Answer
$\left[\begin{array}{ll}
1/8 & 5/8\\
3/8 & 7/8
\end{array}\right]$
Work Step by Step
If $A=\left[\begin{array}{ll}
a & b\\
c & d
\end{array}\right]$, then $A^{-1}=\displaystyle \frac{1}{ad-bc}\left[\begin{array}{ll}
d & -b\\
-c & a
\end{array}\right]$.
----------------------
$I-A=\left[\begin{array}{ll}
1 & 0\\
0 & 1
\end{array}\right]-\left[\begin{array}{ll}
8 & -5\\
-3 & 2
\end{array}\right]=\left[\begin{array}{ll}
-7 & 5\\
3 & -1
\end{array}\right]$
$ad-bc=(-7)(-1)-5(3)=-8$
$(I-A)^{-1}=\displaystyle \frac{1}{-8}\left[\begin{array}{ll}
-1 & -5\\
-3 & -7
\end{array}\right]=\displaystyle \left[\begin{array}{ll}
1/8 & 5/8\\
3/8 & 7/8
\end{array}\right]$
