Answer
The inverse is, $\left[ \begin{matrix}
\frac{1}{8} & \frac{5}{8} \\
\frac{3}{8} & \frac{7}{8} \\
\end{matrix} \right]$.
Work Step by Step
Consider the given matrix $ A=\left[ \begin{matrix}
8 & -5 \\
-3 & 2 \\
\end{matrix} \right]$.
And, identity matrix is $ I=\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]$
Then, $\begin{align}
& \left[ I-A \right]=\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]-\left[ \begin{matrix}
8 & -5 \\
-3 & 2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1-8 & 0+5 \\
0+3 & 1-2 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-7 & 5 \\
3 & -1 \\
\end{matrix} \right]
\end{align}$
Now, the inverse of matrix $\left[ I-A \right]$ is equal to ${{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix}
d & -c \\
-d & a \\
\end{matrix} \right]$
Now, we will compare the matrix to the original matrix, to get, $\begin{align}
& a=-7 \\
& b=5 \\
& c=3 \\
& d=-1
\end{align}$
Now, the inverse is, ${{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix}
d & -c \\
-b & a \\
\end{matrix} \right]$
Substitutes the values to get, $\begin{align}
& {{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix}
d & -c \\
-b & a \\
\end{matrix} \right] \\
& =\frac{1}{-8}\left[ \begin{matrix}
-1 & -5 \\
-3 & -7 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\frac{1}{8} & \frac{5}{8} \\
\frac{3}{8} & \frac{7}{8} \\
\end{matrix} \right]
\end{align}$
Therefore, the inverse of the matrix, ${{\left[ I-A \right]}^{-1}}=\left[ \begin{matrix}
\frac{1}{8} & \frac{5}{8} \\
\frac{3}{8} & \frac{7}{8} \\
\end{matrix} \right]$.
