Question

Find the inverse of the matrix using row column operation of the matrix 4 -2
3 1 ​

Answer 1

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

Given .matrix is

$\rm :\longmapsto\:\bigg[ \begin{matrix}4& - 2 \\ 3&1 \end{matrix} \bigg]$

Let assume that

$\rm :\longmapsto\:A = \bigg[ \begin{matrix}4& - 2 \\ 3&1 \end{matrix} \bigg]$

Using Elementary Row Transformation Method, we have

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

$\rm :\longmapsto\:\bigg[ \begin{matrix}4& - 2 \\ 3&1 \end{matrix} \bigg] = \bigg[ \begin{matrix}1&0 \\ 0&1 \end{matrix} \bigg]A$

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

$\rm :\longmapsto\:\bigg[ \begin{matrix}1& - 3 \\ 3&1 \end{matrix} \bigg] = \bigg[ \begin{matrix}1& - 1 \\ 0&1 \end{matrix} \bigg]A$

$\red{\rm :\longmapsto\:OP \: R_2 \: \to \: R_2 -3 R_1}$

$\rm :\longmapsto\:\bigg[ \begin{matrix}1& - 3 \\ 0&10 \end{matrix} \bigg] = \bigg[ \begin{matrix}1& - 1 \\ - 3&4 \end{matrix} \bigg]A$

$\red{\rm :\longmapsto\:OP \: R_2 \: \to \: \dfrac{1}{10} \: R_2 }$

$\rm :\longmapsto\:\bigg[ \begin{matrix}1& - 3 \\ 0&1 \end{matrix} \bigg] = \bigg[ \begin{matrix}1& - 1 \\ - \dfrac{3}{10} & \dfrac{2}{5} \end{matrix} \bigg]A$

$\red{\rm :\longmapsto\:OP \: R_1 \: \to \: R_1 + 3R_2}$

$\rm :\longmapsto\:\bigg[ \begin{matrix}1&0 \\ 0&1 \end{matrix} \bigg] = \bigg[ \begin{matrix} \dfrac{1}{10} & \dfrac{1}{5} \\ - \dfrac{3}{10} & \dfrac{2}{5} \end{matrix} \bigg]A$

$\rm :\longmapsto\:\bigg[ \begin{matrix}1&0 \\ 0&1 \end{matrix} \bigg] = \dfrac{1}{10}\bigg[ \begin{matrix}1&2 \\ - 3&4 \end{matrix} \bigg] A$

We know,

$\boxed{ \bf{ {AA}^{ - 1} = {A}^{ - 1}A = I }}$

Hence,

$\: \: \: \: \: \: \: \: \: \: \: \: \: \boxed{ \quad \bf{ {A}^{ - 1} = \dfrac{1}{10}\bigg[ \begin{matrix}1&2 \\ - 3&4 \end{matrix} \bigg] \quad}}$

Additional Information :-

$\boxed{ \bf{A \: (adjA) = \: (adjA) \: A = |A| I}}$

$\boxed{ \bf{ |adjA| = { |A| }^{n - 1} }}$

$\boxed{ \bf{ |A(adjA)| = { |A| }^{n} }}$

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

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