Answer
The inverse of the matrix $ A $ is, ${{A}^{-1}}=\left[ \begin{matrix}
3 & 1 \\
2 & 1 \\
\end{matrix} \right]$ and $ A{{A}^{-1}}={{A}^{-1}}A=I $
Work Step by Step
At first we will find, $\begin{align}
& \det A=3\left( 1 \right)-1\left( 2 \right) \\
& =3-2 \\
& =1
\end{align}$
The inverse of the matrix $ A $ is calculated as below:
$\begin{align}
& {{A}^{-1}}=\frac{1}{\det A}\left[ \begin{matrix}
3 & 1 \\
2 & 1 \\
\end{matrix} \right] \\
& =\frac{1}{1}\left[ \begin{matrix}
3 & 1 \\
2 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
3 & 1 \\
2 & 1 \\
\end{matrix} \right]
\end{align}$
Now we will consider, $\begin{align}
& A{{A}^{-1}}=\left[ \begin{matrix}
1 & -1 \\
-2 & 3 \\
\end{matrix} \right]\left[ \begin{matrix}
3 & 1 \\
2 & 1 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1\left( 3 \right)-1\left( 2 \right) & 1\left( 1 \right)-1\left( 1 \right) \\
-2\left( 3 \right)+3\left( 2 \right) & -2\left( 1 \right)+3\left( 1 \right) \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
3-2 & 1-1 \\
-6+6 & -2+3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]
\end{align}$
Next we will consider, $\begin{align}
& {{A}^{-1}}A=\left[ \begin{matrix}
3 & 1 \\
2 & 1 \\
\end{matrix} \right]\left[ \begin{matrix}
1 & -1 \\
-2 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
3\left( 1 \right)+1\left( -2 \right) & 3\left( -1 \right)+1\left( 3 \right) \\
2\left( 1 \right)+1\left( -2 \right) & 2\left( -1 \right)+1\left( 3 \right) \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
3-2 & -3+3 \\
2-2 & -2+3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]
\end{align}$
Clearly, $ A{{A}^{-1}}={{A}^{-1}}A=I $
Hence, ${{A}^{-1}}=\left[ \begin{matrix}
3 & 1 \\
2 & 1 \\
\end{matrix} \right]$.
