Exercise 7.1-Additional Questions - Chapter 7 Matrices and Determinants 11th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
If $A=\left[\begin{array}{ll}1 & 8 \\ 4 & 3\end{array}\right] B=\left[\begin{array}{ll}1 & 3 \\ 7 & 4\end{array}\right] C=\left[\begin{array}{rr}-4 & 6 \\ 3 & -5\end{array}\right]$
Prove that (i) $A B \neq B A$
(ii) $\mathrm{A}(\mathrm{BC})=(\mathrm{AB}) \mathrm{C}$

(iii) $\mathrm{A}(\mathrm{B}+\mathrm{C})=\mathrm{AB}+\mathrm{AC}$
(iv) $\mathrm{AI}=\mathrm{IA}=\mathrm{A}$
Solution:
$(i)$
$
\begin{aligned}
\mathrm{AB} & =\left[\begin{array}{ll}
1 & 8 \\
4 & 3
\end{array}\right]\left[\begin{array}{ll}
1 & 3 \\
7 & 4
\end{array}\right]=\left[\begin{array}{cc}
(1)(1)+(8)(7) & (1)(3)+(8)(4) \\
(4)(1)+(3)(7) & (4)(3)+(3)(4)
\end{array}\right] \\
& =\left[\begin{array}{ll}
1+56 & 3+32 \\
4+21 & 12+12
\end{array}\right]=\left[\begin{array}{ll}
57 & 35 \\
25 & 24
\end{array}\right] \\
\mathrm{BA} & =\left[\begin{array}{ll}
1 & 3 \\
7 & 4
\end{array}\right]\left[\begin{array}{ll}
1 & 8 \\
4 & 3
\end{array}\right]=\left[\begin{array}{cc}
(1)(1)+(3)(4) & (1)(8)+(3)(3) \\
(7)(1)+(4)(4) & (7)(8)+(4)(3)
\end{array}\right] \\
& =\left[\begin{array}{ll}
1+12 & 8+9 \\
7+16 & 56+12
\end{array}\right]=\left[\begin{array}{ll}
13 & 17 \\
23 & 68
\end{array}\right]
\end{aligned}
$
From (1) and (2) we have $\mathrm{AB} \neq \mathrm{BA}$
(ii)
$
\begin{aligned}
(\mathrm{AB}) \mathrm{C} & =\left[\begin{array}{ll}
57 & 35 \\
25 & 24
\end{array}\right]\left[\begin{array}{rr}
-4 & 6 \\
3 & -5
\end{array}\right] \\
& =\left[\begin{array}{ll}
(57)(-4)+(35)(3) & (57)(6)+(35)(-5) \\
(25)(-4)+(24)(3) & (25)(6)+(24)(-5)
\end{array}\right] \\
& =\left[\begin{array}{rr}
-228+105 & 342-175 \\
-100+72 & 150-120
\end{array}\right] \\
\therefore \quad(\mathrm{AB}) \mathrm{C} & =\left[\begin{array}{rr}
-123 & 167 \\
-28 & 30
\end{array}\right]
\end{aligned}
$
...from (1)

$\begin{aligned}
\mathrm{BC}= & {\left[\begin{array}{ll}
1 & 3 \\
7 & 4
\end{array}\right]\left[\begin{array}{rr}
-4 & 6 \\
3 & -5
\end{array}\right] } \\
= & {\left[\begin{array}{ll}
(1)(-4)+(3)(3) & (1)(6)+(3)(-5) \\
(7)(-4)+(4)(3) & (7)(6)+(4)(-5)
\end{array}\right]=\left[\begin{array}{rr}
-4+9 & 6-15 \\
-28+12 & 42-20
\end{array}\right] } \\
\mathrm{BC}= & {\left[\begin{array}{rr}
5 & -9 \\
-16 & 22
\end{array}\right] } \\
\mathrm{A}(\mathrm{BC})= & {\left[\begin{array}{ll}
1 & 8 \\
4 & 3
\end{array}\right]\left[\begin{array}{rr}
5 & -9 \\
-16 & 22
\end{array}\right] } \\
& =\left[\begin{array}{ll}
(1)(5)+(8)(-16) & (1)(-9)+(8)(22) \\
(4)(5)+(3)(-16) & (4)(-9)+(3)(22)
\end{array}\right]=\left[\begin{array}{rr}
5-128 & -9+176 \\
20-48 & -36+66
\end{array}\right]
\end{aligned}$

$
A(B C)=\left[\begin{array}{rr}
-123 & 167 \\
-28 & 30
\end{array}\right]
$
From (3) and (4) we have, $(\mathrm{AB}) \mathrm{C}=\mathrm{A}(\mathrm{BC})$

(iii)
$
\begin{aligned}
\mathrm{B}+\mathrm{C} & =\left[\begin{array}{ll}
1 & 3 \\
7 & 4
\end{array}\right]+\left[\begin{array}{rr}
-4 & 6 \\
3 & -5
\end{array}\right]=\left[\begin{array}{ll}
1-4 & 3+6 \\
7+3 & 4-5
\end{array}\right]=\left[\begin{array}{rr}
-3 & 9 \\
10 & -1
\end{array}\right] \\
\mathrm{A}(\mathrm{B}+\mathrm{C}) & =\left[\begin{array}{ll}
1 & 8 \\
4 & 3
\end{array}\right]\left[\begin{array}{rr}
-3 & 9 \\
10 & -1
\end{array}\right]=\left[\begin{array}{rr}
-3+80 & 9-8 \\
-12+30 & 36-3
\end{array}\right] \\
\mathrm{A}(\mathrm{B}+\mathrm{C}) & =\left[\begin{array}{ll}
77 & 1 \\
18 & 33
\end{array}\right] \\
\mathrm{AB} & =\left[\begin{array}{ll}
57 & 35 \\
25 & 24
\end{array}\right] \\
\mathrm{AC} & =\left[\begin{array}{ll}
1 & 8 \\
4 & 3
\end{array}\right]\left[\begin{array}{rr}
-4 & 6 \\
3 & -5
\end{array}\right]=\left[\begin{array}{rr}
-4+24 & 6-40 \\
-16+9 & 24-15
\end{array}\right]=\left[\begin{array}{rr}
20 & -34 \\
-7 & 9
\end{array}\right] \\
\mathrm{AB}+\mathrm{AC} & =\left[\begin{array}{ll}
57 & 35 \\
25 & 24
\end{array}\right]+\left[\begin{array}{rr}
20 & -34 \\
-7 & 9
\end{array}\right]=\left[\begin{array}{rr}
57+20 & 35-34 \\
25-7 & 24+9
\end{array}\right] \\
& =\left[\begin{array}{ll}
77 & 1 \\
18 & 33
\end{array}\right]
\end{aligned}
$
...from (1)
- From equations (5) and (6) we have $\mathrm{A}(\mathrm{B}+\mathrm{C})=\mathrm{AB}+\mathrm{AC}$

(iv) Since order of $\mathrm{A}$ is $2 \times 2$, take $\mathrm{I}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$.
$
\begin{aligned}
\mathrm{AI} & =\left[\begin{array}{ll}
1 & 8 \\
4 & 3
\end{array}\right]\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{ll}
1(1)+8(0) & 1(0)+8(1) \\
4(1)+3(0) & 4(0)+3(1)
\end{array}\right]=\left[\begin{array}{ll}
1+0 & 0+8 \\
4+0 & 0+3
\end{array}\right] \\
& =\left[\begin{array}{ll}
1 & 8 \\
4 & 3
\end{array}\right]=\mathrm{A} \\
\mathrm{IA} & =\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]\left[\begin{array}{ll}
1 & 8 \\
4 & 3
\end{array}\right]=\left[\begin{array}{ll}
1(1)+0(4) & 1(8)+0(3) \\
0(1)+1(4) & 0(8)+1(3)
\end{array}\right]=\left[\begin{array}{ll}
1+0 & 8+0 \\
0+4 & 0+3
\end{array}\right] \\
& =\left[\begin{array}{ll}
1 & 8 \\
4 & 3
\end{array}\right]=\mathrm{A}
\end{aligned}
$
$\therefore$ From (7) and (8) $\mathrm{AI}=\mathrm{IA}=\mathrm{A}$

Question 2.

If $\mathrm{A}=$ find $\mathrm{A}$
$
\left[\begin{array}{ll}
2 & 3 \\
4 & 5
\end{array}\right] \quad 2-7 \mathrm{~A}-2 \mathrm{I}
$
Solution:
$
\begin{aligned}
& A^2=A A=\left[\begin{array}{ll}
2 & 3 \\
4 & 5
\end{array}\right]\left[\begin{array}{ll}
2 & 3 \\
4 & 5
\end{array}\right]=\left[\begin{array}{ll}
4+12 & 6+15 \\
8+20 & 12+25
\end{array}\right] \\
& A^2=\left[\begin{array}{ll}
16 & 21 \\
28 & 37
\end{array}\right] \\
&-7 A=-7\left[\begin{array}{ll}
2 & 3 \\
4 & 5
\end{array}\right]=\left[\begin{array}{rr}
-14 & -21 \\
-28 & -35
\end{array}\right] \\
&-2 I=-2\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{rr}
-2 & 0 \\
0 & -2
\end{array}\right]
\end{aligned}
$
$(1)+(2)+(3)$ gives $A^2-7 A-2 I=A^2+(-7 A)+(-2 I)$
$
=\left[\begin{array}{ll}
16 & 21 \\
28 & 37
\end{array}\right]+\left[\begin{array}{ll}
-14 & -21 \\
-28 & -35
\end{array}\right]+\left[\begin{array}{rr}
-2 & 0 \\
0 & -2
\end{array}\right]
$
i.e. $A^2-7 A-2 I=\left[\begin{array}{ll}16-14-2 & 21-21+0 \\ 28-28+0 & 37-35-2\end{array}\right]=\left[\begin{array}{ll}0 & 0 \\ 0 & 0\end{array}\right]=0$

Question 3.
Solve:
$
X+2 Y=\left[\begin{array}{rr}
4 & 6 \\
-8 & 10
\end{array}\right] ; X-Y=\left[\begin{array}{rr}
1 & 0 \\
-2 & -2
\end{array}\right]
$

Solution:
Given
$
\begin{aligned}
X+2 Y & =\left[\begin{array}{rr}
4 & 6 \\
-8 & 10
\end{array}\right] \\
X-Y & =\left[\begin{array}{rr}
1 & 0 \\
-2 & -2
\end{array}\right]
\end{aligned}
$
$
\begin{aligned}
(1)-(2) \Rightarrow(X+2 Y)-(X-Y) & =\left[\begin{array}{rr}
4 & 6 \\
-8 & 10
\end{array}\right]-\left[\begin{array}{rr}
1 & 0 \\
-2 & -2
\end{array}\right] \\
3 Y & =\left[\begin{array}{rr}
3 & 6 \\
-6 & 12
\end{array}\right] \Rightarrow Y=\frac{1}{3}\left[\begin{array}{rr}
3 & 6 \\
-6 & 12
\end{array}\right] \\
\Rightarrow Y & =\left[\begin{array}{rr}
1 & 2 \\
-2 & 4
\end{array}\right]
\end{aligned}
$
Substituting matrix $Y$ in equation (1) we have
$
\begin{array}{rlr}
X+2\left[\begin{array}{rr}
1 & 2 \\
-2 & 4
\end{array}\right] & =\left[\begin{array}{rr}
4 & 6 \\
-8 & 10
\end{array}\right] \\
\Rightarrow \quad X+\left[\begin{array}{rr}
2 & 4 \\
-4 & 8
\end{array}\right] & =\left[\begin{array}{rr}
4 & 6 \\
-8 & 10
\end{array}\right] \\
\Rightarrow \quad X & =\left[\begin{array}{rr}
4 & 6 \\
-8 & 10
\end{array}\right]-\left[\begin{array}{rr}
2 & 4 \\
-4 & 8
\end{array}\right]=\left[\begin{array}{rr}
2 & 2 \\
-4 & 2
\end{array}\right]
\end{array}
$

$\therefore$
$
X=\left[\begin{array}{rr}
2 & 2 \\
-4 & 2
\end{array}\right] \text { and } Y=\left[\begin{array}{rr}
1 & 2 \\
-2 & 4
\end{array}\right]
$
Question 4.
Solve: $2 X+Y+\left[\begin{array}{rrr}-2 & 1 & 3 \\ 5 & -7 & 3 \\ 4 & 5 & 4\end{array}\right]=0 ; \mathbf{X}-Y=\left[\begin{array}{rrr}4 & 7 & 0 \\ -1 & 2 & -6 \\ -2 & 8 & -5\end{array}\right]$
Solution:

$
\begin{aligned}
2 X+Y & =-\left[\begin{array}{rrr}
-2 & 1 & 3 \\
5 & -7 & 3 \\
4 & 5 & 4
\end{array}\right]=\left[\begin{array}{rrr}
2 & -1 & -3 \\
-5 & 7 & -3 \\
-4 & -5 & -4
\end{array}\right] \\
X-Y & =\left[\begin{array}{rrr}
4 & 7 & 0 \\
-1 & 2 & -6 \\
-2 & 8 & -5
\end{array}\right] \\
(1)+(2) \Rightarrow 3 X & =\left[\begin{array}{rrr}
2 & -1 & -3 \\
-5 & 7 & -3 \\
-4 & -5 & -4
\end{array}\right]+\left[\begin{array}{rrr}
4 & 7 & 0 \\
-1 & 2 & -6 \\
-2 & 8 & -5
\end{array}\right] \\
& =\left[\begin{array}{rrr}
6 & 6 & -3 \\
-6 & 9 & -9 \\
-6 & 3 & -9
\end{array}\right] \\
\therefore \quad X & =\frac{1}{3}\left[\begin{array}{rrr}
6 & 6 & -3 \\
-6 & 9 & -9 \\
-6 & 3 & -9
\end{array}\right]=\left[\begin{array}{rrr}
2 & 2 & -1 \\
-2 & 3 & -3 \\
-2 & 1 & -3
\end{array}\right] \\
X & =\left[\begin{array}{rrr}
2 & 2 & -1 \\
-2 & 3 & -3 \\
-2 & 1 & -3
\end{array}\right] \text { in }(2) \text { we get, }
\end{aligned}
$
Substituting

$\begin{aligned}
& {\left[\begin{array}{rrr}
2 & 2 & -1 \\
-2 & 3 & -3 \\
-2 & 1 & -3
\end{array}\right]-Y=\left[\begin{array}{rrr}
4 & 7 & 0 \\
-1 & 2 & -6 \\
-2 & 8 & -5
\end{array}\right]} \\
& \therefore \quad\left[\begin{array}{rrr}
2 & 2 & -1 \\
-2 & 3 & -3 \\
-2 & 1 & -3
\end{array}\right]-\left[\begin{array}{rrr}
4 & 7 & 0 \\
-1 & 2 & -6 \\
-2 & 8 & -5
\end{array}\right]=\mathrm{Y} \\
& \text { (i.e.,) } \quad Y=\left[\begin{array}{rrr}
2 & 2 & -1 \\
-2 & 3 & -3 \\
-2 & 1 & -3
\end{array}\right]+\left[\begin{array}{rrr}
-4 & -7 & 0 \\
1 & -2 & 6 \\
2 & -8 & 5
\end{array}\right]=\left[\begin{array}{rrr}
-2 & -5 & -1 \\
-1 & 1 & 3 \\
0 & -7 & 2
\end{array}\right] \\
& \therefore \quad X=\left[\begin{array}{rrr}
2 & 2 & -1 \\
-2 & 3 & -3 \\
-2 & 1 & -3
\end{array}\right] \text { and } Y=\left[\begin{array}{rrr}
-2 & -5 & -1 \\
-1 & 1 & 3 \\
0 & -7 & 2
\end{array}\right] \\
&
\end{aligned}$

Question 5.
If $A=\left[\begin{array}{rr}3 & -5 \\ -4 & 2\end{array}\right]$, show that $A^2-5 \mathrm{~A}-14 \mathrm{I}=0$ where $\mathrm{I}$ is the unit matrix of order 2 .
Solution:
$
\begin{aligned}
A & =\left[\begin{array}{rr}
3 & -5 \\
-4 & 2
\end{array}\right] \\
A^2 & =A \times A=\left[\begin{array}{rr}
3 & -5 \\
-4 & 2
\end{array}\right] \mid\left[\begin{array}{rr}
3 & -5 \\
-4 & 2
\end{array}\right]=\left[\begin{array}{rr}
9+20 & -15-10 \\
-12-8 & 20+4
\end{array}\right] \\
& =\left[\begin{array}{rr}
29 & -25 \\
-20 & 24
\end{array}\right] \\
-5 \mathrm{~A} & =-5\left[\begin{array}{rr}
3 & -5 \\
-4 & 2
\end{array}\right]=\left[\begin{array}{rr}
-15 & 25 \\
20 & -10
\end{array}\right] \\
-14 \mathrm{I} & =-14\left[\begin{array}{rr}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{rr}
-14 & 0 \\
0 & -14
\end{array}\right] \\
\text { Now LHS } & =A^2-5 \mathrm{~A}-14 \mathrm{I}=\left[\begin{array}{rr}
29 & -25 \\
-20 & 24
\end{array}\right]+\left[\begin{array}{rr}
-15 & 25 \\
20 & -10
\end{array}\right]+\left[\begin{array}{rr}
-14 & 0 \\
0 & -14
\end{array}\right] \\
& =\left[\begin{array}{rr}
29-15-14 & -25+25+0 \\
-20+20+0 & 24-10-14
\end{array}\right]=\left[\begin{array}{rr}
0 & 0 \\
0 & 0
\end{array}\right] \\
& =0=\text { RHS }
\end{aligned}
$

Question 6.
If $\mathrm{A}=\left[\begin{array}{ll}3 & -2 \\ 4 & -2\end{array}\right]$, find $\mathrm{k}$ so that $\mathrm{A}^2=\mathrm{kA}-2 \mathrm{I}$
Solution:

$\begin{aligned}
& A=\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right] \\
& A^2=\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right] \\
& =\left[\begin{array}{rr}
9-8 & -6+4 \\
12-8 & -8+4
\end{array}\right]=\left[\begin{array}{ll}
1 & -2 \\
4 & -4
\end{array}\right] \\
& k \mathrm{~A}-2 \mathrm{I}=k\left[\begin{array}{ll}
3 & -2 \\
4 & -2
\end{array}\right]-2\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
3 k & -2 k \\
4 k & -2 k
\end{array}\right]+\left[\begin{array}{rr}
-2 & 0 \\
0 & -2
\end{array}\right] \\
& =\left[\begin{array}{cc}
3 k-2 & -2 k \\
4 k & -2 k-2
\end{array}\right] \\
& \text { Given } \mathrm{A}^2=k \mathrm{~A}-2 \mathrm{I} \\
& \Rightarrow \quad \text { (1) }=\text { (2) } \\
& \Rightarrow \quad\left[\begin{array}{ll}
1 & -2 \\
4 & -4
\end{array}\right]=\left[\begin{array}{cc}
3 k-2 & -2 k \\
4 k & -2 k-2
\end{array}\right] \\
& \Rightarrow \quad 3 k-2=1 \\
& 3 k=3 \\
& k=1 \\
& \text { Verification } 4=4 k=4(1)=4 \quad \therefore k=1 \\
&
\end{aligned}$

Question 7.
If $A=\left[\begin{array}{lll}1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1\end{array}\right]$, show that $A^2-4 A-5 I=0$
Solution:

$\begin{aligned}
\mathrm{A} & =\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right] \\
\mathrm{A}^2=\mathrm{A} \times \mathrm{A} & =\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right] \\
& =\left[\begin{array}{lll}
1+4+4 & 2+2+4 & 2+4+2 \\
2+2+4 & 4+1+4 & 4+2+2 \\
2+4+2 & 4+2+2 & 4+4+1
\end{array}\right] \\
& =\left[\begin{array}{lll}
9 & 8 & 8 \\
8 & 9 & 8 \\
8 & 8 & 9
\end{array}\right] \\
-4 \mathrm{~A} & =-4\left[\begin{array}{lll}
1 & 2 & 2 \\
2 & 1 & 2 \\
2 & 2 & 1
\end{array}\right]=\left[\begin{array}{rrr}
-4 & -8 & -8 \\
-8 & -4 & -8 \\
-8 & -8 & -4
\end{array}\right] \\
-5 \mathrm{I} & =-5\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{rrr}
-5 & 0 & 0 \\
0 & -5 & 0 \\
0 & 0 & -5
\end{array}\right]
\end{aligned}$

$
\text { LHS: } \begin{aligned}
A^2-4 A-5 I & =\left[\begin{array}{lll}
9 & 8 & 8 \\
8 & 9 & 8 \\
8 & 8 & 9
\end{array}\right]+\left[\begin{array}{lll}
-4 & -8 & -8 \\
-8 & -4 & -8 \\
-8 & -8 & -4
\end{array}\right]+\left[\begin{array}{rrr}
-5 & 0 & 0 \\
0 & -5 & 0 \\
0 & 0 & -5
\end{array}\right] \\
& =\left[\begin{array}{lll}
9-4-5 & 8-8+0 & 8-8+0 \\
8-8+0 & 9-4-5 & 8-8+0 \\
8-8+0 & 8-8+0 & 9-4-5
\end{array}\right]=\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right] \\
& =0 \text { R.H.S. }
\end{aligned}
$
Question 8.
If $A=\left[\begin{array}{ll}1 & 2 \\ 2 & 0\end{array}\right], B=\left[\begin{array}{rr}3 & -1 \\ 1 & 0\end{array}\right]$ verify the following:
Solution:

$\begin{aligned}
A & =\left[\begin{array}{ll}
1 & 2 \\
2 & 0
\end{array}\right], \mathrm{B}=\left[\begin{array}{rr}
3 & -1 \\
1 & 0
\end{array}\right] \\
\therefore \quad \mathrm{A}+\mathrm{B} & =\left[\begin{array}{ll}
1 & 2 \\
2 & 0
\end{array}\right]+\left[\begin{array}{rr}
3 & -1 \\
1 & 0
\end{array}\right]=\left[\begin{array}{ll}
4 & 1 \\
3 & 0
\end{array}\right] \\
\mathrm{A}-\mathrm{B} & =\left[\begin{array}{ll}
1 & 2 \\
2 & 0
\end{array}\right]-\left[\begin{array}{rr}
3 & -1 \\
1 & 0
\end{array}\right] \\
& =\left[\begin{array}{ll}
1 & 2 \\
2 & 0
\end{array}\right]+\left[\begin{array}{rr}
-3 & 1 \\
-1 & 0
\end{array}\right]=\left[\begin{array}{rr}
-2 & 3 \\
1 & 0
\end{array}\right]
\end{aligned}$

$=\left[\begin{array}{ll}
16+3 & 4+0 \\
12+0 & 3+0
\end{array}\right]=\left[\begin{array}{ll}
19 & 4 \\
12 & 3
\end{array}\right]$

(i)
$
\begin{aligned}
& \text { LHS }=(A+B)^2=\left[\begin{array}{ll}
19 & 4 \\
12 & 3
\end{array}\right][\text { from (7) }] \\
& \text { RHS }=\mathrm{A}^2+\mathrm{AB}+\mathrm{BA}+\mathrm{B}^2 \\
& =\left[\begin{array}{ll}
5 & 2 \\
2 & 4
\end{array}\right]+\left[\begin{array}{ll}
5 & -1 \\
6 & -2
\end{array}\right]+\left[\begin{array}{ll}
1 & 6 \\
1 & 2
\end{array}\right]+\left[\begin{array}{ll}
8 & -3 \\
3 & -1
\end{array}\right] \\
& \text { (from (3), (5), (6) and (4)) } \\
& =\left[\begin{array}{ll}
5+5+1+8 & 2-1+6-3 \\
2+6+1+3 & 4-2+2-1
\end{array}\right]=\left[\begin{array}{ll}
19 & 4 \\
12 & 3
\end{array}\right] \\
& \text { LHS }=\text { RHS } \Rightarrow(A+B)^2=A^2+A B+B A+B^2 \\
&
\end{aligned}
$

(ii) LHS: $\quad(\mathrm{A}-\mathrm{B})^2=\left[\begin{array}{rr}7 & -6 \\ -2 & 3\end{array}\right][$ from (8).
$
\begin{aligned}
& \cdot \text { RHS: } A^2-A B-B A+B^2=\left[\begin{array}{ll}
5 & 2 \\
2 & 4
\end{array}\right]-\left[\begin{array}{ll}
5 & -1 \\
6 & -2
\end{array}\right]-\left[\begin{array}{ll}
1 & 6 \\
1 & 2
\end{array}\right]+\left[\begin{array}{ll}
8 & -3 \\
3 & -1
\end{array}\right] \\
&=\left[\begin{array}{ll}
5 & 2 \\
2 & 4
\end{array}\right]+\left[\begin{array}{ll}
-5 & 1 \\
-6 & 2
\end{array}\right]+\left[\begin{array}{ll}
-1 & -6 \\
-1 & -2
\end{array}\right]+\left[\begin{array}{ll}
8 & -3 \\
3 & -1
\end{array}\right] \\
&=\left[\begin{array}{ll}
5-5-1+8 & 2+1-6-3 \\
2-6-1+3 & 4+2-2-1
\end{array}\right]=\left[\begin{array}{rr}
7 & -6 \\
-2 & 3
\end{array}\right] \\
& \text { LHS }=\text { RHS } \Rightarrow(A-B)^2=A^2-A B-B A+B^2
\end{aligned}
$
Question 9.
Find matrix $C$ if $A=\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right], B=\left[\begin{array}{rr}-3 & 2 \\ 4 & -1\end{array}\right]$ and $5 C+2 B=A$.
Solution:

$
\begin{aligned}
& 5 \mathrm{C}+2 \mathrm{~B}=\mathrm{A} \\
& \Rightarrow \quad 5 \mathrm{C}=\mathrm{A}-2 \mathrm{~B} \\
& \therefore \quad \mathrm{C}=\frac{1}{5}[\mathrm{~A}-2 \mathrm{~B}] \\
& \therefore A-2 B=\left[\begin{array}{ll}
3 & 7 \\
2 & 5
\end{array}\right]-2\left[\begin{array}{rr}
-3 & 2 \\
4 & -1
\end{array}\right]=\left[\begin{array}{ll}
3 & 7 \\
2 & 5
\end{array}\right]+\left[\begin{array}{rr}
6 & -4 \\
-8 & 2
\end{array}\right] \\
& =\left[\begin{array}{ll}
3+6 & 7-4 \\
2-8 & 5+2
\end{array}\right]=\left[\begin{array}{rr}
9 & 3 \\
-6 & 7
\end{array}\right] \\
& \therefore \quad C=\frac{1}{5}(A-2 B)=\frac{1}{5}\left[\begin{array}{rr}
9 & 3 \\
-6 & 7
\end{array}\right]=\left[\begin{array}{rr}
9 / 5 & 3 / 5 \\
-6 / 5 & 7 / 5
\end{array}\right] \\
&
\end{aligned}
$
Question 10 .
If $A=\left[\begin{array}{rr}1 & -1 \\ 2 & -1\end{array}\right]$ and $B=\left[\begin{array}{rr}x & 1 \\ y & -1\end{array}\right]$ and $(A+B)^2=A^2+B^2$, find $x$ and $y$.
Solution:

$\begin{aligned}
& \mathrm{A}=\left[\begin{array}{rr}
1 & -1 \\
2 & -1
\end{array}\right] ; \mathrm{B}=\left[\begin{array}{rr}
x & 1 \\
y & -1
\end{array}\right] \\
& \therefore \quad \mathrm{A}+\mathrm{B}=\left[\begin{array}{rr}
1 & -1 \\
2 & -1
\end{array}\right]+\left[\begin{array}{rr}
x & 1 \\
y & -1
\end{array}\right]=\left[\begin{array}{rr}
1+x & 0 \\
2+y & -2
\end{array}\right] \\
&
\end{aligned}$

$
\begin{aligned}
& =\left[\begin{array}{cc}
-1 & 0 \\
0 & -1
\end{array}\right] \\
\mathrm{B}^2 & =\left[\begin{array}{cc}
x & 1 \\
y & -1
\end{array}\right]\left[\begin{array}{cc}
x & 1 \\
y & -1
\end{array}\right]=\left[\begin{array}{ll}
x^2+y & x-1 \\
x y-y & y+1
\end{array}\right]
\end{aligned}
$
So, $\quad \begin{aligned} \mathrm{A}^2+\mathrm{B}^2 & =\left[\begin{array}{cc}-1 & 0 \\ 0 & -1\end{array}\right]+\left[\begin{array}{cc}x^2+y & x-1 \\ x y-y & y+1\end{array}\right] \\ & =\left[\begin{array}{cc}x^2+y-1 & x-1 \\ x y-y & y+1-1\end{array}\right]=\left[\begin{array}{cc}x^2+y-1 & x-1 \\ x y-y & y\end{array}\right]\end{aligned}$
Given (1) $=(2) \Rightarrow\left[\begin{array}{cc}(1+x)^2 & 0 \\ (2+y)(1+x)-2(2+y) & 4\end{array}\right]=\left[\begin{array}{cc}x^2+y-1 & x-1 \\ x y-y & y\end{array}\right]$
(i.e.)
$
\begin{aligned}
x-1 & =0 \quad \Rightarrow & x=1 \\
y & =4 \quad & \text { So, } x=1, y=4
\end{aligned}
$