Answer
The inverse matrix of the provided matrix is ${{A}^{-1}}=\left[ \begin{matrix}
\frac{2}{7} & \frac{-3}{7} \\
\frac{1}{7} & \frac{2}{7} \\
\end{matrix} \right]$.
Work Step by Step
Consider the given matrix $ A=\left[ \begin{matrix}
2 & 3 \\
-1 & 2 \\
\end{matrix} \right]$
Now, by using the inverse formula, we get:
${{A}^{-1}}=\frac{1}{\left| ad-bc \right|}\left[ \begin{matrix}
d & -b \\
-c & a \\
\end{matrix} \right]$
Let, $\begin{align}
& a=2 \\
& b=3 \\
& c=-1 \\
& d=2
\end{align}$
Substitute the values to get
$\begin{align}
& {{A}^{-1}}=\frac{1}{\left| ad-bc \right|}\left[ \begin{matrix}
d & -b \\
-c & a \\
\end{matrix} \right] \\
& {{A}^{-1}}=\frac{1}{\left| 2\times 2-3\times \left( -1 \right) \right|}\left[ \begin{matrix}
2 & -3 \\
1 & 2 \\
\end{matrix} \right] \\
& =\frac{1}{4+3}\left[ \begin{matrix}
2 & -3 \\
1 & 2 \\
\end{matrix} \right] \\
& =\frac{1}{7}\left[ \begin{matrix}
2 & -3 \\
1 & 2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\frac{2}{7} & \frac{-3}{7} \\
\frac{1}{7} & \frac{2}{7} \\
\end{matrix} \right]
\end{align}$
So, therefore the inverse of the matrix is given by
${{A}^{-1}}=\left[ \begin{matrix}
\frac{2}{7} & \frac{-3}{7} \\
\frac{1}{7} & \frac{2}{7} \\
\end{matrix} \right]$
Now, check the result for
$ A{{A}^{-1}}={{I}_{2}}$ And ${{A}^{-1}}A={{I}_{2}}$
Here, $ A=\left[ \begin{matrix}
2 & 3 \\
-1 & 2 \\
\end{matrix} \right]$
So, $\begin{align}
& A{{A}^{-1}}=\left[ \begin{matrix}
2 & 3 \\
-1 & 2 \\
\end{matrix} \right]\left[ \begin{matrix}
\frac{2}{7} & \frac{-3}{7} \\
\frac{1}{7} & \frac{2}{7} \\
\end{matrix} \right] \\
& A{{A}^{-1}}=\left[ \begin{matrix}
2\times \frac{2}{7}+3\times \frac{1}{7} & 2\times \frac{-3}{7}+3\times \frac{2}{7} \\
\left( -1 \right)\times \frac{2}{7}+2\times \frac{1}{7} & \left( -1 \right)\times \frac{-3}{7}+2\times \frac{2}{7} \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right] \\
& ={{I}_{2}}
\end{align}$
This implies that
${{A}^{-1}}A={{I}_{2}}$
Now, evaluate the product ${{A}^{-1}}A={{I}_{2}}$.
