}
11 & 11 & 0 \\
16 & 5 & 21 \\
5 & 10 & 9
\end{array}\right]
\end{aligned}\\
&\begin{aligned}
& \text { (iv) }\left[\begin{array}{cc}
\cos ^2 x & \sin ^2 x \\
\sin ^2 x & \cos ^2 x
\end{array}\right]+\left[\begin{array}{cc}
\sin ^2 x & \cos ^2 x \\
\cos ^2 x & \sin ^2 x
\end{array}\right] \\
& =\left[\begin{array}{ll}
\cos ^2 x+\sin ^2 x & \sin ^2 x+\cos ^2 x \\
\sin ^2 x+\cos ^2 x & \cos ^2 x+\sin ^2 x
\end{array}\right] \\
& =\left[\begin{array}{lr}
1 & 1 \\
1 & 1
\end{array}\right] \quad\left(\because \sin ^2 x+\cos ^2 x=1\right)
\end{aligned}
\end{aligned}$

Ex 3.2 Question 3:
Compute the indicated products
(i) $\left[\begin{array}{rr}a & b \\ -b & a\end{array}\right]\left[\begin{array}{rr}a & -b \\ b & a\end{array}\right]$
(ii) $\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]\left[\begin{array}{lll}2 & 3 & 4\end{array}\right]$
(iii) $\left[\begin{array}{rr}1 & -2 \\ 2 & 3\end{array}\right]\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 1\end{array}\right]$
(iv) $\left[\begin{array}{lll}2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6\end{array}\right]\left[\begin{array}{ccc}1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5\end{array}\right]$
(v) $\left[\begin{array}{cc}2 & 1 \\ 3 & 2 \\ -1 & 1\end{array}\right]\left[\begin{array}{rrr}1 & 0 & 1 \\ -1 & 2 & 1\end{array}\right]$
$\left[\begin{array}{rrr}3 & -1 & 3 \\ -1 & 0 & 2\end{array}\right]\left[\begin{array}{rr}2 & -3 \\ 1 & 0 \\ 3 & 1\end{array}\right]$

(vi)
$
\left[\begin{array}{rrr}
3 & -1 & 3 \\
-1 & 0 & 2
\end{array}\right]\left[\begin{array}{rr}
2 & -3 \\
1 & 0 \\
3 & 1
\end{array}\right]
$

Answer
$
\begin{aligned}
& \text { (i) }\left[\begin{array}{rr}
a & b \\
-b & a
\end{array}\right]\left[\begin{array}{rr}
a & -b \\
b & a
\end{array}\right] \\
& {\left[\begin{array}{rr}
a & b \\
-b & a
\end{array}\right]\left[\begin{array}{ll}
a & -b \\
b & a
\end{array}\right]} \\
& =\left[\begin{array}{ll}
a(a)+b(b) & a(-b)+b(a) \\
-b(a)+a(b) & -b(-b)+a(a)
\end{array}\right] \\
& =\left[\begin{array}{ll}
a^2+b^2 & -a b+a b \\
-a b+a b & b^2+a^2
\end{array}\right]=\left[\begin{array}{ll}
a^2+b^2 & 0 \\
0 & a^2+b^2
\end{array}\right]
\end{aligned}
$

$
\text { (ii) }\left[\begin{array}{l}
1 \\
2 \\
3
\end{array}\right]\left[\begin{array}{lll}
2 & 3 & 4
\end{array}\right]=\left[\begin{array}{lll}
1(2) & 1(3) & 1(4) \\
2(2) & 2(3) & 2(4) \\
3(2) & 3(3) & 3(4)
\end{array}\right]=\left[\begin{array}{ccc}
2 & 3 & 4 \\
4 & 6 & 8 \\
6 & 9 & 12
\end{array}\right]
$
(iii) $\left[\begin{array}{rr}1 & -2 \\ 2 & 3\end{array}\right]\left[\begin{array}{lll}1 & 2 & 3 \\ 2 & 3 & 1\end{array}\right]$
$
\begin{aligned}
& =\left[\begin{array}{lll}
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{array}\right] \\
& =\left[\begin{array}{lll}
1-4 & 2-6 & 3-2 \\
2+6 & 4+9 & 6+3
\end{array}\right]=\left[\begin{array}{rcr}
-3 & -4 & 1 \\
8 & 13 & 9
\end{array}\right]
\end{aligned}
$
(iv) $\left[\begin{array}{lll}2 & 3 & 4 \\ 3 & 4 & 5 \\ 4 & 5 & 6\end{array}\right]\left[\begin{array}{ccc}1 & -3 & 5 \\ 0 & 2 & 4 \\ 3 & 0 & 5\end{array}\right]$

$\begin{aligned}
& =\left[\begin{array}{lll}
2(1)+3(0)+4(3) & 2(-3)+3(2)+4(0) & 2(5)+3(4)+4(5) \\
3(1)+4(0)+5(3) & 3(-3)+4(2)+5(0) & 3(5)+4(4)+5(5) \\
4(1)+5(0)+6(3) & 4(-3)+5(2)+6(0) & 4(5)+5(4)+6(5)
\end{array}\right] \\
& =\left[\begin{array}{lll}
2+0+12 & -6+6+0 & 10+12+20 \\
3+0+15 & -9+8+0 & 15+16+25 \\
4+0+18 & -12+10+0 & 20+20+30
\end{array}\right]=\left[\begin{array}{ccc}
14 & 0 & 42 \\
18 & -1 & 56 \\
22 & -2 & 70
\end{array}\right] \\
& \text { (v) }\left[\begin{array}{cr}
2 & 1 \\
3 & 2 \\
-1 & 1
\end{array}\right]\left[\begin{array}{ccc}
1 & 0 & 1 \\
-1 & 2 & 1
\end{array}\right] \\
& =\left[\begin{array}{ccc}
2(1)+1(-1) & 2(0)+1(2) & 2(1)+1(1) \\
3(1)+2(-1) & 3(0)+2(2) & 3(1)+2(1) \\
-1(1)+1(-1) & -1(0)+1(2) & -1(1)+1(1)
\end{array}\right]
\end{aligned}$

Ex 3.2 Question 4:
If $A=\left[\begin{array}{rrr}1 & 2 & -3 \\ 5 & 0 & 2 \\ 1 & -1 & 1\end{array}\right], B=\left[\begin{array}{rrr}3 & -1 & 2 \\ 4 & 2 & 5 \\ 2 & 0 & 3\end{array}\right]$, and $C=\left[\begin{array}{rrr}4 & 1 & 2 \\ 0 & 3 & 2 \\ 1 & -2 & 3\end{array}\right]$, then compute $(A+B)$ and $(B-C)$. Also, verify that $A+(B-C)=(A+B)-C$

Answer
$
\begin{aligned}
A+B & =\left[\begin{array}{ccc}
1 & 2 & -3 \\
5 & 0 & 2 \\
1 & -1 & 1
\end{array}\right]+\left[\begin{array}{rrr}
3 & -1 & 2 \\
4 & 2 & 5 \\
2 & 0 & 3
\end{array}\right] \\
& =\left[\begin{array}{ccc}
1+3 & 2-1 & -3+2 \\
5+4 & 0+2 & 2+5 \\
1+2 & -1+0 & 1+3
\end{array}\right]=\left[\begin{array}{rrr}
4 & 1 & -1 \\
9 & 2 & 7 \\
3 & -1 & 4
\end{array}\right]
\end{aligned}
$

$\begin{aligned}
B-C & =\left[\begin{array}{ccc}
3 & -1 & 2 \\
4 & 2 & 5 \\
2 & 0 & 3
\end{array}\right]-\left[\begin{array}{ccc}
4 & 1 & 2 \\
0 & 3 & 2 \\
1 & -2 & 3
\end{array}\right] \\
& =\left[\begin{array}{ccc}
3-4 & -1-1 & 2-2 \\
4-0 & 2-3 & 5-2 \\
2-1 & 0-(-2) & 3-3
\end{array}\right]=\left[\begin{array}{rrr}
-1 & -2 & 0 \\
4 & -1 & 3 \\
1 & 2 & 0
\end{array}\right]
\end{aligned}$

$\begin{aligned}
A+(B-C) & =\left[\begin{array}{lll}
1 & 2 & -3 \\
5 & 0 & 2 \\
1 & -1 & 1
\end{array}\right]+\left[\begin{array}{rrr}
-1 & -2 & 0 \\
4 & -1 & 3 \\
1 & 2 & 0
\end{array}\right] \\
& =\left[\begin{array}{lll}
1+(-1) & 2+(-2) & -3+0 \\
5+4 & 0+(-1) & 2+3 \\
1+1 & -1+2 & 1+0
\end{array}\right]=\left[\begin{array}{ccc}
0 & 0 & -3 \\
9 & -1 & 5 \\
2 & 1 & 1
\end{array}\right]
\end{aligned}$

$
\begin{aligned}
(A+B)-C & =\left[\begin{array}{ccc}
4 & 1 & -1 \\
9 & 2 & 7 \\
3 & -1 & 4
\end{array}\right]-\left[\begin{array}{ccc}
4 & 1 & 2 \\
0 & 3 & 2 \\
1 & -2 & 3
\end{array}\right] \\
& =\left[\begin{array}{ccc}
4-4 & 1-1 & -1-2 \\
9-0 & 2-3 & 7-2 \\
3-1 & -1-(-2) & 4-3
\end{array}\right]=\left[\begin{array}{ccc}
0 & 0 & -3 \\
9 & -1 & 5 \\
2 & 1 & 1
\end{array}\right]
\end{aligned}
$

Hence, we have verified that $A+(B-C)=(A+B)-C$.

Ex 3.2 Question 5:
$
A=\left[\begin{array}{ccc}
\frac{2}{3} & 1 & \frac{5}{3} \\
\frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\
\frac{7}{3} & 2 & \frac{2}{3}
\end{array}\right]_{\text {and }} B=\left[\begin{array}{ccc}
\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{array}\right]_{\text {then compute } 3 A-5 B .}
$

Answer
$
\begin{aligned}
3 A-5 B & =3\left[\begin{array}{ccc}
\frac{2}{3} & 1 & \frac{5}{3} \\
\frac{1}{3} & \frac{2}{3} & \frac{4}{3} \\
\frac{7}{3} & 2 & \frac{2}{3}
\end{array}\right]-5\left[\begin{array}{ccc}
\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{array}\right] \\
& =\left[\begin{array}{lll}
2 & 3 & 5 \\
1 & 2 & 4 \\
7 & 6 & 2
\end{array}\right]-\left[\begin{array}{lll}
2 & 3 & 5 \\
1 & 2 & 4 \\
7 & 6 & 2
\end{array}\right]=\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]
\end{aligned}
$

Ex 3.2 Question 6:

Simplify
$
\cos \theta\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]+\sin \theta\left[\begin{array}{rr}
\sin \theta & -\cos \theta \\
\cos \theta & \sin \theta
\end{array}\right]
$

Answer
$
\begin{aligned}
& \cos \theta\left[\begin{array}{rr}
\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{array}\right]+\sin \theta\left[\begin{array}{rr}
\sin \theta & -\cos \theta \\
\cos \theta & \sin \theta
\end{array}\right] \\
& =\left[\begin{array}{lc}
\cos ^2 \theta & \cos \theta \sin \theta \\
-\sin \theta \cos \theta & \cos ^2 \theta
\end{array}\right]+\left[\begin{array}{lc}
\sin ^2 \theta & -\sin \theta \cos \theta \\
\sin \theta \cos \theta & \sin ^2 \theta
\end{array}\right] \\
& =\left[\begin{array}{lc}
\cos ^2 \theta+\sin ^2 \theta & \cos \theta \sin \theta-\sin \theta \cos \theta \\
-\sin \theta \cos \theta+\sin \theta \cos \theta & \cos ^2 \theta+\sin ^2 \theta
\end{array}\right] \\
& =\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \quad\left(\because \cos ^2 \theta+\sin ^2 \theta=1\right) \\
&
\end{aligned}
$

Ex 3.2 Question 7:
Find $X$ and $Y$, if
(i)
$
X+Y=\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right] \text { and } X-Y=\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]
$
(ii)
$
2 X+3 Y=\left[\begin{array}{ll}
2 & 3 \\
4 & 0
\end{array}\right] \text { and } 3 X+2 Y=\left[\begin{array}{rr}
2 & -2 \\
-1 & 5
\end{array}\right]
$

Answer
(i)
$
\begin{aligned}
& X+Y=\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right] \\
& X-Y=\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]
\end{aligned}
$

Adding equations (1) and (2), we get:
$
\begin{aligned}
& 2 X=\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]+\left[\begin{array}{ll}
3 & 0 \\
0 & 3
\end{array}\right]=\left[\begin{array}{ll}
7+3 & 0+0 \\
2+0 & 5+3
\end{array}\right]=\left[\begin{array}{ll}
10 & 0 \\
2 & 8
\end{array}\right] \\
& \therefore X=\frac{1}{2}\left[\begin{array}{ll}
10 & 0 \\
2 & 8
\end{array}\right]=\left[\begin{array}{ll}
5 & 0 \\
1 & 4
\end{array}\right]
\end{aligned}
$

Now, $X+Y=\left[\begin{array}{ll}7 & 0 \\ 2 & 5\end{array}\right]$
$
\begin{aligned}
& \Rightarrow\left[\begin{array}{ll}
5 & 0 \\
1 & 4
\end{array}\right]+Y=\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right] \\
& \Rightarrow Y=\left[\begin{array}{ll}
7 & 0 \\
2 & 5
\end{array}\right]-\left[\begin{array}{ll}
5 & 0 \\
1 & 4
\end{array}\right] \\
& \Rightarrow Y=\left[\begin{array}{ll}
7-5 & 0-0 \\
2-1 & 5-4
\end{array}\right] \\
& \therefore Y=\left[\begin{array}{ll}
2 & 0 \\
1 & 1
\end{array}\right]
\end{aligned}
$
(ii)
$
\begin{aligned}
& 2 X+3 Y=\left[\begin{array}{ll}
2 & 3 \\
4 & 0
\end{array}\right] \\
& 3 X+2 Y=\left[\begin{array}{rr}
2 & -2 \\
-1 & 5
\end{array}\right]
\end{aligned}
$

Multiplying equation (3) with (2), we get:
$
\begin{aligned}
& 2(2 X+3 Y)=2\left[\begin{array}{ll}
2 & 3 \\
4 & 0
\end{array}\right] \\
& \Rightarrow 4 X+6 Y=\left[\begin{array}{ll}
4 & 6 \\
8 & 0
\end{array}\right]
\end{aligned}
$

Multiplying equation (4) with (3), we get:

$
\begin{aligned}
& 3(3 X+2 Y)=3\left[\begin{array}{rr}
2 & -2 \\
-1 & 5
\end{array}\right] \\
& \Rightarrow 9 X+6 Y=\left[\begin{array}{rr}
6 & -6 \\
-3 & 15
\end{array}\right]
\end{aligned}
$

From (5) and (6), we have:
$
\begin{aligned}
& (4 X+6 Y)-(9 X+6 Y)=\left[\begin{array}{ll}
4 & 6 \\
8 & 0
\end{array}\right]-\left[\begin{array}{rr}
6 & -6 \\
-3 & 15
\end{array}\right] \\
& \Rightarrow-5 X=\left[\begin{array}{ll}
4-6 & 6-(-6) \\
8-(-3) & 0-15
\end{array}\right]=\left[\begin{array}{ll}
-2 & 12 \\
11 & -15
\end{array}\right] \\
& \therefore X=-\frac{1}{5}\left[\begin{array}{cc}
-2 & 12 \\
11 & -15
\end{array}\right]=\left[\begin{array}{ll}
\frac{2}{5} & -\frac{12}{5} \\
-\frac{11}{5} & 3
\end{array}\right]
\end{aligned}
$

Now,
$
2 X+3 Y=\left[\begin{array}{ll}
2 & 3 \\
4 & 0
\end{array}\right]
$
$
\begin{aligned}
& \Rightarrow 2\left[\begin{array}{ll}
\frac{2}{5} & -\frac{12}{5} \\
-\frac{11}{5} & 3
\end{array}\right]+3 Y=\left[\begin{array}{ll}
2 & 3 \\
4 & 0
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{ll}
\frac{4}{5} & -\frac{24}{5} \\
-\frac{22}{5} & 6
\end{array}\right]+3 Y=\left[\begin{array}{ll}
2 & 3 \\
4 & 0
\end{array}\right] \\
& \Rightarrow 3 Y=\left[\begin{array}{ll}
2 & 3 \\
4 & 0
\end{array}\right]-\left[\begin{array}{ll}
\frac{4}{5} & -\frac{24}{5} \\
-\frac{22}{5} & 6
\end{array}\right]
\end{aligned}
$

$
\therefore Y=\frac{1}{3}\left[\begin{array}{ll}
\frac{6}{5} & \frac{39}{5} \\
\frac{42}{5} & -6
\end{array}\right]=\left[\begin{array}{ll}
\frac{2}{5} & \frac{13}{5} \\
\frac{14}{5} & -2
\end{array}\right]
$

Ex 3.2 Question 8:

Find $X$, if
$
Y=\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right] \text { and } 2 X+Y=\left[\begin{array}{rr}
1 & 0 \\
-3 & 2
\end{array}\right]
$

Answer
$
\begin{aligned}
& 2 X+Y=\left[\begin{array}{rr}
1 & 0 \\
-3 & 2
\end{array}\right] \\
& \Rightarrow 2 X+\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]=\left[\begin{array}{rr}
1 & 0 \\
-3 & 2
\end{array}\right] \\
& \Rightarrow 2 X=\left[\begin{array}{rr}
1 & 0 \\
-3 & 2
\end{array}\right]-\left[\begin{array}{ll}
3 & 2 \\
1 & 4
\end{array}\right]=\left[\begin{array}{ll}
1-3 & 0-2 \\
-3-1 & 2-4
\end{array}\right] \\
& \Rightarrow 2 X=\left[\begin{array}{ll}
-2 & -2 \\
-4 & -2
\end{array}\right] \\
& \therefore X=\frac{1}{2}\left[\begin{array}{ll}
-2 & -2 \\
-4 & -2
\end{array}\right]=\left[\begin{array}{ll}
-1 & -1 \\
-2 & -1
\end{array}\right]
\end{aligned}
$

Ex 3.2 Question 9:
Find $x$ and $y$, if $2\left[\begin{array}{ll}1 & 3 \\ 0 & x\end{array}\right]+\left[\begin{array}{ll}y & 0 \\ 1 & 2\end{array}\right]=\left[\begin{array}{ll}5 & 6 \\ 1 & 8\end{array}\right]$
Answer
$
\begin{aligned}
& 2\left[\begin{array}{ll}
1 & 3 \\
0 & x
\end{array}\right]+\left[\begin{array}{ll}
y & 0 \\
1 & 2
\end{array}\right]=\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cc}
2 & 6 \\
0 & 2 x
\end{array}\right]+\left[\begin{array}{ll}
y & 0 \\
1 & 2
\end{array}\right]=\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{cc}
2+y & 6 \\
1 & 2 x+2
\end{array}\right]=\left[\begin{array}{ll}
5 & 6 \\
1 & 8
\end{array}\right]
\end{aligned}
$

Comparing the corresponding elements of these two matrices, we have:
$
\begin{aligned}
& 2+y=5 \\
& \Rightarrow y=3
\end{aligned}
$
$
\begin{aligned}
& 2 x+2=8 \\
& \Rightarrow x=3 \\
& \therefore x=3 \text { and } y=3
\end{aligned}
$

Ex 3.2 Question 10:
Solve the equation for $x, y, z$ and $t$ if
$
2\left[\begin{array}{ll}
x & z \\
y & t
\end{array}\right]+3\left[\begin{array}{rr}
1 & -1 \\
0 & 2
\end{array}\right]=3\left[\begin{array}{ll}
3 & 5 \\
4 & 6
\end{array}\right]
$

Answer
$
\begin{aligned}
& 2\left[\begin{array}{ll}
x & z \\
y & t
\end{array}\right]+3\left[\begin{array}{lr}
1 & -1 \\
0 & 2
\end{array}\right]=3\left[\begin{array}{ll}
3 & 5 \\
4 & 6
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{ll}
2 x & 2 z \\
2 y & 2 t
\end{array}\right]+\left[\begin{array}{ll}
3 & -3 \\
0 & 6
\end{array}\right]=\left[\begin{array}{ll}
9 & 15 \\
12 & 18
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{ll}
2 x+3 & 2 z-3 \\
2 y & 2 t+6
\end{array}\right]=\left[\begin{array}{ll}
9 & 15 \\
12 & 18
\end{array}\right]
\end{aligned}
$

Comparing the corresponding elements of these two matrices, we get:

$
\begin{aligned}
& 2 x+3=9 \\
& \Rightarrow 2 x=6 \\
& \Rightarrow x=3
\end{aligned}
$
$
\begin{aligned}
& 2 y=12 \\
& \Rightarrow y=6 \\
& 2 z-3=15 \\
& \Rightarrow 2 z=18 \\
& \Rightarrow z=9
\end{aligned}
$
$
\begin{aligned}
& 2 z-3=15 \\
& \Rightarrow 2 z=18 \\
& \Rightarrow z=9
\end{aligned}
$
$
\begin{aligned}
& 2 t+6=18 \\
& \Rightarrow 2 t=12 \\
& \Rightarrow t=6
\end{aligned}
$
$\therefore x=3, y=6, z=9$, and $t=6$

Ex 3.2 Question 11:
If $x\left[\begin{array}{l}2 \\ 3\end{array}\right]+y\left[\begin{array}{c}-1 \\ 1\end{array}\right]=\left[\begin{array}{c}10 \\ 5\end{array}\right]$, find values of $x$ and $y$.
Answer
$
x\left[\begin{array}{l}
2 \\
3
\end{array}\right]+y\left[\begin{array}{c}
-1 \\
1
\end{array}\right]=\left[\begin{array}{l}
10 \\
5
\end{array}\right]
$

$\begin{aligned}
& \Rightarrow\left[\begin{array}{l}
2 x \\
3 x
\end{array}\right]+\left[\begin{array}{c}
-y \\
y
\end{array}\right]=\left[\begin{array}{l}
10 \\
5
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{l}
2 x-y \\
3 x+y
\end{array}\right]=\left[\begin{array}{l}
10 \\
5
\end{array}\right]
\end{aligned}$

Comparing the corresponding elements of these two matrices, we get:
$
2 x-y=10 \text { and } 3 x+y=5
$

Adding these two equations, we have:
$
\begin{aligned}
& 5 x=15 \\
& \Rightarrow x=3
\end{aligned}
$

Now, $3 x+y=5$
$
\begin{aligned}
& \Rightarrow y=5-3 x \\
& \Rightarrow y=5-9=-4 \\
& \therefore x=3 \text { and } y=-4
\end{aligned}
$

Ex 3.2 Question 12:
Given $3\left[\begin{array}{ll}x & y \\ z & w\end{array}\right]=\left[\begin{array}{cc}x & 6 \\ -1 & 2 w\end{array}\right]+\left[\begin{array}{cc}4 & x+y \\ z+w & 3\end{array}\right]$, find the values of $x, y, z$ and w.

Answer
$
\begin{aligned}
& 3\left[\begin{array}{ll}
x & y \\
z & w
\end{array}\right]=\left[\begin{array}{cc}
x & 6 \\
-1 & 2 w
\end{array}\right]+\left[\begin{array}{cc}
4 & x+y \\
z+w & 3
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{ll}
3 x & 3 y \\
3 z & 3 w
\end{array}\right]=\left[\begin{array}{ll}
x+4 & 6+x+y \\
-1+z+w & 2 w+3
\end{array}\right]
\end{aligned}
$

Comparing the corresponding elements of these two matrices, we get:
$
\begin{aligned}
& 3 x=x+4 \\
& \Rightarrow 2 x=4 \\
& \Rightarrow x=2
\end{aligned}
$

$
\begin{aligned}
& 3 y=6+x+y \\
& \Rightarrow 2 y=6+x=6+2=8 \\
& \Rightarrow y=4
\end{aligned}
$
$
\begin{aligned}
& 3 w=2 w+3 \\
& \Rightarrow w=3
\end{aligned}
$
$
\begin{aligned}
& 3 z=-1+z+w \\
& \Rightarrow 2 z=-1+w=-1+3=2 \\
& \Rightarrow z=1
\end{aligned}
$
$\therefore x=2, y=4, z=1$, and $w=3$

Ex 3.2 Question 13:
$
F(x)=\left[\begin{array}{ccc}
\cos x & -\sin x & 0 \\
\sin x & \cos x & 0 \\
0 & 0 & 1
\end{array}\right] \text {, show that } F(x) F(y)=F(x+y) \text {. }
$

Answer
$
\begin{aligned}
& F(x)=\left[\begin{array}{ccc}
\cos x & -\sin x & 0 \\
\sin x & \cos x & 0 \\
0 & 0 & 1
\end{array}\right], F(y)=\left[\begin{array}{ccc}
\cos y & -\sin y & 0 \\
\sin y & \cos y & 0 \\
0 & 0 & 1
\end{array}\right] \\
& F(x+y)=\left[\begin{array}{ccc}
\cos (x+y) & -\sin (x+y) & 0 \\
\sin (x+y) & \cos (x+y) & 0 \\
0 & 0 & 1
\end{array}\right]
\end{aligned}
$

$\begin{aligned}
& F(x) F(y) \\
& =\left[\begin{array}{ccc}
\cos x & -\sin x & 0 \\
\sin x & \cos x & 0 \\
0 & 0 & 1
\end{array}\right]\left[\begin{array}{ccc}
\cos y & -\sin y & 0 \\
\sin y & \cos y & 0 \\
0 & 0 & 1
\end{array}\right] \\
& =\left[\begin{array}{ccc}
\cos x \cos y-\sin x \sin y+0 & -\cos x \sin y-\sin x \cos y+0 & 0 \\
\sin x \cos y+\cos x \sin y+0 & -\sin x \sin y+\cos x \cos y+0 & 0 \\
0 & 0 & 0
\end{array}\right] \\
& =\left[\begin{array}{ccc}
\cos (x+y) & -\sin (x+y) & 0 \\
\sin (x+y) & \cos (x+y) & 0 \\
0 & 0 & 1
\end{array}\right] \\
& =F(x+y) \\
& \therefore F(x) F(y)=F(x+y)
\end{aligned}$

Ex 3.2 Question 14:
Show that
(i) $\left[\begin{array}{rr}5 & -1 \\ 6 & 7\end{array}\right]\left[\begin{array}{ll}2 & 1 \\ 3 & 4\end{array}\right] \neq\left[\begin{array}{ll}2 & 1 \\ 3 & 4\end{array}\right]\left[\begin{array}{rr}5 & -1 \\ 6 & 7\end{array}\right]$
(ii) $\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0\end{array}\right]\left[\begin{array}{rrr}-1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4\end{array}\right] \neq\left[\begin{array}{rrr}-1 & 1 & 0 \\ 0 & -1 & 1 \\ 2 & 3 & 4\end{array}\right]\left[\begin{array}{lll}1 & 2 & 3 \\ 0 & 1 & 0 \\ 1 & 1 & 0\end{array}\right]$

Answer
(i)
$
\begin{aligned}
& {\left[\begin{array}{rr}
5 & -1 \\
6 & 7
\end{array}\right]\left[\begin{array}{rr}
2 & 1 \\
3 & 4
\end{array}\right]} \\
& =\left[\begin{array}{ll}
5(2)-1(3) & 5(1)-1(4) \\
6(2)+7(3) & 6(1)+7(4)
\end{array}\right] \\
& =\left[\begin{array}{ll}
10-3 & 5-4 \\
12+21 & 6+28
\end{array}\right]=\left[\begin{array}{ll}
7 & 1 \\
33 & 34
\end{array}\right]
\end{aligned}
$
$
\begin{aligned}
& {\left[\begin{array}{ll}
2 & 1 \\
3 & 4
\end{array}\right]\left[\begin{array}{rr}
5 & -1 \\
6 & 7
\end{array}\right]} \\
& =\left[\begin{array}{ll}
2(5)+1(6) & 2(-1)+1(7) \\
3(5)+4(6) & 3(-1)+4(7)
\end{array}\right] \\
& =\left[\begin{array}{ll}
10+6 & -2+7 \\
15+24 & -3+28
\end{array}\right]=\left[\begin{array}{ll}
16 & 5 \\
39 & 25
\end{array}\right] \\
& \therefore\left[\begin{array}{rr}
5 & -1 \\
6 & 7
\end{array}\right]\left[\begin{array}{ll}
2 & 1 \\
3 & 4
\end{array}\right] \neq\left[\begin{array}{ll}
2 & 1 \\
3 & 4
\end{array}\right]\left[\begin{array}{rr}
5 & -1 \\
6 & 7
\end{array}\right]
\end{aligned}
$
(ii)
$
\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{array}\right]\left[\begin{array}{rrr}
-1 & 1 & 0 \\
0 & -1 & 1 \\
2 & 3 & 4
\end{array}\right]
$

$\begin{aligned}
& =\left[\begin{array}{lll}
1(-1)+2(0)+3(2) & 1(1)+2(-1)+3(3) & 1(0)+2(1)+3(4) \\
0(-1)+1(0)+0(2) & 0(1)+1(-1)+0(3) & 0(0)+1(1)+0(4) \\
1(-1)+1(0)+0(2) & 1(1)+1(-1)+0(3) & 1(0)+1(1)+0(4)
\end{array}\right] \\
& =\left[\begin{array}{ccc}
5 & 8 & 14 \\
0 & -1 & 1 \\
-1 & 0 & 1
\end{array}\right]
\end{aligned}$

$\begin{aligned}
& {\left[\begin{array}{rrr}
-1 & 1 & 0 \\
0 & -1 & 1 \\
2 & 3 & 4
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{array}\right]} \\
& =\left[\begin{array}{lll}
-1(1)+1(0)+0(1) & -1(2)+1(1)+0(1) & -1(3)+1(0)+0(0) \\
0(1)+(-1)(0)+1(1) & 0(2)+(-1)(1)+1(1) & 0(3)+(-1)(0)+1(0) \\
2(1)+3(0)+4(1) & 2(2)+3(1)+4(1) & 2(3)+3(0)+4(0)
\end{array}\right] \\
& =\left[\begin{array}{rrr}
-1 & -1 & -3 \\
1 & 0 & 0 \\
6 & 11 & 6
\end{array}\right] \\
& \therefore\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{array}\right]\left[\begin{array}{rrr}
-1 & 1 & 0 \\
0 & -1 & 1 \\
2 & 3 & 4
\end{array}\right] \neq\left[\begin{array}{rrr}
-1 & 1 & 0 \\
0 & -1 & 1 \\
2 & 3 & 4
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 3 \\
0 & 1 & 0 \\
1 & 1 & 0
\end{array}\right]
\end{aligned}$

Ex 3.2 Question 15:
$
A=\left[\begin{array}{rrr}
2 & 0 & 1 \\
2 & 1 & 3 \\
1 & -1 & 0
\end{array}\right]
$

Find $A^2-5 A+6 I$ if
Answer
We have $A^2=A \times A$
$
\begin{aligned}
& A^2=A A=\left[\begin{array}{ccc}
2 & 0 & 1 \\
2 & 1 & 3 \\
1 & -1 & 0
\end{array}\right]\left[\begin{array}{ccc}
2 & 0 & 1 \\
2 & 1 & 3 \\
1 & -1 & 0
\end{array}\right] \\
& =\left[\begin{array}{lll}
2(2)+0(2)+1(1) & 2(0)+0(1)+1(-1) & 2(1)+0(3)+1(0) \\
2(2)+1(2)+3(1) & 2(0)+1(1)+3(-1) & 2(1)+1(3)+3(0) \\
1(2)+(-1)(2)+0(1) & 1(0)+(-1)(1)+0(-1) & 1(1)+(-1)(3)+0(0)
\end{array}\right]
\end{aligned}
$

$\begin{aligned}
& =\left[\begin{array}{lll}
4+0+1 & 0+0-1 & 2+0+0 \\
4+2+3 & 0+1-3 & 2+3+0 \\
2-2+0 & 0-1+0 & 1-3+0
\end{array}\right] \\
& =\left[\begin{array}{rrr}
5 & -1 & 2 \\
9 & -2 & 5 \\
0 & -1 & -2
\end{array}\right] \\
& \therefore A^2-5 A+6 I
\end{aligned}$

$\begin{aligned}
& =\left[\begin{array}{rrr}
5 & -1 & 2 \\
9 & -2 & 5 \\
0 & -1 & -2
\end{array}\right]-5\left[\begin{array}{rrr}
2 & 0 & 1 \\
2 & 1 & 3 \\
1 & -1 & 0
\end{array}\right]+6\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \\
& =\left[\begin{array}{rrr}
5 & -1 & 2 \\
9 & -2 & 5 \\
0 & -1 & -2
\end{array}\right]-\left[\begin{array}{rrr}
10 & 0 & 5 \\
10 & 5 & 15 \\
5 & -5 & 0
\end{array}\right]+\left[\begin{array}{lll}
6 & 0 & 0 \\
0 & 6 & 0 \\
0 & 0 & 6
\end{array}\right]
\end{aligned}$

$\begin{aligned}
& =\left[\begin{array}{ccc}
5-10 & -1-0 & 2-5 \\
9-10 & -2-5 & 5-15 \\
0-5 & -1+5 & -2-0
\end{array}\right]+\left[\begin{array}{lll}
6 & 0 & 0 \\
0 & 6 & 0 \\
0 & 0 & 6
\end{array}\right] \\
& =\left[\begin{array}{ccc}
-5 & -1 & -3 \\
-1 & -7 & -10 \\
-5 & 4 & -2
\end{array}\right]+\left[\begin{array}{ccc}
6 & 0 & 0 \\
0 & 6 & 0 \\
0 & 0 & 6
\end{array}\right] \\
& =\left[\begin{array}{ccc}
-5+6 & -1+0 & -3+0 \\
-1+0 & -7+6 & -10+0 \\
-5+0 & 4+0 & -2+6
\end{array}\right] \\
& =\left[\begin{array}{ccc}
1 & -1 & -3 \\
-1 & -1 & -10 \\
-5 & 4 & 4
\end{array}\right]
\end{aligned}$

Ex 3.2 Question 16:
$
A=\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right] \text {, prove that } A^3-6 A^2+7 A+2 I=O
$

Answer
$
\begin{aligned}
A^2=A A & =\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right]\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right] \\
& =\left[\begin{array}{lll}
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{array}\right]=\left[\begin{array}{lll}
5 & 0 & 8 \\
2 & 4 & 5 \\
8 & 0 & 13
\end{array}\right]
\end{aligned}
$

$\begin{aligned}
& \text { Now } A^3=A^2 \cdot A \\
&=\left[\begin{array}{lll}
5 & 0 & 8 \\
2 & 4 & 5 \\
8 & 0 & 13
\end{array}\right]\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right] \\
&=\left[\begin{array}{lll}
5+0+16 & 0+0+0 & 10+0+24 \\
2+0+10 & 0+8+0 & 4+4+15 \\
8+0+26 & 0+0+0 & 16+0+39
\end{array}\right] \\
&=\left[\begin{array}{lll}
21 & 0 & 34 \\
12 & 8 & 23 \\
34 & 0 & 55
\end{array}\right] \\
& \therefore A^3-6 A^2+7 A+2 I
\end{aligned}$

$\begin{aligned}
& =\left[\begin{array}{lll}
21 & 0 & 34 \\
12 & 8 & 23 \\
34 & 0 & 55
\end{array}\right]-6\left[\begin{array}{lll}
5 & 0 & 8 \\
2 & 4 & 5 \\
8 & 0 & 13
\end{array}\right]+7\left[\begin{array}{lll}
1 & 0 & 2 \\
0 & 2 & 1 \\
2 & 0 & 3
\end{array}\right]+2\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \\
& =\left[\begin{array}{lll}
21 & 0 & 34 \\
12 & 8 & 23 \\
34 & 0 & 55
\end{array}\right]-\left[\begin{array}{lll}
30 & 0 & 48 \\
12 & 24 & 30 \\
48 & 0 & 78
\end{array}\right]+\left[\begin{array}{lll}
7 & 0 & 14 \\
0 & 14 & 7 \\
14 & 0 & 21
\end{array}\right]+\left[\begin{array}{lll}
2 & 0 & 0 \\
0 & 2 & 0 \\
0 & 0 & 2
\end{array}\right] \\
& =\left[\begin{array}{lll}
21+7+2 & 0+0+0 & 34+14+0 \\
12+0+0 & 8+14+2 & 23+7+0 \\
34+14+0 & 0+0+0 & 55+21+2
\end{array}\right]-\left[\begin{array}{lll}
30 & 0 & 48 \\
12 & 24 & 30 \\
48 & 0 & 78
\end{array}\right]
\end{aligned}$

$\begin{aligned}
& =\left[\begin{array}{lll}
30 & 0 & 48 \\
12 & 24 & 30 \\
48 & 0 & 78
\end{array}\right]-\left[\begin{array}{lll}
30 & 0 & 48 \\
12 & 24 & 30 \\
48 & 0 & 78
\end{array}\right] \\
& =\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]=O \\
& \therefore A^3-6 A^2+7 A+2 I=O
\end{aligned}$

Ex 3.2 Question 17:
$
\text { If } A=\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right] \text { and } I=\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right] \text {, find } k \text { so that } A^2=k A-2 I
$

Answer
$
\begin{aligned}
A^2=A \cdot A & =\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right] \\
& =\left[\begin{array}{ll}
3(3)+(-2)(4) & 3(-2)+(-2)(-2) \\
4(3)+(-2)(4) & 4(-2)+(-2)(-2)
\end{array}\right]=\left[\begin{array}{ll}
1 & -2 \\
4 & -4
\end{array}\right]
\end{aligned}
$
$
\begin{aligned}
& \text { Now } A^2=k A-2 I \\
& \Rightarrow\left[\begin{array}{ll}
1 & -2 \\
4 & -4
\end{array}\right]=k\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]-2\left[\begin{array}{cc}
1 & 0 \\
0 & 1
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{ll}
1 & -2 \\
4 & -4
\end{array}\right]=\left[\begin{array}{ll}
3 k & -2 k \\
4 k & -2 k
\end{array}\right]-\left[\begin{array}{cc}
2 & 0 \\
0 & 2
\end{array}\right] \\
& \Rightarrow\left[\begin{array}{ll}
1 & -2 \\
4 & -4
\end{array}\right]=\left[\begin{array}{ll}
3 k-2 & -2 k \\
4 k & -2 k-2
\end{array}\right]
\end{aligned}
$

Comparing the corresponding elements, we have:

$
\begin{aligned}
& 3 k-2=1 \\
& \Rightarrow 3 k=3 \\
& \Rightarrow k=1
\end{aligned}
$

Thus, the value of $k$ is 1 .

Ex 3.2 Question 18:
$
\begin{aligned}
& A=\left[\begin{array}{cc}
0 & -\tan \frac{\alpha}{2} \\
\tan \frac{\alpha}{2} & 0
\end{array}\right] \text { and } I \text { is the identity matrix of order 2, show that } \\
& I+A=(I-A)\left[\begin{array}{rr}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]
\end{aligned}
$

Answer

On the L.H.S.
$
\begin{aligned}
& I+A \\
& =\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]+\left[\begin{array}{cc}
0 & -\tan \frac{\alpha}{2} \\
\tan \frac{\alpha}{2} & 0
\end{array}\right] \\
& =\left[\begin{array}{cc}
1 & -\tan \frac{\alpha}{2} \\
\tan \frac{\alpha}{2} & 1
\end{array}\right]
\end{aligned}
$

On the R.H.S.
$
(I-A)\left[\begin{array}{rr}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]
$

$=\left(\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]-\left[\begin{array}{lr}
0 & -\tan \frac{\alpha}{2} \\
\tan \frac{\alpha}{2} & 0
\end{array}\right]\right)\left[\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right]$

$\begin{aligned}
& =\left[\begin{array}{rr}
1 & \tan \frac{\alpha}{2} \\
-\tan \frac{\alpha}{2} & 1
\end{array}\right]\left[\begin{array}{rr}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}\right] \\
& =\left[\begin{array}{cc}
\cos \alpha+\sin \alpha \tan \frac{\alpha}{2} & -\sin \alpha+\cos \alpha \tan \frac{\alpha}{2} \\
-\cos \alpha \tan \frac{\alpha}{2}+\sin \alpha & \sin \alpha \tan \frac{\alpha}{2}+\cos \alpha
\end{array}\right]
\end{aligned}$

$
\begin{aligned}
& =\left[\begin{array}{lc}
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}+\left(2 \cos ^2 \frac{\alpha}{2}-1\right) \tan \frac{\alpha}{2} \\
-\left(2 \cos ^2 \frac{\alpha}{2}-1\right) \tan \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}+1-2 \sin ^2 \frac{\alpha}{2}
\end{array}\right] \\
& =\left[\begin{array}{cc}
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{array}\right] \\
& =\left[\begin{array}{cc}
1 & -\tan \frac{\alpha}{2} \\
\tan \frac{\alpha}{2} & 1
\end{array}\right]
\end{aligned}
$

Thus, from (1) and (2), we get L.H.S. = R.H.S.

Ex 3.2 Question 19:
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:
(a) Rs 1,800
(b) Rs 2,000

Answer
(a) Let Rs $x$ be invested in the first bond. Then, the sum of money invested in the second bond will be Rs $(30000-x)$.
It is given that the first bond pays $5 \%$ interest per year and the second bond pays $7 \%$ interest per year.
Therefore, in order to obtain an annual total interest of Rs 1800 , we have:
$
\left[\begin{array}{ll}
x & (30000-x)
\end{array}\right]\left[\begin{array}{c}
\frac{5}{100} \\
\frac{7}{100}
\end{array}\right]=1800 \quad\left[\text { S.I. for } 1 \text { year }=\frac{\text { Principal } \times \text { Rate }}{100}\right]
$

$
\begin{aligned}
& \Rightarrow \frac{5 x}{100}+\frac{7(30000-x)}{100}=1800 \\
& \Rightarrow 5 x+210000-7 x=180000 \\
& \Rightarrow 210000-2 x=180000 \\
& \Rightarrow 2 x=210000-180000 \\
& \Rightarrow 2 x=30000 \\
& \Rightarrow x=15000
\end{aligned}
$

Thus, in order to obtain an annual total interest of Rs 1800 , the trust fund should invest Rs 15000 in the first bond and the remaining Rs 15000 in the second bond.
(b) Let Rs $x$ be invested in the first bond. Then, the sum of money invested in the second bond will be Rs $(30000-x)$.
Therefore, in order to obtain an annual total interest of Rs 2000 , we have:

$
\begin{aligned}
& {\left[\begin{array}{ll}
x & (30000-x)
\end{array}\right]\left[\begin{array}{c}
\frac{5}{100} \\
\frac{7}{100}
\end{array}\right]=2000} \\
& \Rightarrow \frac{5 x}{100}+\frac{7(30000-x)}{100}=2000 \\
& \Rightarrow 5 x+210000-7 x=200000 \\
& \Rightarrow 210000-2 x=200000 \\
& \Rightarrow 2 x=210000-200000 \\
& \Rightarrow 2 x=10000 \\
& \Rightarrow x=5000
\end{aligned}
$

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

Ex 3.2 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, and 10 dozen economics books.
The selling prices of a chemistry book, a physics book, and an economics book are respectively given as Rs 80 , Rs 60 , and Rs 40 .
The total amount of money that will be received from the sale of all these books can be represented in the form of a matrix as:
$
12\left[\begin{array}{lll}
10 & 8 & 10
\end{array}\right]\left[\begin{array}{l}
80 \\
60 \\
40
\end{array}\right]
$

$
\begin{aligned}
& =12[10 \times 80+8 \times 60+10 \times 40] \\
& =12(800+480+400) \\
& =12(1680) \\
& =20160
\end{aligned}
$

Thus, the bookshop will receive Rs 20160 from the sale of all these books.

Ex 3.2 Question 21:
Assume $X, Y, Z, W$ and $P$ are matrices of order $2 \times n, 3 \times k, 2 \times p, n \times 3$, and $p \times k$ respectively. The restriction on $n, k$ and $p$ so that $P Y+W Y$ 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
Matrices $P$ and $Y$ are of the orders $p \times k$ and $3 \times k$ respectively.
Therefore, matrix $P Y$ will be defined if $k=3$. Consequently, $P Y$ will be of the order $p \times k$.
Matrices $W$ and $Y$ are of the orders $n \times 3$ and $3 \times k$ respectively.
Since the number of columns in $W$ is equal to the number of rows in $Y$, matrix $W Y$ is well-defined and is of the order $n \times k$.
Matrices $P Y$ and $W Y$ can be added only when their orders are the same.
However, $P Y$ is of the order $p \times k$ and $W Y$ is of the order $n \times k$. Therefore, we must have $p=n$.
Thus, $k=3$ and $p=n$ are the restrictions on $n, k$, and $p$ so that $P Y+W Y$ will be defined.

Ex 3.2 Question 22:
Assume $X, Y, Z, W$ and $P$ are matrices of order $2 \times n, 3 \times k, 2 \times p, n \times 3$, and $p \times k$ respectively. If $n=p$, then the order of the matrix $7 X-5 Z$ is
$\mathbf{A} p \times 2 \mathbf{B} 2 \times n \mathbf{C} n \times 3 \mathbf{D} p \times n$
Answer
The correct answer is B.
Matrix $X$ is of the order $2 \times n$.
Therefore, matrix $7 X$ is also of the same order.
Matrix $Z$ is of the order $2 \times p$, i.e., $2 \times n$ [Since $n=p$ ]
Therefore, matrix $5 Z$ is also of the same order.
Now, both the matrices $7 X$ and $5 Z$ are of the order $2 \times n$.
Thus, matrix $7 X-5 Z$ is well-defined and is of the order $2 \times n$.