Solve the Following Question.(2 Marks)

Question 12 Marks

Find the inverse of the matrix $\left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right]$

Answer

Let $A=\left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right]$
$\therefore| A |=\left|\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right|=4-3=1 \neq 0 $
$ \therefore A ^{-1} \text { exists. } $
$ \therefore AA ^{-1}= I $
$ {\left[\begin{array}{cc}2 & -3 \\ -1 & 2\end{array}\right] A^{-1}=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]} $
$ R _1 \rightarrow R _1+ R _2 $
$ {\left[\begin{array}{cc}1 & -1 \\ -1 & 2\end{array}\right] A^{-1}=\left[\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right]} $
$ R _2 \rightarrow R _2+ R _1 $
$ {\left[\begin{array}{cc}1 & -1 \\ 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{ll}1 & 1 \\ 1 & 2\end{array}\right]} $
$ R_1 \rightarrow R_1+R_2 $
${\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right] A^{-1}=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right]} $
$ \therefore A ^{-1}=\left[\begin{array}{ll}2 & 3 \\ 1 & 2\end{array}\right] $

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MCQ 22 Marks

The inverse of the matrix $\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{array}\right]$ is

A
$\left[\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{array}\right]$
B
$\left[\begin{array}{ccc}-\frac{1}{2} & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & 1\end{array}\right]$
C
$\left[\begin{array}{ccc}-1 & 0 & 0 \\ 0 & -\frac{1}{2} & 0 \\ 0 & 0 & \frac{1}{2}\end{array}\right]$
D
$-\frac{1}{2}\left[\begin{array}{ccc}-\frac{1}{2} & 0 & 0 \\ 0 & -1 & 0 \\ 0 & 0 & -1\end{array}\right]$

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Question 32 Marks

If $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$ and $A X=I$, then find $X$ by using elementary transformations.

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Question 42 Marks

Find the cofactors of the elements for the matrix $\left[\begin{array}{ll}-1 & 2 \\ -3 & 4\end{array}\right]$

Answer

Let $A=\left[\begin{array}{ll}-1 & 2 \\ -3 & 4\end{array}\right]=\left[\begin{array}{ll}a_{11} & a_{12} \\ a_{21} & a_{22}\end{array}\right]$
Here,
$A_{11}=(-1)^2(4)=4=\text { cofactor of } a_{11}$
$A_{12}=(-1)^3(-3)=3=\text { cofactor of } a_{12}$
$A_{21}=(-1)^3(2)=-2=\text { cofactor of } a_{21}$
$A_{22}=(-1)^4(-1)=-1=\text { cofactor of } a_{22}$
Hence cofactors are
$A_{11}=4, A_{12}=3$
$A_{21}=-2, A_{22}=-1$

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Question 52 Marks

Find the matrix of co-factors for the matrix $\left[\begin{array}{cc}1 & 3 \\ 4 & -1\end{array}\right]$

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MCQ 62 Marks

If $A=\left[\begin{array}{cc}2 & -3 \\ 3 & 5\end{array}\right]$, then adjoint of matrix $A$ is .....

A
$\left[\begin{array}{cc}1 & 3 \\ -4 & 1\end{array}\right]$
B
$\left[\begin{array}{cc}1 & -3 \\ -4 & 2\end{array}\right]$
C
$\left[\begin{array}{cc}1 & 3 \\ 4 & -2\end{array}\right]$
D
$\left[\begin{array}{cc}-1 & -3 \\ -4 & 2\end{array}\right]$

Answer

$A=\left[\begin{array}{cc}2 & -3 \\ 4 & 1\end{array}\right]$
then adjoint of matrix A
$\operatorname{adj} A=\left[\begin{array}{cc}1 & 3 \\ -4 & 2\end{array}\right]$

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MCQ 72 Marks

The inverse of the matrix $\left[\begin{array}{ll}-1 & 5 \\ -3 & 2\end{array}\right]$ is

A
$\frac{1}{13}\left[\begin{array}{ll}2 & -5 \\ 3 & -1\end{array}\right]$
B
$\frac{1}{13}\left[\begin{array}{ll}-1 & 5 \\ -3 & 2\end{array}\right]$
C
$\frac{1}{13}\left[\begin{array}{cc}-1 & -3 \\ 5 & 2\end{array}\right]$
D
$\frac{1}{13}\left[\begin{array}{cc}1 & 5 \\ 3 & -2\end{array}\right]$

Answer

(A) $1 / 13\left[\begin{array}{ll}2 & -5 \\ 3 & -1\end{array}\right]$
$|A|=-2+15=13$
$A^{-1}=\frac{\operatorname{adj} A}{|A|}=\frac{\left[\begin{array}{lr}2 & -5 \\ 3 & -1\end{array}\right]}{13}$
$A^{-1}=\frac{1}{13}\left[\begin{array}{ll}2 & -5 \\ 3 & -1\end{array}\right]$

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Question 82 Marks

Find $(A B)^{-1}$ if $A=\left[\begin{array}{ccc}1 & 2 & 3 \\ 1 & -2 & -3\end{array}\right], B=\left[\begin{array}{cc}1 & -1 \\ 1 & 2 \\ 1 & -2\end{array}\right]$

Answer

$A=\left[\begin{array}{ccc}1 & 2 & 3 \\ 1 & -2 & -3\end{array}\right], B=\left[\begin{array}{cc}1 & -1 \\ 1 & 2 \\ 1 & -2\end{array}\right]$
$A B=\left[\begin{array}{ccc}1 & 2 & 3 \\ 1 & -2 & -3\end{array}\right]\left[\begin{array}{cc}1 & -1 \\ 1 & 2 \\ 1 & -2\end{array}\right]$
$=\left[\begin{array}{cc}6 & -3 \\ -4 & 1\end{array}\right]$
$(A B)^{-1}(A B)=I$
$(A B)^{-1}\left[\begin{array}{cc}6 & -3 \\ -4 & 1\end{array}\right]=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$
Using $R_1 \rightarrow \frac{1}{6} R_1$
$(A B)^{-1}\left[\begin{array}{cc}1 & -\frac{1}{2} \\ -4 & 1\end{array}\right]=\left[\begin{array}{ll}\frac{1}{6} & 0 \\ 0 & 1\end{array}\right]$
Using $R_2 \rightarrow R_2+4 R_1$
$(A B)^{-1}\left[\begin{array}{ll}1 & -\frac{1}{2} \\ 0 & -1\end{array}\right]=\left[\begin{array}{ll}\frac{1}{6} & 0 \\ \frac{2}{3} & 1\end{array}\right]$
using $R_2 \rightarrow(-1) R_2$
$(A B)^{-1}\left[\begin{array}{cc}1 & -\frac{1}{2} \\ 0 & 1\end{array}\right]=\left[\begin{array}{cc}\frac{1}{6} & 0 \\ -\frac{2}{3} & -1\end{array}\right]$
Using $R_1 \rightarrow R_1+\left(\frac{1}{2}\right) R_2$
$(A B)^{-1}\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]=\left[\begin{array}{ll}-\frac{1}{6} & -\frac{1}{2} \\ -\frac{2}{3} & -1\end{array}\right]$
$(A B)^{-1} I=\left[\begin{array}{ll}-\frac{1}{6} & -\frac{1}{2} \\ -\frac{2}{3} & -1\end{array}\right]$
$(A B)^{-1}=\left[\begin{array}{ll}-\frac{1}{6} & -\frac{1}{2} \\ -\frac{2}{3} & -1\end{array}\right]$

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MCQ 92 Marks

If $A=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$, then $A^6=......$

A
$6A$
B
$12A$
C
$16A$
✓
$32A$

Answer

Correct option: D.

$32A$

Given that
$A=\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 2\end{array}\right]$
we can write
$A=2\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
Here $2$ is scalar multiple
Therefore ,$A=2 \times I,$ where $I= \left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
thus, $A^6=\left\{2\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\right\}^6$
$=2^6 \times\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]^6$
$=2^6 \times I^6$,
$=2^6 \times I, $
since $I^6=I$
$=2^{ 5 }\times A [\because A=2I]$
$=32A$

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MCQ 102 Marks

If $A =\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$ and $A (\operatorname{adj} A )= KI$, then the value of ' $K$ ' is .....

A
2
B
-2
C
10
D
-10

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Question 112 Marks

If $A=\left[\begin{array}{cc}2 & -2 \\ 4 & 3\end{array}\right]$ then find $A^{-1}$ by adjoint method.

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MCQ 122 Marks

The inverse of the matrix $\left[\begin{array}{cc}1 & -1 \\ 2 & 3\end{array}\right]$ is ......

A
$\frac{1}{5}\left[\begin{array}{cc}3 & -1 \\ -2 & 1\end{array}\right]$
B
$\frac{1}{5}\left[\begin{array}{cc}3 & 1 \\ -2 & 1\end{array}\right]$
C
$\frac{1}{5}\left[\begin{array}{ll}-3 & 1 \\ -2 & 1\end{array}\right]$
D
$\frac{1}{5}\left[\begin{array}{ll}3 & -1 \\ 2 & -1\end{array}\right]$.

Answer

(B) $\frac{1}{5}\left[\begin{array}{cc}3 & 1 \\ -2 & 1\end{array}\right]$
if $A=\left[\begin{array}{ll}a & b \\ c & d\end{array}\right]$, then $A^{-1}=\frac{1}{a d-b c}\left[\begin{array}{cc}d & -b \\ -c & a\end{array}\right]$
$\therefore A=\frac{1}{5}\left[\begin{array}{cc}3 & 1 \\ -2 & 1\end{array}\right]$

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Question 132 Marks

If $A=\left[\begin{array}{cc}2 & -3 \\ 3 & 5\end{array}\right]$ then find $A^{-1}$ by adjoint method.

Answer

Given $A=\left[\begin{array}{cc}2 & -3 \\ 3 & 5\end{array}\right]$
$\therefore|A|=\left|\begin{array}{cc}2 & -3 \\ 3 & 5\end{array}\right|=10+9=19 \neq 0$
$\therefore A ^{-1}$ exist
$A_{11}=(-1)^{1+1} \cdot M_{11}=5$
$A_{12}=(-1)^{1+2} \cdot M_{12}=-3$
$A_{21}=(-1)^{2+1} \cdot M_{21}=-(-3)=3$
$A_{22}=(-1)^{2+2} \cdot M_{22}=2$
Hence, matrix of the co-factors is

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Maths Solve the Following Question.(2 Marks)

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