Answer
$A^{-1}=\left[\begin{array}{rr}
1/3 & 0\\
0 & 2
\end{array}\right]$
Work Step by Step
$1.\quad $Start with $[A|I]$,
$2.\quad $ Row reduce,
$3.\quad $ If the reduced form is $[I|B], $then $B=A^{-1}.$
If not, then A is singular (has no inverse)
----
$\left[\begin{array}{lllll}
3 & 0 & | & 1 & 0\\
0 & 1/2 & | & 0 & 1
\end{array}\right]\left\{\begin{array}{l}
\div 3.\\
\times 2.
\end{array}\right.$
$\rightarrow\left[\begin{array}{lllll}
1 & 0 & | & 1/3 & 0\\
0 & 1 & | & 0 & 2
\end{array}\right]$
$A^{-1}=\left[\begin{array}{rr}
1/3 & 0\\
0 & 2
\end{array}\right]$
Check: $\left[\begin{array}{ll}
3 & 0\\
0 & 1/2
\end{array}\right]\left[\begin{array}{ll}
1/3 & 0\\
0 & 2
\end{array}\right]=\left[\begin{array}{ll}
1+0 & 0+0\\
0+0 & 0+1
\end{array}\right]=I_{2}$