Answer
$A^{-1}=\left[\begin{array}{ll}
1/6 & 1/4\\
1/3 & 0
\end{array}\right]$
Work Step by Step
$ad-bc=0-12=-12$
$A^{-1}=-\displaystyle \frac{1}{12}\left[\begin{array}{ll}
-2 & -3\\
-4 & 0
\end{array}\right]=\displaystyle \left[\begin{array}{ll}
1/6 & 1/4\\
1/3 & 0
\end{array}\right]$
Checking,
$AA^{-1}=\left[\begin{array}{ll}
0(\frac{1}{6})+3(\frac{1}{3}) & 0(\frac{1}{4})+3(0)\\
4(\frac{1}{6})-2(\frac{1}{3}) & 4(\frac{1}{4})-2(0)
\end{array}\right]=\left[\begin{array}{ll}
1 & 0\\
0 & 1
\end{array}\right]=I_{2}$
$A^{-1}A=\left[\begin{array}{ll}
\frac{1}{6}(0)+\frac{1}{4}(4) & \frac{1}{6}(3)+\frac{1}{4}(-2)\\
\frac{1}{3}(0)+0(4) & \frac{1}{3}(3)+0(-2)
\end{array}\right]=\left[\begin{array}{ll}
1 & 0\\
0 & 1
\end{array}\right]=I_{2}$
