$\left[ {\begin{array}{cccccccccccccccccccc}3&{10}\\2&7\end{array}} \right]$

.:

Let us take A =$\left[ {\begin{array}{cccccccccccccccccccc}3&{10}\\2&7\end{array}} \right]$

We know that, A = IA
$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}3&{10}\\2&7\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&0\\0&1\end{array}} \right]A$

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

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

Applying ${R_1} \to {R_1} - 3{R_2}$
$\left[ {\begin{array}{cccccccccccccccccccc}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}7&{ - 10}\\{ - 2}&3\end{array}} \right]A$

Hence, ${A^{ - 1}} = \left[ {\begin{array}{cccccccccccccccccccc}7&{ - 10}\\{ - 2}&3\end{array}} \right]$