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

.:

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

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

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