Exercise 1.1 - Chapter 1 Applications of Matrices and Determinants 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

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On April 24, 2024, 11:35 AM

Applications of Matrices and Determinants
Ex $1.1$
Question 1.

 Find the adjoint of the following:
(i) $\left[\begin{array}{cc}-3 & 4 \\ 6 & 2\end{array}\right]$
(ii) $\left[\begin{array}{lll}2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2\end{array}\right]$
(iii) $\frac{1}{3}\left[\begin{array}{ccc}2 & 2 & 1 \\ -2 & 1 & 2 \\ 1 & -2 & 2\end{array}\right]$
Solution:
(i) $\left[\begin{array}{cc}-3 & 4 \\ 6 & 2\end{array}\right]$
If matrix $\quad \mathrm{A}=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)$,
Then
$\operatorname{adj} \mathrm{A}=\left(\mathrm{A}_{i j}\right)^{\mathrm{T}}=\left(\begin{array}{rr}d & -b \\ -c & a\end{array}\right)$
Here
$
A=\left(\begin{array}{rr}
-3 & 4 \\
6 & 2
\end{array}\right), \text { So adj } A=\left(\begin{array}{rr}
2 & -4 \\
-6 & -3
\end{array}\right)
$

(ii) $\left[\begin{array}{lll}2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2\end{array}\right]$
Let $A=\left[\begin{array}{lll}2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2\end{array}\right]$ $\operatorname{adj} \mathrm{A}=\left(\mathrm{A}_{i j}\right)^{\mathrm{T}}$ where $\left(\mathrm{A}_{i j}\right)=$ Co-factor matrix of $\mathrm{A}$

$\begin{aligned}
& =\left[\begin{array}{lll}
+(8-7) & -(6-3) & +(21-12) \\
-(6-7) & +(4-3) & -(14-9) \\
+(3-4) & -(2-3) & +(8-9)
\end{array}\right]=\left(\begin{array}{ccc}
1 & -3 & 9 \\
1 & 1 & -5 \\
-1 & 1 & -1
\end{array}\right) \\
& \therefore \quad\left(A_{i i}\right)^{\mathrm{T}}=\left(\begin{array}{rrr}
1 & -3 & 9 \\
1 & 1 & -5 \\
-1 & 1 & -1
\end{array}\right)^{\mathrm{T}}=\left(\begin{array}{rrr}
1 & 1 & -1 \\
-3 & 1 & 1 \\
9 & -5 & -1
\end{array}\right) \\
& \operatorname{adj} A=\left(\begin{array}{rrr}
1 & 1 & -1 \\
-3 & 1 & 1 \\
9 & -5 & -1
\end{array}\right) \\
&
\end{aligned}$

$\text { (iii) } \begin{aligned}
& \frac{1}{3}\left[\begin{array}{ccc}
2 & 2 & 1 \\
-2 & 1 & 2 \\
1 & -2 & 2
\end{array}\right] \\
& \mathrm{A}=\frac{1}{3}\left(\begin{array}{rrr}
2 & 2 & 1 \\
-2 & 1 & 2 \\
1 & -2 & 2
\end{array}\right) \text { (i.e) } \mathrm{A}=\left(\begin{array}{rrr}
2 / 3 & 2 / 3 & 1 / 3 \\
-2 / 3 & 1 / 3 & 2 / 3 \\
1 / 3 & -2 / 3 & 2 / 3
\end{array}\right)
\end{aligned}$

$
\begin{aligned}
& =\frac{1}{9}\left(\begin{array}{rrr}
+(2+4) & -(-4-2) & +(4-1) \\
-(4+2) & +(4-1) & -(-4-2) \\
+(4-1) & -(4+2) & +(2+4)
\end{array}\right) \\
& =\frac{1}{9}\left(\begin{array}{rrr}
6 & 6 & 3 \\
-6 & 3 & 6 \\
3 & -6 & 6
\end{array}\right)=\frac{1}{3}\left(\begin{array}{rrr}
2 & 2 & 1 \\
-2 & 1 & 2 \\
1 & -2 & 2
\end{array}\right) \\
\operatorname{adj~} A & =\left(\mathrm{A}_{i j}\right)^{\mathrm{T}}=\frac{1}{3}\left(\begin{array}{rrr}
2 & -2 & 1 \\
2 & 1 & -2 \\
1 & 2 & 2
\end{array}\right)
\end{aligned}
$
Question 2.

Find the inverse (if it exists) of the following:
(i) $\left[\begin{array}{cc}-2 & 4 \\ 1 & -3\end{array}\right]$
(ii) $\left[\begin{array}{lll}5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5\end{array}\right]$
(iii) $\left[\begin{array}{lll}2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2\end{array}\right]$
Solution:
For a matrix $A, A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)$. Where $|A| \neq 0$. If $|\mathrm{A}|=0$ then $\mathrm{A}$ is called a singular matrix and so $\mathrm{A}^{-1}$ does not exist.

Let
$
\begin{aligned}
A & =\left[\begin{array}{cc}
-2 & 4 \\
1 & -3
\end{array}\right] \\
|A| & =\left[\begin{array}{cc}
-2 & 4 \\
1 & -3
\end{array}\right]=6-4=2 \neq 0
\end{aligned}
$
So $\mathrm{A}$ is a non singular matrix and so $\mathrm{A}^{-1}$ exists.
$
\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|}(\operatorname{adj} \mathrm{A})
$
For a matrix $\mathrm{A}=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right), \operatorname{adj} \mathrm{A}=\left(\begin{array}{cc}d & -b \\ -c & a\end{array}\right)$
So $\operatorname{adj} A=\left(\begin{array}{ll}-3 & -4 \\ -1 & -2\end{array}\right) \quad \therefore A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{2}\left(\begin{array}{ll}-3 & -4 \\ -1 & -2\end{array}\right)$
(ii) $\left[\begin{array}{lll}5 & 1 & 1 \\ 1 & 5 & 1 \\ 1 & 1 & 5\end{array}\right]$
Let
$
\begin{aligned}
A & =\left[\begin{array}{lll}
5 & 1 & 1 \\
1 & 5 & 1 \\
1 & 1 & 5
\end{array}\right] \\
Now |A| & =\left[\begin{array}{lll}
5 & 1 & 1 \\
1 & 5 & 1 \\
1 & 1 & 5
\end{array} \mid=5(25-1)-1(5-1)+1(1-5)\right. \\
& =5(24)-4-4=120-8=112 \neq 0
\end{aligned}
$
$\Rightarrow \mathrm{A}$ is a non singular matrix and so $\mathrm{A}^{-1}$ exists.

$\begin{aligned}
\mathrm{A}^{-1} & =\frac{1}{|\mathrm{~A}|}(\operatorname{adj} \mathrm{A}) \\
\operatorname{adj} \mathrm{A} & =\left(\mathrm{A}_{i j}\right)^{\mathrm{T}}
\end{aligned}$

$
=\left(\begin{array}{rrr}
24 & -4 & -4 \\
-4 & 24 & -4 \\
-4 & -4 & 24
\end{array}\right)
$
$\operatorname{adj} A=\left(A_{i j}\right)^T=\left(\begin{array}{ccc}24 & -4 & -4 \\ -4 & 24 & -4 \\ -4 & -4 & 24\end{array}\right)$
$
A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{112}\left(\begin{array}{ccc}
24 & -4 & -4 \\
-4 & 24 & -4 \\
-4 & -4 & 24
\end{array}\right)=\frac{1}{28}\left(\begin{array}{ccc}
6 & -1 & -1 \\
-1 & 6 & -1 \\
-1 & -1 & 6
\end{array}\right)
$
(iii) $\left[\begin{array}{lll}2 & 3 & 1 \\ 3 & 4 & 1 \\ 3 & 7 & 2\end{array}\right]$
Let
$
\begin{aligned}
A & =\left[\begin{array}{lll}
2 & 3 & 1 \\
3 & 4 & 1 \\
3 & 7 & 2
\end{array}\right] \\
|A| & =\left[\begin{array}{lll}
2 & 3 & 1 \\
3 & 4 & 1 \\
3 & 7 & 2
\end{array} \mid=2(8-7)-3(6-3)+1(21-12)\right. \\
& =2(1)-3(3)+1(9)=2-9+9=2 \neq 0
\end{aligned}
$
So $\mathrm{A}^{-1}$ exists
Now
$
\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|}(\operatorname{adj} \mathrm{A})
$

$\begin{aligned}
\operatorname{adj} A & =\left(\begin{array}{rrr}
1 & 1 & -1 \\
-3 & 1 & 1 \\
9 & -5 & -1
\end{array}\right) \\
A^{-1} & =\frac{1}{2}\left(\begin{array}{rrr}
1 & 1 & -1 \\
-3 & 1 & 1 \\
9 & -5 & -1
\end{array}\right)
\end{aligned}$

Question 3.
If $\mathrm{F}(\alpha)=\left[\begin{array}{ccc}\cos \alpha & 0 & \sin \alpha \\ 0 & 1 & 0 \\ -\sin \alpha & 0 & \cos \alpha\end{array}\right]$ show that $[\mathrm{F}(\alpha)]^{-1}=\mathrm{F}(-\alpha)$
Solution:
Let $\mathrm{A}=\mathrm{F}(\alpha)$
So $[\mathrm{F}(\alpha)]^{-1}=\mathrm{A}^{-1}$
Now
$
\begin{aligned}
A & =\left[\begin{array}{ccc}
\cos \alpha & 0 & \sin \alpha \\
0 & 1 & 0 \\
-\sin \alpha & 0 & \cos \alpha
\end{array}\right] \\
|A| & =\left[\begin{array}{ccc}
\cos \alpha & 0 & \sin \alpha \\
0 & 1 & 0 \\
-\sin \alpha & 0 & \cos \alpha
\end{array} \mid\right.
\end{aligned}
$
Expanding the determinant - along $\mathrm{R}_2$ We get
$
-0()+1\left[\cos ^2 \alpha+\sin ^{-2} \alpha\right]-0()=1 \neq 0
$
So $\mathrm{A}^{-1}$ exists
Now
$
A^{-1}=\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{1}(\operatorname{adj} A)=\operatorname{adj} A
$
To Find adj $\mathbf{A}: \operatorname{adj} \mathrm{A}=\left(\mathrm{A}_i\right)^{\mathrm{T}}$

$\begin{aligned}
&= {\left[\begin{array}{ccc}
+(\cos \alpha) & -(0) & +(\sin \alpha) \\
-(0) & +(1) & -(0) \\
+(-\sin \alpha) & -(0) & +(\cos \alpha)
\end{array}\right]=\left(\begin{array}{ccc}
\cos \alpha & 0 & \sin \alpha \\
0 & 1 & 0 \\
-\sin \alpha & 0 & \cos \alpha
\end{array}\right) } \\
& \operatorname{adj} A=\left(\mathrm{A}^{-1}\right)=\left(\mathrm{A}_{i j}\right)^{\mathrm{T}}=\left(\begin{array}{ccc}
\cos \alpha & 0 & -\sin \alpha \\
0 & 1 & 0 \\
\sin \alpha & 0 & \cos \alpha
\end{array}\right)
\end{aligned}$

(i.e)
Given
So
$
\begin{aligned}
& \mathrm{A}^{-1}=[\mathrm{F}(\alpha)]^{-1}=\left(\begin{array}{ccc}
\cos \alpha & 0 & -\sin \alpha \\
0 & 1 & 0 \\
\sin \alpha & 0 & \cos \alpha
\end{array}\right) \\
& F(\alpha)=\left[\begin{array}{ccc}
\cos \alpha & 0 & \sin \alpha \\
0 & 1 & 0 \\
-\sin \alpha & 0 & \cos \alpha
\end{array}\right] \\
& F(-\alpha)=\left[\begin{array}{ccc}
\cos (-\alpha) & 0 & \sin (-\alpha) \\
0 & 1 & 0 \\
-\sin (-\alpha) & 0 & \cos (-\alpha)
\end{array}\right] \\
& =\left[\begin{array}{ccc}
\cos \alpha & 0 & -\sin \alpha \\
0 & 1 & 0 \\
\sin \alpha & 0 & \cos \alpha
\end{array}\right] \\
& (\therefore \cos (-\theta)=\cos \theta \text { and } \sin (-\theta)=-\sin \theta) \\
&
\end{aligned}
$
Here
(1) $=(2)$
$\Rightarrow[F(\alpha)]^{-1}=F(-\alpha)$

Question 4.

If $A=$ show that $A$
$
\left[\begin{array}{cc}
5 & 3 \\
-1 & -2
\end{array}\right] \quad 2-3 \mathrm{~A}-7 \mathrm{I}_2=\mathrm{O}_2 \text {. Hence find } \mathrm{A}^{-1} .
$
Solution:
$
A=\left[\begin{array}{cc}
5 & 3 \\
-1 & -2
\end{array}\right]
$

$\begin{aligned}
\text { LHS } & =\mathrm{A}^2-3 \mathrm{~A}-7 \mathrm{I}_2=\left(\begin{array}{ll}
22 & 9 \\
-3 & 1
\end{array}\right)-3\left(\begin{array}{cc}
5 & 3 \\
-1 & -2
\end{array}\right)-7\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right) \\
& =\left(\begin{array}{ll}
22 & 9 \\
-3 & 1
\end{array}\right)-\left(\begin{array}{rr}
15 & 9 \\
-3 & 6
\end{array}\right)+\left(\begin{array}{rr}
-7 & 0 \\
0 & -7
\end{array}\right) \\
& =\left(\begin{array}{rr}
22 & 9 \\
-3 & 1
\end{array}\right)+\left(\begin{array}{rr}
-15 & -9 \\
3 & 6
\end{array}\right)+\left(\begin{array}{rr}
-7 & 0 \\
0 & -7
\end{array}\right)=\left(\begin{array}{rr}
22-15-7 & 9-9+0 \\
-3+3+0 & 1+6-7
\end{array}\right) \\
& =\left(\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right)=\mathrm{O}_2=\text { RHS }
\end{aligned}$

To Find A-1
Now we have proved that $\mathrm{A}^2-3 \mathrm{~A}-7 \mathrm{I}_2=\mathrm{O}_2$
Post multiply by $\mathrm{A}^{-1}$ we get
$\mathrm{A}-3 \mathrm{I}-7 \mathrm{~A}^{-1}=\mathrm{O}_2$

$\begin{aligned}
& 7 A^{-1}=A-3 I=\left(\begin{array}{rr}
5 & 3 \\
-1 & -2
\end{array}\right)-3\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)=\left(\begin{array}{rr}
5 & 3 \\
-1 & -2
\end{array}\right)+\left(\begin{array}{rr}
-3 & 0 \\
0 & -3
\end{array}\right)=\left(\begin{array}{cc}
2 & 3 \\
-1 & -5
\end{array}\right) \\
& 7 A^{-1}=\left(\begin{array}{cc}
2 & 3 \\
-1 & -5
\end{array}\right) \quad \Rightarrow A^{-1}=\frac{1}{7}\left(\begin{array}{cc}
2 & 3 \\
-1 & -5
\end{array}\right) \\
&
\end{aligned}$

Question 5.
If $\mathbf{A}=\frac{1}{9}\left[\begin{array}{ccc}-8 & 1 & 4 \\ 4 & 4 & 7 \\ 1 & -8 & 4\end{array}\right]$ prove that $\mathrm{A}^{-1}=\mathrm{A}^{\mathrm{T}}$
Solution:
$
\begin{aligned}
& A=\frac{1}{9}\left(\begin{array}{ccc}
-8 & 1 & 4 \\
4 & 4 & 7 \\
1 & -8 & 4
\end{array}\right) \\
& |A|=\frac{1}{9^3}\left|\begin{array}{ccc}
-8 & 1 & 4 \\
4 & 4 & 7 \\
1 & -8 & 4
\end{array}\right|=\frac{1}{729}[-8(16+56)-1(16-7)+4(-32-4)] \\
& =\frac{1}{729}[-8(72)-1(9)+4(-36)]=\frac{1}{729}[9(-8 \times 8-1-4 \times 4)] \\
& =\frac{1}{81}(-64-1-16)=\frac{1}{81}(-81)=-1 \neq 0 \\
& \therefore \mathrm{A}^{-1} \text { exists } \quad \mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|}(\operatorname{adj} \mathrm{A}) \\
& \text { To find adj } A: \quad \text { adj } A=\left(A_{i j}\right)^{\mathrm{T}} \\
&
\end{aligned}
$

$\begin{aligned}
A^{-1} & =\frac{1}{|A|}(\operatorname{adj} A)=\frac{1}{-1}\left[\frac{1}{9}\left(\begin{array}{rrr}
8 & -4 & -1 \\
-1 & -4 & 8 \\
-4 & -7 & -4
\end{array}\right)\right] \\
& =\frac{1}{9}\left(\begin{array}{ccc}
-8 & 4 & 1 \\
1 & 4 & -8 \\
4 & 7 & 4
\end{array}\right) \\
A & =\frac{1}{9}\left(\begin{array}{rrr}
-8 & 1 & 4 \\
4 & 4 & 7 \\
1 & -8 & 4
\end{array}\right) \\
A^{\mathrm{T}} & =\frac{1}{9}\left(\begin{array}{ccc}
-8 & 4 & 1 \\
1 & 4 & -8 \\
4 & 7 & 4
\end{array}\right) \Rightarrow A^{-1}=A^{\mathrm{T}} \\
(1) & =(2)
\end{aligned}$

Question 6.
If $\mathbf{A}=\left[\begin{array}{rr}8 & -4 \\ -5 & 3\end{array}\right]$, verify that $\mathrm{A}(\operatorname{adj} \mathrm{A})=(\operatorname{adj} \mathrm{A}) \mathrm{A}=|\mathrm{A}| \mathrm{I}_2$

Solution:
$
\begin{aligned}
& A=\left(\begin{array}{rr}
8 & -4 \\
-5 & 3
\end{array}\right) \quad \therefore|\mathrm{A}|=\left|\begin{array}{rr}
8 & -4 \\
-5 & 3
\end{array}\right|=24-20=4 \\
& A=\left(\begin{array}{rr}
8 & -4 \\
-5 & 3
\end{array}\right) \\
& \operatorname{adj} A=\left(\begin{array}{ll}
3 & 4 \\
5 & 8
\end{array}\right) . \\
& \mathrm{A}(\operatorname{adj~A})=\left(\begin{array}{rr}
8 & -4 \\
-5 & 3
\end{array}\right)\left(\begin{array}{ll}
3 & 4 \\
5 & 8
\end{array}\right)=\left(\begin{array}{cc}
24-20 & 32-32 \\
-15+15 & -20+24
\end{array}\right) \\
&
\end{aligned}
$

$\begin{aligned}
& =\left(\begin{array}{ll}
4 & 0 \\
0 & 4
\end{array}\right)=4\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)=|A| I_2 \\
& (\operatorname{adj} A) A=\left(\begin{array}{ll}
3 & 4 \\
5 & 8
\end{array}\right) \cdot\left(\begin{array}{rr}
8 & -4 \\
-5 & 3
\end{array}\right)=\left(\begin{array}{cc}
24-20 & -12+12 \\
40-40 & -20+24
\end{array}\right) \\
& =\left(\begin{array}{ll}
4 & 0 \\
0 & 4
\end{array}\right)=4\left(\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right)=|A| I_2 \\
& (1)=(2) \quad \Rightarrow A(\operatorname{adj} A)=(\operatorname{adj} A) A=|A| I_2 \\
&
\end{aligned}$

Question 7.
If $\mathbf{A}=\left[\begin{array}{ll}3 & 2 \\ 7 & 5\end{array}\right]$, and $\mathbf{B}=\left[\begin{array}{cc}-1 & -3 \\ 5 & 2\end{array}\right]$ verify that $(\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}$.
Solution:
Given
$
A=\left(\begin{array}{ll}
3 & 2 \\
7 & 5
\end{array}\right) \text { and } B=\left(\begin{array}{cc}
-1 & -3 \\
5 & 2
\end{array}\right)
$
To prove
$
\begin{aligned}
(\mathrm{AB})^{-1} & =\mathrm{B}^{-1} \mathrm{~A}^{-1} \\
\mathrm{AB} & =\left(\begin{array}{ll}
3 & 2 \\
7 & 5
\end{array}\right) \downarrow\left(\begin{array}{cc}
-1 & -3 \\
5 & 2
\end{array}\right)=\left(\begin{array}{cc}
-3+10 & -9+4 \\
-7+25 & -21+10
\end{array}\right)=\left(\begin{array}{cc}
7 & -5 \\
18 & -11
\end{array}\right) \\
\mathrm{AB} & =\left(\begin{array}{cc}
7 & -5 \\
18 & -11
\end{array}\right)
\end{aligned}
$

To find $(\mathbf{A B})^{-1}:|A B|=\left|\begin{array}{cc}7 & -5 \\ 18 & -11\end{array}\right|=-77+90=13$
$
\begin{aligned}
& (\operatorname{adj} A B)=\left(\begin{array}{ll}
-11 & 5 \\
-18 & 7
\end{array}\right) \\
&
\end{aligned}
$
So
$(A B)^{-1}=\frac{1}{13}(\operatorname{adj} A B)=\frac{1}{13}\left(\begin{array}{rr}-11 & 5 \\ -18 & 7\end{array}\right)$
Here
$A=\left(\begin{array}{ll}3 & 2 \\ 7 & 5\end{array}\right),|A|=\left|\begin{array}{ll}3 & 2 \\ 7 & 5\end{array}\right|=15-14=1$
$\operatorname{adj} A=\left(\begin{array}{cc}5 & -2 \\ -7 & 3\end{array}\right)$
So
$\mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|}$ adj $\mathrm{A}=\frac{1}{1}\left(\begin{array}{cc}5 & -2 \\ -7 & 3\end{array}\right)=\left(\begin{array}{cc}5 & -2 \\ -7 & 3\end{array}\right)$
$\mathrm{B}=\left(\begin{array}{cc}-1 & -3 \\ 5 & 2\end{array}\right),|\mathrm{B}|=\left|\begin{array}{cc}-1 & -3 \\ 5 & 2\end{array}\right|=-2+15=13$

$\begin{aligned}
\operatorname{adj} \mathrm{B} & =\left(\begin{array}{cc}
2 & 3 \\
-5 & -1
\end{array}\right) \\
\mathrm{B}^{-1} & =\frac{1}{|\mathrm{~B}|}(\operatorname{adj} \mathrm{B})=\frac{1}{13}\left(\begin{array}{cc}
2 & 3 \\
-5 & -1
\end{array}\right) \\
\mathrm{B}^{-1} \mathrm{~A}^{-1} & =\frac{1}{13}\left(\begin{array}{cc}
2 & 3 \\
-5 & -1
\end{array}\right)\left(\begin{array}{cc}
5 & -2 \\
-7 & 3
\end{array}\right) \\
& =\frac{1}{13}\left(\begin{array}{cc}
10-21 & -4+9 \\
-25+7 & 10-3
\end{array}\right)=\frac{1}{13}\left(\begin{array}{cc}
-11 & 5 \\
-18 & 7
\end{array}\right) \\
(1) & =(2) \quad \Rightarrow(\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1}
\end{aligned}$

Question 8.
If $\operatorname{adj}(A)=\left[\begin{array}{ccc}2 & -4 & 2 \\ -3 & 12 & -7 \\ -2 & 0 & 2\end{array}\right]$ find $A$
Solution:
Given
$
\operatorname{adj} A=\left(\begin{array}{ccc}
2 & -4 & 2 \\
-3 & 12 & -7 \\
-2 & 0 & 2
\end{array}\right) \quad \therefore|\operatorname{adj} A|=\left|\begin{array}{ccc}
2 & -4 & 2 \\
-3 & 12 & -7 \\
-2 & 0 & 2
\end{array}\right|
$
Expanding along $\mathrm{R}_3=-2(28-24)-0()+2(24-12)=-2(4)+2(12)=-8+24=16 \neq 0$ To find adj (adj A):
Let
$
B=\operatorname{adj} A
$
(i.e) $\mathrm{B}=\left(\begin{array}{ccc}2 & -4 & 2 \\ -3 & 12 & -7 \\ -2 & 0 & 2\end{array}\right)$
To find adj (adj B):
$
\operatorname{adj} \mathrm{B}=\left(\mathrm{B}_{i j}\right)^{\mathrm{T}}
$

Question 9.
If $\operatorname{adj}(A)=\left[\begin{array}{ccc}0 & -2 & 0 \\ 6 & 2 & -6 \\ -3 & 0 & 6\end{array}\right]$ find $A^{-1}$
Solution:
$
\operatorname{adj} A=\left[\begin{array}{ccc}
0 & -2 & 0 \\
6 & 2 & -6 \\
-3 & 0 & 6
\end{array}\right] \quad \therefore|\operatorname{adj} A|=\left|\begin{array}{ccc}
0 & -2 & 0 \\
6 & 2 & -6 \\
-3 & 0 & 6
\end{array}\right|
$
expanding along $R_1$
$
\begin{aligned}
& -(-2)[36-18]=2(18)=36 \\
& \mathrm{~A}^{-1}=\pm \frac{1}{\sqrt{|\operatorname{adj} \mathrm{A}|}} \operatorname{adj} \mathrm{A}=\pm \frac{1}{\sqrt{36}}\left(\begin{array}{ccc}
0 & -2 & 0 \\
6 & 2 & -6 \\
-3 & 0 & 6
\end{array}\right)=\pm \frac{1}{6}\left(\begin{array}{ccc}
0 & -2 & 0 \\
6 & 2 & -6 \\
-3 & 0 & 6
\end{array}\right) \\
&
\end{aligned}
$

Question 10.
Find $\operatorname{adj}(\operatorname{adj}(A))$ if adj $A=\left[\begin{array}{ccc}1 & 0 & 1 \\ 0 & 2 & 0 \\ -1 & 0 & 1\end{array}\right]$
Solution:
Given $\operatorname{adj} A=\left(\begin{array}{ccc}1 & 0 & 1 \\ 0 & 2 & 0 \\ -1 & 0 & 1\end{array}\right)$ to find $\operatorname{adj}(\operatorname{adj} A)$
Let $\operatorname{adj} A=B$ $\therefore \quad \operatorname{adj}(\operatorname{adj} \mathrm{A})=\operatorname{adj} \mathrm{B}$ (i.e) $\mathrm{B}=\left(\begin{array}{ccc}1 & 0 & 1 \\ 0 & 2 & 0 \\ -1 & 0 & 1\end{array}\right)$
Now adj $B=\left(B_{i j}\right)^{\top}$

$\begin{aligned}
= & {\left[\begin{array}{ccc}
+(2-0) & -(0-0) & +(0+2) \\
-(0) & +(1+1) & -(0) \\
+(0-2) & -(0) & +(2-0)
\end{array}\right]=\left(\begin{array}{ccc}
2 & 0 & 2 \\
0 & 2 & 0 \\
-2 & 0 & 2
\end{array}\right) } \\
\left(\mathrm{B}_{i j}\right)^{\mathrm{T}} & =\left(\begin{array}{ccc}
2 & 0 & -2 \\
0 & 2 & 0 \\
2 & 0 & 2
\end{array}\right) \quad \text { (i.e) } \operatorname{adj} \mathrm{B}=\operatorname{adj}(\operatorname{adj} \mathrm{A})=\left(\begin{array}{ccc}
2 & 0 & -2 \\
0 & 2 & 0 \\
2 & 0 & 2
\end{array}\right)
\end{aligned}$

Question 11.
$
\mathbf{A}=\left[\begin{array}{cc}
1 & \tan x \\
-\tan x & 1
\end{array}\right] \text { show that } \mathbf{A}^{\mathrm{T}} \mathbf{A}^{-1}=\left[\begin{array}{cc}
\cos 2 x & -\sin 2 x \\
\sin 2 x & \cos 2 x
\end{array}\right]
$
Solution:
$
\begin{aligned}
\mathrm{A} & =\left[\begin{array}{cc}
1 & \tan x \\
-\tan x & 1
\end{array}\right] \quad \therefore \mathrm{A}^{\mathrm{T}}=\left(\begin{array}{cc}
1 & -\tan x \\
\tan x & 1
\end{array}\right) \\
|\mathrm{A}| & =\left[\begin{array}{cc}
1 & \tan x \\
-\tan x & 1
\end{array}\right]=1+\tan ^2 x \\
\operatorname{adj} \mathrm{A} & =\left(\begin{array}{cc}
1 & -\tan x \\
\tan x & 1
\end{array}\right) \therefore \mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|}(\operatorname{adj} \mathrm{A})=\frac{1}{1+\tan ^2 x}\left(\begin{array}{cc}
1 & -\tan x \\
\tan x & 1
\end{array}\right)
\end{aligned}
$

$\begin{aligned}
\mathrm{A}^{\mathrm{T}} \mathrm{A}^{-1} & =\frac{1}{1+\tan ^2 x}\left[\begin{array}{cc}
1 & -\tan x \\
\tan x & 1
\end{array}\right]\left[\begin{array}{cc}
1 & -\tan x \\
\tan x & 1
\end{array}\right] \\
& =\frac{1}{1+\tan ^2 x}\left[\begin{array}{cc}
1-\tan ^2 x & -2 \tan x \\
2 \tan x & 1-\tan ^2 x
\end{array}\right] \\
& =\left(\begin{array}{cc}
\frac{1-\tan ^2 x}{1+\tan ^2 x} & \frac{-2 \tan x}{1+\tan ^2 x} \\
\frac{2 \tan x}{1+\tan ^2 x} & \frac{1-\tan ^2 x}{1+\tan ^2 x}
\end{array}\right)=\left(\begin{array}{cc}
\cos 2 x & -\sin 2 x \\
\sin 2 x & \cos 2 x
\end{array}\right)=\text { RHS }
\end{aligned}$

Question 12.
Find the matrix A for which $\mathrm{A}\left[\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right]=\left[\begin{array}{cc}14 & 7 \\ 7 & 7\end{array}\right]$
Solution:
Given A $\left[\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right]=\left[\begin{array}{cc}14 & 7 \\ 7 & 7\end{array}\right]$
Let $\mathrm{B}=\left(\begin{array}{cc}5 & 3 \\ -1 & -2\end{array}\right)$ and $\mathrm{C}=\left(\begin{array}{cc}14 & 7 \\ 7 & 7\end{array}\right)$
Given $A B=C$, To find $A$
Now $\mathrm{AB}=\mathrm{C}$
Post multiply by $\mathrm{B}^{-1}$ on both sides
$\mathrm{ABB}^{-1}=\mathrm{CB}^{-1}$ (i.e) $\mathrm{A}\left(\mathrm{BB}^{-1}\right)=\mathrm{CB}^{-1}$
$\Rightarrow \mathrm{A}(\mathrm{I})=\mathrm{CB}^{-1}$ (i.e) $\mathrm{A}=\mathrm{CB}^{-1}$

To find $\mathrm{B}^{-1}$ :
$
\begin{aligned}
& B=\left(\begin{array}{cc}
5 & 3 \\
-1 & -2
\end{array}\right) \\
& |\mathrm{B}|=\left|\begin{array}{cc}
5 & 3 \\
-1 & -2
\end{array}\right|=-10+3=-7 \neq 0 \\
& \operatorname{adj} B=\left(\begin{array}{cc}
-2 & -3 \\
1 & 5
\end{array}\right) \\
& \mathrm{B}^{-1}=\frac{1}{|\mathrm{~B}|}(\operatorname{adj} \mathrm{B})=\frac{1}{-7}\left(\begin{array}{cc}
-2 & -3 \\
1 & 5
\end{array}\right)=\frac{1}{7}\left(\begin{array}{cc}
2 & 3 \\
-1 & -5
\end{array}\right) \\
& \mathrm{A}=\mathrm{CB}^{-1}=\frac{1}{7}\left(\begin{array}{cc}
14 & 7 \\
7 & 7
\end{array}\right)\left(\begin{array}{cc}
2 & 3 \\
-1 & -5
\end{array}\right)=\frac{1}{7}(7)\left(\begin{array}{ll}
2 & 1 \\
1 & 1
\end{array}\right)\left(\begin{array}{cc}
2 & 3 \\
-1 & -5
\end{array}\right) \\
& =\overrightarrow{\left(\begin{array}{ll}
2 & 1 \\
1 & 1
\end{array}\right)} \mid\left(\begin{array}{cc}
2 & 3 \\
-1 & -5
\end{array}\right)=\left(\begin{array}{cc}
4-1 & 6-5 \\
2-1 & 3-5
\end{array}\right)=\left(\begin{array}{cc}
3 & 1 \\
1 & -2
\end{array}\right) \\
& A=\left(\begin{array}{cc}
3 & 1 \\
1 & -2
\end{array}\right) \\
&
\end{aligned}
$

Question 13.
Given $\mathbf{A}=\left[\begin{array}{cc}1 & -1 \\ 2 & 0\end{array}\right], \mathbf{B}=\left[\begin{array}{cc}3 & -2 \\ 1 & 1\end{array}\right]$ and $\mathbf{C}\left[\begin{array}{ll}1 & 1 \\ 2 & 2\end{array}\right]$, find a matrix $\mathrm{X}$ such that $\mathrm{AXB}=\mathrm{C}$.
Solution:
$
\mathrm{A} \times \mathrm{B}=\mathrm{C}
$
Pre multiply by $A^{-1}$ and post multiply by $B^{-1}$ we get
$
\mathrm{A}^{-1} \mathrm{~A} \times \mathrm{BB}^{-1}=\mathrm{A}^{-1} \mathrm{CB}^{-1} \text { (i.e) } \mathrm{X}=\mathrm{A}^{-1} \mathrm{CB}^{-1}
$

$\begin{aligned}
& \mathrm{A}=\left(\begin{array}{cc}
1 & -1 \\
2 & 0
\end{array}\right) \quad \therefore|\mathrm{A}|=\left|\begin{array}{cc}
1 & -1 \\
2 & 0
\end{array}\right|=0+2=2 \\
& \operatorname{adj} A=\left(\begin{array}{cc}
0 & 1 \\
-2 & 1
\end{array}\right) \quad \therefore A^{-1}=\left(\frac{1}{|A|} \operatorname{adj} A\right)=\frac{1}{2}\left(\begin{array}{cc}
0 & 1 \\
-2 & 1
\end{array}\right) \\
& \mathrm{B}=\left(\begin{array}{cc}
3 & -2 \\
1 & 1
\end{array}\right) \quad|\mathrm{B}|=\left|\begin{array}{cc}
3 & -2 \\
1 & 1
\end{array}\right|=3+2=5 \\
& \operatorname{adj} \mathrm{B}=\left(\begin{array}{cc}
1 & 2 \\
-1 & 3
\end{array}\right) \quad \therefore \mathrm{B}^{-1}=\left(\frac{1}{|\mathrm{~B}|} \operatorname{adj} \mathrm{B}\right)=\frac{1}{5}\left(\begin{array}{cc}
1 & 2 \\
-1 & 3
\end{array}\right) \\
& \mathrm{X}=\mathrm{A}^{-1} \mathrm{C} \mathrm{B}^{-1}=\frac{1}{2}\left(\begin{array}{cc}
0 & 1 \\
-2 & 1
\end{array}\right)\left(\begin{array}{ll}
1 & 1 \\
2 & 2
\end{array}\right) \frac{1}{5}\left(\begin{array}{cc}
1 & 2 \\
-1 & 3
\end{array}\right) \\
& =\frac{1}{10}\left(\begin{array}{rr}
0 & 1 \\
-2 & 1
\end{array}\right)\left[\left(\begin{array}{ll}
1 & 1 \\
2 & 2
\end{array}\right) \|\left(\begin{array}{rr}
1 & 2 \\
-1 & 3
\end{array}\right)\right] \\
& =\frac{1}{10}\left(\begin{array}{rr}
0 & 1 \\
-2 & 1
\end{array}\right)\left[\begin{array}{cc}
1-1 & 2+3 \\
2-2 & 4+6
\end{array}\right]=\frac{1}{10}\left(\begin{array}{rc}
0 & 1 \\
-2 & 1
\end{array}\right)\left(\begin{array}{cc}
0 & 5 \\
0 & 10
\end{array}\right) \\
& =\frac{1}{10}\left[\begin{array}{cc}
0+0 & 0+10 \\
0+0 & -10+10
\end{array}\right]=\frac{1}{10}\left(\begin{array}{cc}
0 & 10 \\
0 & 0
\end{array}\right)=\left(\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right) \\
& \mathrm{X}=\left(\begin{array}{ll}
0 & 1 \\
0 & 0
\end{array}\right) \\
&
\end{aligned}$

Question 14.
If $\mathbf{A}=\left[\begin{array}{lll}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{array}\right]$, show that $A^{-1}=\frac{1}{2}\left(A^2-3 I\right)$.
Solution:

$\begin{aligned}
A & =\left(\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array}\right) \\
|A| & =\left(\begin{array}{lll}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{array} \mid=0()-1(0-1)+1(1-0)=0+1+1=2\right. \\
\operatorname{adj} A & =\left(A_{i j}\right)^T
\end{aligned}$

$\begin{aligned}
A^2-3 I & =\left(\begin{array}{lll}
2 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 2
\end{array}\right)-3\left(\begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right) \\
& =\left(\begin{array}{lll}
2 & 1 & 1 \\
1 & 2 & 1 \\
1 & 1 & 2
\end{array}\right)+\left(\begin{array}{ccc}
-3 & 0 & 0 \\
0 & -3 & 0 \\
0 & 0 & -3
\end{array}\right)=\left(\begin{array}{ccc}
-1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 1 & -1
\end{array}\right) \\
\frac{1}{2}\left(A^2-3 I\right) & =\frac{1}{2}\left(\begin{array}{ccc}
-1 & 1 & 1 \\
1 & -1 & 1 \\
1 & 1 & -1
\end{array}\right) \\
(1) & =(2)
\end{aligned}$

Question 15.
Decrypt the received encoded message $\left[\begin{array}{lll}2 & -3\end{array}\right]\left[\begin{array}{ll}20 & 4\end{array}\right]$ with the encryption matrix $\left[\begin{array}{cc}-1 & -1 \\ 2 & 1\end{array}\right]$ and the decryption matrix as its inverse, where the system of codes are described by the numbers 1-26 to the letters $A-Z$ respectively, and the number 0 to a blank space.
Solution:
Let the encoding matrix be $\left[\begin{array}{cc}-1 & -1 \\ 2 & 1\end{array}\right]$
Let
Now
$
\begin{aligned}
A & =\left(\begin{array}{cc}
-1 & -1 \\
2 & 1
\end{array}\right) \\
|A| & =\left|\begin{array}{cc}
-1 & -1 \\
2 & 1
\end{array}\right|=-1+2=1
\end{aligned}
$
So $\operatorname{adj} A=\left(\begin{array}{cc}1 & 1 \\ -2 & -1\end{array}\right)$
$
A^{-1}=\frac{1}{1}\left(\begin{array}{cc}
1 & 1 \\
-2 & -1
\end{array}\right)=\left(\begin{array}{cc}
1 & 1 \\
-2 & -1
\end{array}\right)
$

So the sequence of decoded matrices is [8 5], [12 16 $]$. Thus the receivers read this message as HELP.