Answer
$\left[\begin{array}{ll}
1/4 & 5/8\\
1/2 & 3/4
\end{array}\right]$
Work Step by Step
If $A=\left[\begin{array}{ll}
a & b\\
c & d
\end{array}\right]$, then $A^{-1}=\displaystyle \frac{1}{ad-bc}\left[\begin{array}{ll}
d & -b\\
-c & a
\end{array}\right]$.
----------------------
$I-A=\left[\begin{array}{ll}
1 & 0\\
0 & 1
\end{array}\right]-\left[\begin{array}{ll}
7 & -5\\
-4 & 3
\end{array}\right]=\left[\begin{array}{ll}
-6 & 5\\
4 & -2
\end{array}\right]$
$ad-bc=(-6)(-2)-5(4)=-8$
$(I-A)^{-1}=\displaystyle \frac{1}{-8}\left[\begin{array}{ll}
-2 & -5\\
-4 & -6
\end{array}\right]=\displaystyle \left[\begin{array}{ll}
1/4 & 5/8\\
1/2 & 3/4
\end{array}\right]$
