Answer
See the proof below.
Work Step by Step
Consider the given expression $ A=\left[ \begin{matrix}
a & b \\
c & d \\
\end{matrix} \right]$
For the given expression, the values are, $\begin{align}
& a=2 \\
& b=1 \\
& c=3 \\
& d=1 \\
\end{align}$
So, $\begin{align}
& ad-bc=2\times 1-1\times 3 \\
& =-1 \\
& \ne 0
\end{align}$
That is ${{A}^{-1}}$ exists and the inverse is, $\begin{align}
& {{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix}
d & -b \\
-c & a \\
\end{matrix} \right] \\
& =\frac{1}{2\times 1-1\times 3}\left[ \begin{matrix}
1 & -1 \\
-3 & 2 \\
\end{matrix} \right] \\
& =\frac{1}{-1}\left[ \begin{matrix}
1 & -1 \\
-3 & 2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & -1 \\
-3 & 2 \\
\end{matrix} \right]
\end{align}$
