Question

Solve it.


Irrevelent answers will be deleted. ​

Answer 1

$\large\underline{\sf{Solution-}}$

Given matrix is

$\rm :\longmapsto\:A = \bigg[ \begin{matrix}2&1 \\ 1&1 \end{matrix} \bigg]$

Using Elementary Row Transformation Method, we have

$\rm :\longmapsto\:A = IA$

$\rm :\longmapsto\:\bigg[ \begin{matrix}2&1 \\ 1&1 \end{matrix} \bigg] = \bigg[ \begin{matrix}1&0 \\ 0&1 \end{matrix} \bigg]A$

$\red{\rm :\longmapsto\:OP \: R_1 \: \leftrightarrow \: R_2}$

$\rm :\longmapsto\:\bigg[ \begin{matrix}1&1 \\ 2&1 \end{matrix} \bigg] = \bigg[ \begin{matrix}0&1 \\ 1&0 \end{matrix} \bigg]A$

$\red{\rm :\longmapsto\:OP \: R_2 \: \rightarrow \: R_2 - 2R_1}$

$\rm :\longmapsto\:\bigg[ \begin{matrix}1&1 \\ 0& - 1 \end{matrix} \bigg] = \bigg[ \begin{matrix}0&1 \\ 1& - 2 \end{matrix} \bigg]A$

$\red{\rm :\longmapsto\:OP \: R_2 \: \rightarrow \: - \: R_2}$

$\rm :\longmapsto\:\bigg[ \begin{matrix}1&1 \\ 0&1 \end{matrix} \bigg] = \bigg[ \begin{matrix}0&1 \\ - 1& 2 \end{matrix} \bigg]A$

$\red{\rm :\longmapsto\:OP \: R_1 \: \rightarrow \:R_1 \: - \: R_2}$

$\rm :\longmapsto\:\bigg[ \begin{matrix}1&0 \\ 0&1 \end{matrix} \bigg] = \bigg[ \begin{matrix}1& - 1 \\ - 1& 2 \end{matrix} \bigg]A$

Since, we know

$\red{\boxed{\tt{ {A \: A}^{ - 1} \: = I \: }}}$

$\bf\implies \: {A}^{ - 1} = \bigg[ \begin{matrix}1& - 1 \\ - 1&2 \end{matrix} \bigg]$

▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬▬

Learn More :-

If A and R are square matrix of order n, then

$\boxed{\tt{ {(AR)}^{ - 1} = {R}^{ - 1} {A}^{ - 1} \: }}$

$\boxed{\tt{ | {A}^{ - 1} | = \frac{1}{ |A| } \: }}$

$\boxed{\tt{ |A'| = |A| \: }}$

$\boxed{\tt{ |kA| = {k}^{n} |A| \: }}$

$\boxed{\tt{ |adj \: A| = { |A| }^{n - 1} \: }}$