$\left[ {\begin{array}{cccccccccccccccccccc}3&{ - 1}\\{ - 4}&2\end{array}} \right]$

.:

Let us take A = $\left[ {\begin{array}{cccccccccccccccccccc}3&{ - 1}\\{ - 4}&2\end{array}} \right]$
We know that, A = IA
$\Rightarrow$ $\left[ {\begin{array}{cccccccccccccccccccc}3&{ - 1}\\{ - 4}&2\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&0\\0&1\end{array}} \right]A$

Applying ${R_1} \to {R_1} + {R_2}$
$\left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&1\\{ - 4}&2\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&1\\0&1\end{array}} \right]A$

Applying ${R_1} \to - {R_1}$
$\left[ {\begin{array}{cccccccccccccccccccc}1&{ - 1}\\{ - 4}&2\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&{ - 1}\\0&1\end{array}} \right]A$

Applying ${R_2} \to {R_2} + 4{R_1}$
$\left[ {\begin{array}{cccccccccccccccccccc}1&{ - 1}\\0&{ - 2}\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&{ - 1}\\{ - 4}&{ - 3}\end{array}} \right] = A$

Applying ${R_2} \to - {R_2}$
$\left[ {\begin{array}{cccccccccccccccccccc}1&{ - 1}\\0&2\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}{ - 1}&{ - 1}\\4&3\end{array}} \right] = A$

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

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

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