$\left[ {\begin{array}{cccccccccccccccccccc}1&{ - 1}\\2&3\end{array}} \right]$

.:

Let us take $A = \left[ {\begin{array}{cccccccccccccccccccc}1&{ - 1}\\2&3\end{array}} \right]$

We know that, A = IA
$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}1&{ - 1}\\2&3\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&0\\0&1\end{array}} \right]A$

Applying ${R_2} \to {R_2} - 2{R_1}$
$\left[ {\begin{array}{cccccccccccccccccccc}1&{ - 1}\\0&5\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&0\\{ - 2}&1\end{array}} \right]A$

Applying ${R_2} \to \cfrac{1}{5}{R_2}$
$\left[ {\begin{array}{cccccccccccccccccccc}0&{ - 1}\\0&1\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&0\\{\cfrac{{ - 2}}{5}}&{\cfrac{1}{5}}\end{array}} \right]A$

Applying ${R_1} \to {R_1} + {R_2}$
$\left[ {\begin{array}{cccccccccccccccccccc}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{\cfrac{3}{5}}&{\cfrac{1}{5}}\\{\cfrac{{ - 2}}{5}}&{\cfrac{1}{5}}\end{array}} \right]A$

Hence, ${A^{ - 1}} = \left[ {\begin{array}{cccccccccccccccccccc}{\cfrac{3}{5}}&{\cfrac{1}{5}}\\{\cfrac{{ - 2}}{5}}&{\cfrac{1}{5}}\end{array}} \right]$