Question

find the inverse of matrix by using elementary operation [1 -2. 2 1]​

Answer 1

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

Given matrix is

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

We know,

By Elementary Row Transformation Method,

$\rm :\longmapsto\:A = I \: A$

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

$\rm :\longmapsto\:\boxed{ \tt{ \: OP \: R_2 \: \to \: R_2 - 2R_1 \: }}$

$\rm :\longmapsto\:\bigg[ \begin{matrix}1& - 2 \\ 0&5 \end{matrix} \bigg] = \bigg[ \begin{matrix}1& 0 \\ - 2&1 \end{matrix} \bigg] \: A$

$\rm :\longmapsto\:\boxed{ \tt{ \: OP \: R_2 \: \to \: \frac{1}{5} \: R_2 \: }}$

$\rm :\longmapsto\:\bigg[ \begin{matrix}1& - 2 \\ 0&1 \end{matrix} \bigg] = \bigg[ \begin{matrix}1& 0 \\ - \dfrac{2}{5} & \dfrac{1}{5} \end{matrix} \bigg] \: A$

$\rm :\longmapsto\:\boxed{ \tt{ \: OP \: R_1 \: \to \: R_1 + 2R_2 \: }}$

$\rm :\longmapsto\:\bigg[ \begin{matrix}1& 0 \\ 0&1 \end{matrix} \bigg] = \bigg[ \begin{matrix} \dfrac{1}{5} & \dfrac{2}{5} \\ - \dfrac{2}{5} & \dfrac{1}{5} \end{matrix} \bigg] \: A$

We know,

$\rm :\longmapsto\:\boxed{ \tt{ \: {AA}^{ - 1} \: = \: I \: }}$

On comparing, we get

$\rm :\longmapsto\:\boxed{ \tt{ \: {A}^{ - 1} \: = \: \bigg[ \begin{matrix} \dfrac{1}{5} & \dfrac{2}{5} \\ - \dfrac{2}{5} & \dfrac{1}{5} \end{matrix} \bigg] \: }}$

Verification :-

$\rm :\longmapsto\: {AA}^{ - 1}$

$\rm \:  =  \:\bigg[ \begin{matrix}1& - 2 \\ 2&1 \end{matrix} \bigg] \times \bigg[ \begin{matrix} \dfrac{1}{5} & \dfrac{2}{5} \\ - \dfrac{2}{5} & \dfrac{1}{5} \end{matrix} \bigg]$

$\rm \:  =  \:\bigg[ \begin{matrix}1& - 2 \\ 2&1 \end{matrix} \bigg] \times \dfrac{1}{5} \times \bigg[ \begin{matrix}1&2 \\ - 2&1 \end{matrix} \bigg]$

$\rm \:  =  \:\dfrac{1}{5}\bigg[ \begin{matrix}5& 0 \\ 0&5 \end{matrix} \bigg]$

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

Hence, Verified