$\left[ {\begin{array}{cccccccccccccccccccc}2&3\\5&7\end{array}} \right]$

.:

Let us take $A = \left[ {\begin{array}{cccccccccccccccccccc}2&3\\5&7\end{array}} \right]$

We know that, A = IA
$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}2&3\\5&7\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&0\\0&1\end{array}} \right]A$

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

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

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

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

Applying ${R_1} \to {R_1} - 2{R_2}$
$\left[ {\begin{array}{cccccccccccccccccccc}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{ - 7}&3\\5&{ - 2}\end{array}} \right]A$

Hence, ${A^{ - 1}} = \left[ {\begin{array}{cccccccccccccccccccc}{ - 7}&3\\5&{ - 2}\end{array}} \right]$