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