Answer
$A^{-1}=\left[\begin{array}{rr}
3 & 5\\
-2 & -3
\end{array}\right]$
Work Step by Step
$A=\displaystyle \left[\begin{array}{ll}
a & b\\
c & d
\end{array}\right] \Rightarrow A^{-1}=\frac{1}{ad-bc}\left[\begin{array}{ll}
d & -b\\
-c & a
\end{array}\right]$
($A^{-1}$ exists if and only if $ad-bc\neq 0)$
----
$\left[\begin{array}{ll}
a & b\\
c & d
\end{array}\right]=\left[\begin{array}{ll}
-3 & -5\\
2 & 3
\end{array}\right]$
$ad-bc=-3(3)-(-5)(2)=1\neq 0$
$A^{-1}=(1)\cdot\left[\begin{array}{ll}
3 & 5\\
-2 & -3
\end{array}\right]$
$A^{-1}=\left[\begin{array}{ll}
3 & 5\\
-2 & -3
\end{array}\right]$
