$\left[ {\begin{array}{cccccccccccccccccccc}6&{ - 3}\\{ - 2}&1\end{array}} \right]$

.:

Let us take A = $\left[ {\begin{array}{cccccccccccccccccccc}6&{ - 3}\\{ - 2}&1\end{array}} \right]$

We know that, A = IA
$\left[ {\begin{array}{cccccccccccccccccccc}6&{ - 3}\\{ - 2}&1\end{array}} \right] = \left[ {\begin{array}{cccccccccccccccccccc}1&0\\0&1\end{array}} \right]A$

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

We have all zeroes in the first row of the left hand side matrix of the above equation.
$\therefore$ ${A^{ - 1}}$ does not exist.