Answer
The inverse is $=\left[ \begin{matrix}
\frac{1}{4} & \frac{5}{8} \\
\frac{1}{2} & \frac{3}{4} \\
\end{matrix} \right]$
Work Step by Step
Consider the matrix $ A=\left[ \begin{matrix}
7 & -5 \\
-4 & 3 \\
\end{matrix} \right]$.
And, identity matrix is $ I=\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]$
Then, $\begin{align}
& \left[ I-A \right]=\left[ \begin{matrix}
1 & 0 \\
0 & 1 \\
\end{matrix} \right]-\left[ \begin{matrix}
7 & -5 \\
-4 & 3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
1-7 & 0+5 \\
0+4 & 1-3 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
-6 & 5 \\
4 & -2 \\
\end{matrix} \right]
\end{align}$
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, $\begin{align}
& a=-6 \\
& b=5 \\
& c=4 \\
& d=-2
\end{align}$
Now, the inverse is, ${{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix}
d & -c \\
-b & a \\
\end{matrix} \right]$
Substitute the values to get, $\begin{align}
& {{A}^{-1}}=\frac{1}{ad-bc}\left[ \begin{matrix}
d & -c \\
-b & a \\
\end{matrix} \right] \\
& =\frac{1}{-8}\left[ \begin{matrix}
-2 & -5 \\
-4 & -6 \\
\end{matrix} \right] \\
& =\left[ \begin{matrix}
\frac{1}{4} & \frac{5}{8} \\
\frac{1}{2} & \frac{3}{4} \\
\end{matrix} \right]
\end{align}$
Therefore, the inverse of the matrix, ${{\left[ I-A \right]}^{-1}}=\left[ \begin{matrix}
\frac{1}{4} & \frac{5}{8} \\
\frac{1}{2} & \frac{3}{4} \\
\end{matrix} \right]$.
