$\left[ {\begin{array}{cccccccccccccccccccc}4&5\\3&4\end{array}} \right]$

.:

Let us take A$= \left[ {\begin{array}{cccccccccccccccccccc}4&5\\3&4\end{array}} \right]$

We know that, A = IA
$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}4&5\\3&4\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&0\\0&1\end{array}} \right]A$

Applying ${R_1} \to {R_1} - {R_2}$
$\left[ {\begin{array}{cccccccccccccccccccc}1&1\\3&4\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&{ - 1}\\0&1\end{array}} \right]A$

Applying ${R_2} \to {R_2} - 3{R_1}$
$\left[ {\begin{array}{cccccccccccccccccccc}1&1\\0&1\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&{ - 1}\\{ - 3}&4\end{array}} \right]A$

Applying ${R_1} \to {R_1} - {R_2}$
$\left[ {\begin{array}{cccccccccccccccccccc}1&0\\0&1\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}4&{ - 5}\\{ - 3}&4\end{array}} \right]A$

Hence, ${A^{ - 1}} = \left[ {\begin{array}{cccccccccccccccccccc}4&{ - 5}\\{ - 3}&4\end{array}} \right]$