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

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Additional Problems
Question 1.
Using elementary transformations find the inverse of the following matrix
Solution:
, Let
Now
$
\begin{aligned}
& A=\left[\begin{array}{ll}
4 & 7 \\
3 & 6
\end{array}\right] \\
& A=I A
\end{aligned}
$
$\therefore \quad\left[\begin{array}{ll}4 & 7 \\ 3 & 6\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \mathrm{A}$
Operating $R_1 \rightarrow R_1-R_2 \quad\left[\begin{array}{ll}1 & 1 \\ 3 & 6\end{array}\right]=\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right] A$
Operating $R_2 \rightarrow R_2-3 R_1\left[\begin{array}{ll}1 & 1 \\ 0 & 3\end{array}\right]=\left[\begin{array}{cc}1 & -1 \\ -3 & 4\end{array}\right] A$
Operating $R_2 \rightarrow \frac{1}{3} R_2 \quad\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}1 & -1 \\ -1 & 4 / 3\end{array}\right] A$
Operating $R_1 \rightarrow R_1-R_2\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}2 & -7 / 3 \\ -1 & 4 / 3\end{array}\right] A$
Thus
$
A^{-1}=\left[\begin{array}{cc}
2 & -7 / 3 \\
-1 & 4 / 3
\end{array}\right]=\frac{1}{3}\left[\begin{array}{cc}
6 & -7 \\
-3 & 4
\end{array}\right]
$

Question 2.
Using elementary transformations find the inverse of the matrix

Solution:
$
\left[\begin{array}{ccc}
1 & -1 & 1 \\
2 & 1 & -3 \\
1 & 1 & 1
\end{array}\right]
$
Let
$
A=\left[\begin{array}{ccc}
1 & -1 & 1 \\
2 & 1 & -3 \\
1 & 1 & 1
\end{array}\right]
$
Now
$
A=I A
$
Operating $R_2 \rightarrow R_2-2 R_2$ and $R_3 \rightarrow R_3-R_1$
$
\therefore\left[\begin{array}{ccc}
1 & -1 & 1 \\
2 & 1 & -3 \\
1 & 1 & 1
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \mathrm{A}
$
$
\left[\begin{array}{rrr}
1 & -1 & 1 \\
0 & 3 & -5 \\
0 & 2 & 0
\end{array}\right]=\left[\begin{array}{rrr}
1 & 0 & 0 \\
-2 & 1 & 0 \\
-1 & 0 & 1
\end{array}\right] \mathrm{A}
$
Operating $R_2 \rightarrow \frac{1}{3} R_2$
$
\left[\begin{array}{ccc}
1 & -1 & 1 \\
0 & 1 & -5 / 3 \\
0 & 2 & 0
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0 \\
-2 / 3 & 1 / 3 & 0 \\
-1 & 0 & 1
\end{array}\right] \mathrm{A}
$

Operating $R_3 \rightarrow R_3-2 R_2$
$
\left[\begin{array}{ccc}
1 & -1 & 1 \\
0 & 1 & -5 / 3 \\
0 & 0 & 10 / 3
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0 \\
-2 / 3 & 1 / 3 & 0 \\
1 / 3 & -2 / 3 & 1
\end{array}\right] \mathrm{A}
$
Operating $\mathrm{R}_3 \rightarrow \frac{3}{10} \mathrm{R}_3$
$
\left[\begin{array}{ccc}
1 & -1 & 1 \\
0 & 1 & -5 / 3 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & 0 \\
-2 / 3 & 1 / 3 & 0 \\
1 / 10 & -1 / 5 & 3 / 10
\end{array}\right] \mathrm{A}
$
Operating $R_1 \rightarrow R_1+R_2$
$
\left[\begin{array}{ccc}
1 & 0 & -2 / 3 \\
0 & 1 & -5 / 3 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
1 / 3 & 1 / 3 & 0 \\
-2 / 3 & 1 / 3 & 0 \\
1 / 10 & -1 / 5 & 3 / 10
\end{array}\right] \mathrm{A}
$
Operating $\mathrm{R}_1 \rightarrow \mathrm{R}_1+\frac{2}{3} \mathrm{R}_3$ and $\mathrm{R}_2 \rightarrow \mathrm{R}_2+\frac{5}{3} \mathrm{R}_3$
$
\begin{aligned}
{\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] } & =\left[\begin{array}{ccc}
2 / 5 & 1 / 5 & 1 / 5 \\
-1 / 2 & 0 & 1 / 2 \\
1 / 10 & -1 / 5 & 3 / 10
\end{array}\right] \mathrm{A} \\
\mathrm{A}^{-1} & =\left[\begin{array}{ccc}
2 / 5 & 1 / 5 & 1 / 5 \\
-1 / 2 & 0 & 1 / 2 \\
1 / 10 & -1 / 5 & 3 / 10
\end{array}\right] .
\end{aligned}
$

Question 3.
$
\left[\begin{array}{cc}
3 & -1 \\
-4 & 2
\end{array}\right]
$
Using elementary transformation find the inverse of the matrix

Solution:

Let
Now
$
A=\left[\begin{array}{cc}
3 & -1 \\
-4 & 2
\end{array}\right]
$
Applying $R_1 \leftrightarrow-R_2$
$\mathrm{A}=\mathrm{IA}$
$\Rightarrow\left[\begin{array}{cc}3 & -1 \\ -4 & 2\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] \mathbf{A}$
$\Rightarrow\left[\begin{array}{cc}4 & -2 \\ 3 & -1\end{array}\right]=\left[\begin{array}{cc}0 & -1 \\ 1 & 0\end{array}\right]$
Applying $\mathrm{R}_1 \rightarrow \mathrm{R}_1-\mathrm{R}_2$
$\Rightarrow\left[\begin{array}{ll}1 & -1 \\ 3 & -1\end{array}\right]=\left[\begin{array}{cc}-1 & -1 \\ 1 & 0\end{array}\right] \mathrm{A}$
Applying $R_2 \rightarrow R_2-3 R_1$
$\Rightarrow\left[\begin{array}{cc}1 & -1 \\ 0 & 2\end{array}\right]=\left[\begin{array}{cc}-1 & -1 \\ 4 & 3\end{array}\right] A$
Applying $\mathrm{R}_2 \rightarrow \frac{1}{2} \mathrm{R}_2 \quad \Rightarrow\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}-1 & -1 \\ 2 & 3 / 2\end{array}\right] \mathrm{A}$
Applying $\mathrm{R}_1 \rightarrow \mathrm{R}_1+\mathrm{R}_2 \quad \Rightarrow\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 1 / 2 \\ 2 & 3 / 2\end{array}\right] \mathrm{A}$
Thus
$
\mathrm{A}^{-1}=\left[\begin{array}{ll}
1 & 1 / 2 \\
2 & 3 / 2
\end{array}\right]
$
Question 4.
$
\left[\begin{array}{ccc}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right]
$
Solution:

Let
Now
$
A=\left[\begin{array}{ccc}
1 & 3 & -2 \\
-3 & 0 & -5 \\
2 & 5 & 0
\end{array}\right]
$
Applying $R_2 \rightarrow R_2+3 R_1$ and $R_3 \rightarrow R_3-2 R_1 \Rightarrow\left[\begin{array}{ccc}1 & 3 & -2 \\ 0 & 9 & -11 \\ 0 & -1 & 4\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ 3 & 1 & 0 \\ -2 & 0 & 1\end{array}\right] A$
Applying $\mathrm{R}_2 \rightarrow \mathrm{R}_2+8 \mathrm{R}_3$
Applying $\mathrm{R}_3 \rightarrow \mathrm{R}_3+\mathrm{R}_2$
Applying $\mathrm{R}_3 \rightarrow \frac{1}{25} \mathrm{R}_3$ $\Rightarrow\left[\begin{array}{ccc}1 & 3 & -2 \\ 0 & 1 & 21 \\ 0 & -1 & 4\end{array}\right]=\left[\begin{array}{rrr}1 & 0 & 0 \\ -13 & 1 & 8 \\ -2 & 0 & 1\end{array}\right]$ A $\Rightarrow\left[\begin{array}{ccc}1 & 3 & -2 \\ 0 & 1 & 21 \\ 0 & 0 & 25\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ -13 & 1 & 8 \\ -15 & 1 & 9\end{array}\right] \mathrm{A}$
Applying $R_1 \rightarrow R_1-3 R_2$ $\Rightarrow\left[\begin{array}{ccc}1 & 3 & -2 \\ 0 & 1 & 21 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}1 & 0 & 0 \\ -13 & 1 & 8 \\ -3 / 5 & 1 / 25 & 9 / 25\end{array}\right] \mathrm{A}$
Applying $\mathrm{R}_1 \rightarrow \mathrm{R}_1+65 \mathrm{R}_3$ $\Rightarrow\left[\begin{array}{rrr}1 & 0 & -65 \\ 0 & 1 & 21 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}40 & -3 & -24 \\ -13 & 1 & 8 \\ -3 / 5 & 1 / 25 & 9 / 25\end{array}\right] A$ $\Rightarrow\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 21 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}1 & -2 / 5 & -3 / 5 \\ -13 & 1 & 8 \\ -3 / 5 & 1 / 25 & 9 / 25\end{array}\right] \mathrm{A}$
Applying $R_2 \rightarrow R_2-21 R_3$ $\Rightarrow\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]=\left[\begin{array}{ccc}1 & -2 / 5 & -3 / 5 \\ -2 / 5 & 4 / 25 & 11 / 25 \\ -3 / 5 & 1 / 25 & 9 / 25\end{array}\right] \mathrm{A}$

Thus
$
A^{-1}=\left[\begin{array}{ccc}
1 & -2 / 5 & -3 / 5 \\
-2 / 5 & 4 / 25 & 11 / 25 \\
-3 / 5 & 1 / 25 & 9 / 25
\end{array}\right]
$
Question 5.
$
\left[\begin{array}{lll}
2 & 5 & 3 \\
3 & 4 & 1 \\
1 & 6 & 2
\end{array}\right] .
$
Using elementary transformation, find the inverse of the following matrix Solution:
Here
$
A=\left[\begin{array}{lll}
2 & 5 & 3 \\
3 & 4 & 1 \\
1 & 6 & 2
\end{array}\right]
$
Now
$
\mathrm{A}=\mathrm{IA}
$
$
\therefore \quad\left[\begin{array}{lll}
2 & 5 & 3 \\
3 & 4 & 1 \\
1 & 6 & 2
\end{array}\right]=\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \mathrm{A}
$
Operating $\mathrm{R}_1 \rightarrow \mathrm{R}_1-\mathrm{R}_3$
$
\left[\begin{array}{ccc}
1 & -1 & 1 \\
3 & 4 & 1 \\
1 & 6 & 2
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & -1 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \mathrm{A}
$
Operating $\mathrm{R}_2 \rightarrow \mathrm{R}_2-3 \mathrm{R}_1$ and $\mathrm{R}_3 \rightarrow \mathrm{R}_3-\mathrm{R}_1$
$
\left[\begin{array}{ccc}
1 & -1 & 1 \\
0 & 7 & -2 \\
0 & 7 & 1
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & -1 \\
-3 & 1 & 3 \\
-1 & 0 & 2
\end{array}\right] \mathrm{A}
$

Operating $\mathrm{R}_2 \rightarrow \frac{1}{7} \mathrm{R}_2$
$
\left[\begin{array}{ccc}
1 & -1 & 1 \\
0 & 1 & -2 / 7 \\
0 & 7 & 1
\end{array}\right]=\left[\begin{array}{ccc}
1 & 0 & -1 \\
-3 / 7 & 1 / 7 & 3 / 7 \\
-1 & 0 & 2
\end{array}\right] \mathrm{A}
$
Operating $R_1 \rightarrow R_1+R_2$ and $R_3 \rightarrow R_3-7 R_2$
$
\left[\begin{array}{ccc}
1 & 0 & 5 / 7 \\
0 & 1 & -2 / 7 \\
0 & 0 & 3
\end{array}\right]=\left[\begin{array}{ccc}
4 / 7 & 1 / 7 & -4 / 7 \\
-3 / 7 & 1 / 7 & 3 / 7 \\
2 & -1 & -1
\end{array}\right] \mathrm{A}
$
Operating $R_3 \rightarrow \frac{1}{3} \mathrm{R}_3$
$
\left[\begin{array}{ccc}
1 & 0 & 5 / 7 \\
0 & 1 & -2 / 7 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
4 / 7 & 1 / 7 & -4 / 7 \\
-3 / 7 & 1 / 7 & 3 / 7 \\
2 / 3 & -1 / 3 & -1 / 3
\end{array}\right] \mathrm{A}
$
Operating $\mathrm{R}_1 \rightarrow \mathrm{R}_1-\frac{5}{7} \mathrm{R}_3$ and $\mathrm{R}_2 \rightarrow \mathrm{R}_2+\frac{2}{7} \mathrm{R}_3$
$
\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]=\left[\begin{array}{ccc}
2 / 21 & 8 / 21 & -1 / 3 \\
-5 / 21 & 1 / 21 & 1 / 3 \\
2 / 3 & -1 / 3 & -1 / 3
\end{array}\right] \mathrm{A}
$
Thus
$
A^{-1}=\left[\begin{array}{ccc}
2 / 21 & 8 / 21 & -1 / 3 \\
-5 / 21 & 1 / 21 & 1 / 3 \\
2 / 3 & -1 / 3 & -1 / 3
\end{array}\right]=\frac{1}{21}\left[\begin{array}{ccc}
2 & 8 & -7 \\
-5 & 1 & 7 \\
14 & -7 & 7
\end{array}\right]
$
Question 6.
Given $\mathbf{A}=\left[\begin{array}{ccc}1 & -1 & 2 \\ 3 & 0 & -2 \\ 1 & 0 & 3\end{array}\right]$ verify that $A(\operatorname{adj} A)=(\operatorname{adjA}) \mathbf{A}=|\mathbf{A}| \mathbf{I}_3$.
Solution:

Let
$
\begin{aligned}
& A=\left[\begin{array}{ccc}
1 & -1 & 2 \\
3 & 0 & -2 \\
1 & 0 & 3
\end{array}\right] \\
& |A|=\left|\begin{array}{ccc}
1 & -1 & 2 \\
3 & 0 & -2 \\
1 & 0 & 3
\end{array}\right|=1(0-0)-(-1)(9+2)+2(0-0)=0+11+0=11 \\
& \therefore \quad A_{11}=(0-0)=0, \quad A_{12}=-(9+2)=-11, A_{13}=(0-0)=0 \\
& A_{21}=-(3-0)=3, A_{22}=(3-2)=1, \quad A_{23}=-(0+1)=-1 \\
& A_{31}=(2-0)=2, \quad A_{32}=-(-2-6)=8, \quad A_{33}=(0+3)=3 \\
& \operatorname{adj} A=\left[\begin{array}{ccc}
0 & -11 & 0 \\
3 & 1 & -1 \\
2 & 8 & 3
\end{array}\right]^T=\left[\begin{array}{ccc}
0 & 3 & 2 \\
-11 & 1 & 8 \\
0 & -1 & 3
\end{array}\right] \\
&
\end{aligned}
$

Question 7.
Let $\mathbf{A}=\left[\begin{array}{ll}3 & 7 \\ 2 & 5\end{array}\right]$ and $\mathbf{B}=\left[\begin{array}{ll}6 & 8 \\ 7 & 9\end{array}\right]$, verify that $(\mathbf{A B})^{-\mathbf{1}}=\mathbf{B}^{-\mathbf{1}} \mathbf{A}^{-\mathbf{1}}$.
Solution:

Here
$
\begin{aligned}
& A=\left[\begin{array}{ll}
3 & 7 \\
2 & 5
\end{array}\right] \quad \therefore|A|=\left|\begin{array}{ll}
3 & 7 \\
2 & 5
\end{array}\right|=15-4=1 \\
& A_{11}=5, A_{12}=-2, A_{21}=-7, A_{22}=3 \\
& \operatorname{adj} A=\left[\begin{array}{cc}
5 & -2 \\
-7 & 3
\end{array}\right]^T=\left[\begin{array}{cc}
5 & -7 \\
-2 & 3
\end{array}\right] \\
& \mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj} \mathrm{A}=\frac{1}{1}\left[\begin{array}{cc}
5 & -7 \\
-2 & 3
\end{array}\right]=\left[\begin{array}{cc}
5 & -7 \\
-2 & 3
\end{array}\right] \\
& B=\left[\begin{array}{ll}
6 & 8 \\
7 & 9
\end{array}\right] \\
& \therefore|B|=\left|\begin{array}{ll}
6 & 8 \\
7 & 9
\end{array}\right|=54-56=-2 \\
& \mathrm{~B}_{11}=9, \mathrm{~B}_{12}=-7, \mathrm{~B}_{21}=-8, \mathrm{~B}_{22}=6 \\
& \operatorname{adj} B=\left[\begin{array}{cc}
9 & -7 \\
-8 & 6
\end{array}\right]^{\mathrm{T}}\left[\begin{array}{cc}
9 & -8 \\
-7 & 6
\end{array}\right] \\
&
\end{aligned}
$

$\begin{aligned}
& \therefore \quad \mathrm{B}^{-1}=\frac{1}{|\mathrm{~B}|} \operatorname{adj} \mathrm{B}=\frac{1}{-2}\left[\begin{array}{cc}
9 & -8 \\
-7 & 6
\end{array}\right]=\frac{1}{2}\left[\begin{array}{cc}
-9 & 8 \\
7 & -6
\end{array}\right] \\
& \text { Now } \\
& \mathrm{AB}=\left[\begin{array}{ll}
3 & 7 \\
2 & 5
\end{array}\right]\left[\begin{array}{ll}
6 & 8 \\
7 & 9
\end{array}\right]=\left[\begin{array}{ll}
18+49 & 24+63 \\
12+35 & 16+45
\end{array}\right]=\left[\begin{array}{ll}
67 & 87 \\
47 & 61
\end{array}\right] \\
& \therefore \quad|\mathrm{AB}|=\left[\begin{array}{ll}
67 & 87 \\
47 & 61
\end{array}\right]=4087-4089=-2 \\
& \mathrm{AB}_{11}=61, \mathrm{AB}_{12}=-47, \mathrm{AB}_{21}=-87, \mathrm{AB}_{22}=67 \\
& \therefore \quad \operatorname{adj}(\mathrm{AB})=\left[\begin{array}{cc}
61 & -47 \\
-87 & 67
\end{array}\right]^{\mathrm{T}}=\left[\begin{array}{cc}
61 & -87 \\
-47 & 67
\end{array}\right] \\
& \therefore \quad(\mathrm{AB})^{-1}=\frac{1}{|\mathrm{AB}|} \operatorname{adj}(\mathrm{AB})=\frac{1}{-2}\left[\begin{array}{cc}
61 & -87 \\
-47 & 67
\end{array}\right]=\frac{1}{2}\left[\begin{array}{cc}
-61 & 87 \\
47 & -67
\end{array}\right] \\
& (\mathrm{AB})^{-1}=\mathrm{B}^{-1} \mathrm{~A}^{-1} \\
&
\end{aligned}$

Question 8.
If $\mathbf{A}=\left[\begin{array}{cc}3 & 1 \\ -1 & 2\end{array}\right]$ show that $\mathbf{A}^{\mathbf{2}}-\mathbf{5 A}+\mathbf{7} \mathbf{I}=\mathbf{0}$. Hence find $\mathbf{A}^{-\mathbf{1}}$.
Solution:
Here
$
\begin{aligned}
\therefore \quad \mathrm{A}^2 & =\mathrm{A} \cdot \mathrm{A}=\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]=\left[\begin{array}{cc}
9-1 & 3+2 \\
-3-2 & -1+4
\end{array}\right]=\left[\begin{array}{cc}
8 & 5 \\
-5 & 3
\end{array}\right] \\
\therefore \quad \mathrm{A}^2-5 \mathrm{~A}+7 \mathrm{I} & =\left[\begin{array}{cc}
8 & 5 \\
-5 & 3
\end{array}\right]-5\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]+7\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]=\left[\begin{array}{cc}
8 & 5 \\
-5 & 3
\end{array}\right]-\left[\begin{array}{cc}
15 & 5 \\
-5 & 10
\end{array}\right]+\left[\begin{array}{ll}
7 & 0 \\
0 & 7
\end{array}\right] \\
& =\left[\begin{array}{cc}
8-15+7 & 5-5+0 \\
-5+5+0 & 3-10+7
\end{array}\right]=\left[\begin{array}{ll}
0 & 0 \\
0 & 0
\end{array}\right]=0
\end{aligned}
$
Thus $\quad \mathrm{A}^2-5 \mathrm{~A}+7 \mathrm{I}=0$
Pre-multiplying by $\mathrm{A}^{-1}$ on both sides, we get
$
\begin{aligned}
& A^{-1}\left(A^2-5 A+7 I\right)=A^{-1} .0 \quad \Rightarrow A^{-1} A^2-5 A^{-1} A+7 A^{-1} I=0 \\
& \Rightarrow \quad \mathrm{A}-5 \mathrm{I}+7 \mathrm{~A}^{-1}=0 \\
& \Rightarrow \quad A^{-1}=\frac{1}{7}(5 \mathrm{I}-\mathrm{A})=\frac{1}{7}\left(5\left[\begin{array}{ll}
1 & 0 \\
0 & 1
\end{array}\right]-\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\right)=\frac{1}{7}\left(\left[\begin{array}{ll}
5 & 0 \\
0 & 5
\end{array}\right]-\left[\begin{array}{cc}
3 & 1 \\
-1 & 2
\end{array}\right]\right) \\
& =\frac{1}{7}\left[\begin{array}{cc}
2 & -1 \\
1 & 3
\end{array}\right] \\
& \mathrm{A}^{-1}=\frac{1}{7}\left[\begin{array}{cc}
2 & -1 \\
1 & 3
\end{array}\right] . \\
&
\end{aligned}
$

Question 9.
If $A=\left[\begin{array}{rrr}2 & -1 & 1 \\ -1 & 2 & -1 \\ 1 & -1 & 2\end{array}\right]$, verify that $A^3-6 A^2+9 A-4 I=0$ and hence find $A^{-1}$.
Solution:

Here
$
\text { Here } \begin{aligned}
A & =\left[\begin{array}{rrr}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right] \\
\therefore \quad A^2 & =A \cdot A=\left[\begin{array}{rrr}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right]\left[\begin{array}{rrr}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right] \\
& =\left[\begin{array}{rrr}
4+1+1 & -2-2-1 & 2+1+2 \\
-2-2-1 & 1+4+1 & -1-2-2 \\
2+1+2 & -1-2-2 & 1+1+4
\end{array}\right]=\left[\begin{array}{rrr}
6 & -5 & 5 \\
-5 & 6 & -5 \\
5 & -5 & 6
\end{array}\right] \\
A^3=A^2 \cdot A & =\left[\begin{array}{rrr}
6 & -5 & 5 \\
-5 & 6 & -5 \\
5 & -5 & 6
\end{array}\right]\left[\begin{array}{rrr}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right] \\
& =\left[\begin{array}{ccc}
12+5+5 & -6-10-5 & 6+5+10 \\
-10-6-5 & 5+12+5 & -5-6-10 \\
10+5+6 & -5-10-6 & 5+5+12
\end{array}\right]=\left[\begin{array}{ccc}
22 & -21 & 21 \\
-21 & 22 & -21 \\
21 & -21 & 22
\end{array}\right]
\end{aligned}
$
Now $A^3-6 A^2+9 A-4 I$
$
\begin{aligned}
& =\left[\begin{array}{ccc}
22 & -21 & 21 \\
-21 & 22 & -21 \\
21 & -21 & 22
\end{array}\right]-6\left[\begin{array}{ccc}
6 & -5 & 5 \\
-5 & 6 & -5 \\
5 & -5 & 6
\end{array}\right]+9\left[\begin{array}{rrr}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right]-4\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right] \\
& =\left[\begin{array}{ccc}
22 & -21 & 21 \\
-21 & 22 & -21 \\
21 & -21 & 22
\end{array}\right]-\left[\begin{array}{ccc}
36 & -30 & 30 \\
-30 & 36 & -30 \\
30 & -30 & 36
\end{array}\right]+\left[\begin{array}{ccc}
18 & -9 & 9 \\
-9 & 18 & -9 \\
9 & -9 & 18
\end{array}\right]-\left[\begin{array}{ccc}
4 & 0 & 0 \\
0 & 4 & 0 \\
0 & 0 & 4
\end{array}\right] \\
& =\left[\begin{array}{lll}
22-36+18-4 & -21+30-9-0 & 21-30+9-0 \\
-21+30-9-0 & 22-36+18-4 & -21+30-9-0 \\
21-30+9-0 & -21+30-9-0 & 22-36+18-4
\end{array}\right]=\left[\begin{array}{lll}
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0
\end{array}\right]=0 \\
&
\end{aligned}
$

Thus $\mathrm{A}^3-6 \mathrm{~A}^2+9 \mathrm{~A}-4 \mathrm{I}=0$.
Pre-multiplying by $\mathrm{A}^{-1}$ on both sides, we get
$
\begin{aligned}
& \mathrm{A}^{-1}\left(\mathrm{~A}^3-6 \mathrm{~A}^2+9 \mathrm{~A}-4 \mathrm{I}\right)=\mathrm{A}^{-1} \cdot 0 \quad \Rightarrow \mathrm{A}^{-1} \mathrm{~A}^3-6 \mathrm{~A}^{-1} \mathrm{~A}^2+9 \mathrm{~A}^{-1} \mathrm{~A}-4 \mathrm{~A}^{-1} \mathrm{I}=0 \\
& \Rightarrow A^2-6 A+9 I-4 A^{-1}=0 \\
& \Rightarrow \quad A^{-1}=\frac{1}{4}\left(A^2-6 A+9 I\right) \\
& =\frac{1}{4}\left(\left[\begin{array}{rrr}
6 & -5 & 5 \\
-5 & 6 & -5 \\
5 & -5 & 6
\end{array}\right]-6\left[\begin{array}{rrr}
2 & -1 & 1 \\
-1 & 2 & -1 \\
1 & -1 & 2
\end{array}\right]+9\left[\begin{array}{lll}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{array}\right]\right) \\
& =\frac{1}{4}\left(\left[\begin{array}{rrr}
6 & -5 & 5 \\
-5 & 6 & -5 \\
5 & -5 & 6
\end{array}\right]-\left[\begin{array}{ccc}
12 & -6 & 6 \\
-6 & 12 & -6 \\
6 & -6 & 12
\end{array}\right]+\left[\begin{array}{lll}
9 & 0 & 0 \\
0 & 9 & 0 \\
0 & 0 & 9
\end{array}\right]\right) \\
& =\frac{1}{4}\left[\begin{array}{ccc}
6-12+9 & -5+6+0 & 5-6+0 \\
-5+6+0 & 6-12+9 & -5+6+0 \\
5-6+0 & -5+6+0 & 6-12+9
\end{array}\right]=\frac{1}{4}\left[\begin{array}{rrr}
3 & 1 & -1 \\
1 & 3 & 1 \\
-1 & 1 & 3
\end{array}\right] \\
& A^{-1}=\frac{1}{4}\left[\begin{array}{rrr}
3 & 1 & -1 \\
1 & 3 & 1 \\
-1 & 1 & 3
\end{array}\right] \\
&
\end{aligned}
$
Question 10.
Find the inverse of the matrices $\left[\begin{array}{rrr}1 & -1 & 2 \\ 0 & 2 & -3 \\ 3 & -2 & 4\end{array}\right]$.

Solution:
Let
$
\begin{aligned}
& A=\left[\begin{array}{rrr}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right] \\
& A=\left|\begin{array}{rrr}
1 & -1 & 2 \\
0 & 2 & -3 \\
3 & -2 & 4
\end{array}\right|=1(8-6)-(-1)(0+9)+2(0-6)=2+9-12=-1 \\
& A_{11}=(8-6)=2, \quad A_{12}=-(0+9)=-9, \quad A_{13}=(0-6)=-6 \\
& A_{21}=(-4+4)=0, A_{22}=(4-6)=-2, \quad A_{23}=(-2+3)=-1 \\
& A_{31}=(3-4)=-1, A_{32}=-(-3-0)=3, \quad A_{33}=(2-0)=2 \\
& \operatorname{adj} A=\left[\begin{array}{ccc}
2 & -9 & -6 \\
0 & -2 & -1 \\
-1 & 3 & 2
\end{array}\right]=\left[\begin{array}{ccc}
2 & 0 & -1 \\
-9 & -2 & 3 \\
-6 & -1 & 2
\end{array}\right] \\
& \mathrm{A}^{-1}=\frac{1}{|\mathrm{~A}|} \operatorname{adj} A=\frac{1}{-1}\left[\begin{array}{ccc}
2 & 0 & -1 \\
-9 & -2 & 3 \\
-6 & -1 & 2
\end{array}\right]=\left[\begin{array}{rrr}
-2 & 0 & 1 \\
9 & 2 & -3 \\
6 & 1 & -2
\end{array}\right] \\
&
\end{aligned}
$