Answer
$A^{-1}=\left[\begin{array}{rr}
12 & -1\\
-15 & 3/2
\end{array}\right]$
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 $ad-bc\neq 0)$
----
$\left[\begin{array}{ll}
a & b\\
c & d
\end{array}\right]=\left[\begin{array}{ll}
1/2 & 1/3\\
5 & 4
\end{array}\right]$
$ad-bc=\displaystyle \frac{1}{2}(4)-(\frac{1}{3})(5)=\frac{1}{3}\neq 0$
$\displaystyle \frac{1}{ad-bc}=3$
$A^{-1}=3\left[\begin{array}{ll}
4 & -1/3\\
-5 & 1/2
\end{array}\right]$
$A^{-1}=\left[\begin{array}{ll}
12 & -1\\
-15 & 3/2
\end{array}\right]$
