$\left[ {\begin{array}{cccccccccccccccccccc}1&3\\2&7\end{array}} \right]$

.:

Let us take $A = \left[ {\begin{array}{cccccccccccccccccccc}1&3\\2&7\end{array}} \right]$

We know that, A = IA

$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}1&3\\2&7\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&0\\0&1\end{array}} \right]A$

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

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

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