Question

$\large{\underline{\underline{\red{\bf{Question}}}}}$

☯ Using elementary transformation, find the inverse of the given matrix.
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⠀⠀⠀➝ $\bigg[ \begin{matrix}2&5 \\ 1&3 \end{matrix} \bigg]$
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❏ No spamming
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Answer 1

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

Given matrix is

$\rm :\longmapsto\:\bigg[ \begin{matrix}2&5 \\ 1&3 \end{matrix} \bigg]$

Let assume that

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

Using Elementary Row Transformation Method, we have

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

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

$\bf :\longmapsto\:OP \: R_1 \: \to \: R_1 - R_2$

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

$\bf :\longmapsto\:OP \: R_2 \: \to \: R_2 - R_1$

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

$\bf :\longmapsto\:OP \: R_1 \: \to \: R_1 - 2R_2$

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

Now, we know,

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

Hence,

$\bf :\longmapsto\: {A}^{ - 1} \: = \: \bigg[ \begin{matrix}3& - 5 \\ - 1&2 \end{matrix} \bigg]$

Additional Information :-

$\boxed{ \rm{ \: | {A}^{ - 1} | = \dfrac{1}{ |A| } }}$

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

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

$\boxed{ \rm{ |I| \: = \: 1 }}$

$\boxed{ \rm{ |AB| \: = \: |A| \: |B| }}$

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