Answer
The inverse is,
${{C}^{-1}}=\frac{1}{5}\left[ \begin{array}{*{35}{r}}
3 & -2 \\
1 & 1 \\
\end{array} \right]$
Work Step by Step
Consider the inverse of the matrix $C$
${{C}^{-1}}=\frac{1}{\left| C \right|}\text{adj}\left( C \right)$
Consider the determinant of the matrix
$\begin{align}
& \left| C \right|=\left| \begin{array}{*{35}{r}}
1 & 2 \\
-1 & 3 \\
\end{array} \right| \\
& =3+2 \\
& =5
\end{align}$
Consider the adjoin of the matrix
$\begin{align}
& \text{adj}\left( C \right)={{\left[ \begin{matrix}
+\left( 3 \right) & -\left( -1 \right) \\
-\left( 2 \right) & +\left( 1 \right) \\
\end{matrix} \right]}^{t}} \\
& ={{\left[ \begin{array}{*{35}{r}}
3 & 1 \\
-2 & 1 \\
\end{array} \right]}^{t}} \\
& =\left[ \begin{array}{*{35}{r}}
3 & -2 \\
1 & 1 \\
\end{array} \right]
\end{align}$
Therefore, the inverse of the matrix $C$ is given by,
${{C}^{-1}}=\frac{1}{5}\left[ \begin{array}{*{35}{r}}
3 & -2 \\
1 & 1 \\
\end{array} \right]$
Therefore, the inverse of the matrix is ${{C}^{-1}}=\frac{1}{5}\left[ \begin{array}{*{35}{r}}
3 & -2 \\
1 & 1 \\
\end{array} \right]$.
