Solve the Following Question.(2 Marks)

Question 12 Marks

if $A=\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right]$ and $X$ is a $2 \times 2$ matrix such that $A X=1$, then find $X$.

Answer

We will reduce the matrix $A$ to the identity matrix by using row transformations. During this proccess, I will be converted to the matrix $X.$
We have $AX = I.$
$\therefore\left[\begin{array}{ll}1 & 2 \\ 3 & 4\end{array}\right] X=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]$By $R_2-3 R_1$, we get,
$\left[\begin{array}{rr}1 & 2 \\0 & -2\end{array}\right] X=\left[\begin{array}{rr}1 & 0 \\-3 & 1\end{array}\right]$
By $\left(-\frac{1}{2}\right) R_2$, we get,
$\left[\begin{array}{ll}1 & 2 \\0 & 1\end{array}\right] X=\left[\begin{array}{rr}1 & 0 \\\frac{3}{2} & -\frac{1}{2}\end{array}\right]$
By $R_1-2 R_2$, we get, Maharash
$\begin{array}{l}{\left[\begin{array}{ll}1 & 0 \\0 & 1\end{array}\right]X=\left[\begin{array}{rr}-2 & 1 \\\frac{3}{2} & -\frac{1}{2}\end{array}\right]} \\ \therefore X=\left[\begin{array}{rr}-2 & 1 \\\frac{3}{2} & -\frac{1}{2}\end{array}\right]\end{array}$

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Question 22 Marks

If $A=\left[\begin{array}{lll}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right]$ is a nonsingular matrix then find $A^{-1}$ by elementary row transformations. Hence, find the inverse of $\left[\begin{array}{ccc}2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{array}\right]$

Answer

Since $A$ is a non-singular matrix, then find $A^{-1}$ by using elementary row transformations.
We write $AA^{-1} = I$
$\therefore\left[\begin{array}{lll}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right] A^{-1}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
By $\left(\frac{1}{x}\right) \mathrm{R}_1,\left(\frac{1}{y}\right) \mathrm{R}_2$ and $\left(\frac{1}{z}\right) \mathrm{R}_3$, we get,
${\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right] \mathrm{A}^{-1}=\left[\begin{array}{lll}\frac{1}{x} & 0 & 0 \\ 0 & \frac{1}{y} & 0 \\ 0 & 0 & \frac{1}{z}\end{array}\right]} \\ \therefore A^{-1}=\left[\begin{array}{ccc}\frac{1}{x} & 0 & 0 \\ 0 & \frac{1}{y} & 0 \\ 0 & 0 & \frac{1}{z}\end{array}\right] .$
Comparing $\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{array}\right]$ with $\left[\begin{array}{lll}x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z\end{array}\right]$,
we get, $x = 2, y = 1, z = -1$
$\therefore \frac{1}{x}=\frac{1}{2}, \frac{1}{y}=\frac{1}{1}=1, \frac{1}{z}=\frac{1}{-1}=-1$
$\left[\begin{array}{lll}2 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{array}\right]$ is $\left(\begin{array}{ccc}\frac{1}{2} & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1\end{array}\right)$.

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Question 32 Marks

Find $A B$, if $A=\left[\begin{array}{ccc}1 & 2 & 3 \\ 1 & -2 & -3\end{array}\right]$ and $B=\left[\begin{array}{cc}1 & -1 \\ 1 & 2 \\ 1 & -2\end{array}\right]$ Examine whether $A B$ has inverse or not.

Answer

$A B=\left[\begin{array}{rrr}1 & 2 & 3 \\ 1 & -2 & -3\end{array}\right] \times\left[\begin{array}{rr}1 & -1 \\ 1 & 2 \\ 1 & -2\end{array}\right]$
$=\left[\begin{array}{rr}1(1)+2(1)+3(1) & 1(-1)+2(2)+3(-2) \\ 1(1)+(-2)(1)+(-3)(1) & 1(-1)+(-2)(2)+(-3)(-2)\end{array}\right]$
$=\left[\begin{array}{rr}1+2+3 & -1+4-6 \\ 1-2-3 & -1-4+6\end{array}\right]$
$=\left[\begin{array}{rr}6 & -3 \\ -4 & 1\end{array}\right] \\ \therefore|A B|=\left|\begin{array}{rr}6 & -3 \\ -4 & 1\end{array}\right|=6-12=-6 \neq 0$
$\therefore A$ is a non$-$singular matrix.
Hence, $(AB)^{-1}$ exist.

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Question 42 Marks

Solve the following equations by the method of reduction. $x + 3y + 3z = 12, x + 4y + 4z = 15$ and $x + 3y + 4z = 13.$

Answer

The above equations can be written in the form $AX = B$
$ \text { i.c. } \quad\left[\begin{array}{lll} 1 & 3 & 3 \\ 1 & 4 & 4 \\ 1 & 3 & 4 \end{array}\right]\left[\begin{array}{l} x \\ y \\ z \end{array}\right]=\left[\begin{array}{c} 12 \\ 15 \\ 13 \end{array}\right]$
using $R _2 \rightarrow R _2- R _1$ and $R _3 \rightarrow R _3- R _1$
we get $\left[\begin{array}{lll}1 & 3 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{c}12 \\ 3 \\ 1\end{array}\right]$
Again using $R _1 \rightarrow R _1-3 R _2$ and $ R _2 \rightarrow R _2- R _3$
We get $\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}3 \\ 2 \\ 1\end{array}\right]$
Hence the required solution is $x=3, y=2, z=1. ($verify$)$

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Question 52 Marks

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

Answer

For given matrix $A,$ we get,$
M _{11}=3, A _{11}=(-1)^{1+1}(3)=3$
$M _{12}=4, A _{12}=(-1)^{1+2}(4)=-4$
$M _{21}=-2, A _{21}=(-1)^{2+1}(-2)=2$
$M _{22}=2, A _{22}=(-1)^{2+2}(2)=2 $
$ \therefore \text { adj } A =\left[\begin{array}{cc}3 & 2 \\ -4 & 2 \end{array}\right] $ and $| A |=\left[\begin{array}{cc} 2 & -2 \\ 4 & 3 \end{array}\right]=6+8=14 \neq 0 $
$\therefore $ using $ A^{-1}=\frac{1}{| A |}(\operatorname{adj} A ) $
$ A^{-1}=\frac{1}{14}\left[\begin{array}{cc} 3 & 2 \\ -4 & 2 \end{array}\right]$

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Question 62 Marks

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

Answer

$\therefore M_{11}=1$
$\therefore A_{11}=(-1)^{1+1}(1)=1$
$\therefore M_{12}=4$
$\therefore A_{12}=(-1)^{1+2}(4)=-4$
$\therefore M_{21}=-3$
$\therefore A_{21}=(-1)^{2+1}(-3)=3$
$\therefore M_{22}=2$
$\therefore A_{22}=(-1)^{2+2}(2)=2$
Here $a_{11}=2$
$a _{12}=-3$
$a_{21}=4$
$a_{22}=1$
$\therefore \quad \text { the matrix }\left[A_{i i}\right]_{ D 2}=\left[\begin{array}{cc} 1 & 4 \\ 3 & 2\end{array}\right]$
$\therefore \quad\left[A_{i j}\right]_{2 \infty 2}^{ T }=\left[\begin{array}{cc} 1 & 3 \\ -4 & 2\end{array}\right]$
$\therefore \quad \text { adj } A \quad=\left[\begin{array}{cc} 1 & 3 \\ -4 & 2\end{array}\right]$

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Question 72 Marks

Find the co-factors of the elements of the following matrices : $\left[\begin{array}{ll}-1 & 2 \\ -3 & 4\end{array}\right]$

Answer

Let $A=\left[\begin{array}{ll}-1 & 2 \\ -3 & 4\end{array}\right]$
$\text { Here, } a_{11}=-11, M_{11}=4$
$\therefore A_{11}=(-1)^{1+1}(4)=4$
$a_{12}=2, M_{12}=-3$
$\therefore A_{12}=(-1)^{1+2}(-3)=3$
$a_{21}=-3, M_{21}=-2$
$\therefore A_{21}=(-1)^{2+1}(2)=-2$
$a_{22}=4, M_{22}=-1$
$\therefore A_{22}=(-1)^{2+2}(-1)=-1 .$

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Question 82 Marks

Find the adjoint of the following matrices $:\left[\begin{array}{cc}2 & -3 \\ 3 & 5\end{array}\right]$

Answer

Let $A=2 -3$
Here, $a_{11}=2, M_{11}=5$
$\therefore A_{11}=(-1)^{1+1}(5)=5$
$a_{12}=-3, M_{12}=3$
$\therefore A_{12}=(-1)^{1+2}(3)=-3$
$a_{21}=3, M_{21}=-3$
$\therefore A A_{21}=(-1)^{2+1}(-3)=3$
$a_{22}=5, M_{22}=2$
$\therefore A_{22}=(-1)^{2+1}=2$
$\therefore$ the co$-$factor matrix $=\left[\begin{array}{ll}A_{11} & A_{12} \\ A_{21} & A_{22}\end{array}\right]$
$ =\left[\begin{array}{rr}5 & -3 \\ 3 & 2\end{array}\right] $
$ \therefore \operatorname{adj~} \mathrm{A}=\left(\begin{array}{rr}5 & 3 \\ -3 & 2\end{array}\right)$

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Maths Solve the Following Question.(2 Marks)

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