Answer
$B^{-1}=\left[\begin{array}{rrr}
-9/2 & -1 & 4\\
3 & 1 & -3\\
7/2 & 1 & -3
\end{array}\right]$
Work Step by Step
Set up a matrix $[B|I]$, and using row operations, transform into $[I|B^{-1}].$
$\left[\begin{array}{rrr|rrr}
0 & -2 & 2 & 1 & 0 & 0\\
3 & 1 & 3 & 0 & 1 & 0\\
1 & -2 & 3 & 0 & 0 & 1
\end{array}\right]\left\{\begin{array}{l}
R_{1}\leftrightarrow R_{3}.\\
.\\
.
\end{array}\right. $
$\left[\begin{array}{rrr|rrr}
1 & -2 & 3 & 0 & 0 & 1\\
3 & 1 & 3 & 0 & 1 & 0\\
0 & -2 & 2 & 1 & 0 & 0
\end{array}\right]\left\{\begin{array}{l}
.\\
-3R_{1}.\\
.
\end{array}\right.$
$\left[\begin{array}{rrr|rrr}
1 & -2 & 3 & 0 & 0 & 1\\
0 & 7 & -6 & 0 & 1 & -3\\
0 & -2 & 2 & 1 & 0 & 0
\end{array}\right]\left\{\begin{array}{l}
.\\
+3R_{3}.\\
.
\end{array}\right.$
$\left[\begin{array}{rrr|rrr}
1 & -2 & 3 & 0 & 0 & 1\\
0 & 1 & 0 & 3 & 1 & -3\\
0 & -2 & 2 & 1 & 0 & 0
\end{array}\right]\left\{\begin{array}{l}
+2R_{2}.\\
.\\
+2R_{2}.
\end{array}\right.$
$\left[\begin{array}{rrr|rrr}
1 & 0 & 3 & 6 & 2 & -5\\
0 & 1 & 0 & 3 & 1 & -3\\
0 & 0 & 2 & 7 & 2 & -6
\end{array}\right]\left\{\begin{array}{l}
.\\
.\\
\div 2.
\end{array}\right.$
$\left[\begin{array}{rrr|rrr}
1 & 0 & 3 & 6 & 2 & -5\\
0 & 1 & 0 & 3 & 1 & -3\\
0 & 0 & 1 & 7/2 & 1 & -3
\end{array}\right]\left\{\begin{array}{l}
-3R_{3}.\\
.\\
.
\end{array}\right.$
$\left[\begin{array}{rrr|rrr}
1 & 0 & 0 & -9/2 & -1 & 4\\
0 & 1 & 0 & 3 & 1 & -3\\
0 & 0 & 1 & 7/2 & 1 & -3
\end{array}\right]$
$B^{-1}=\left[\begin{array}{lll}
-9/2 & -1 & 4\\
3 & 1 & -3\\
7/2 & 1 & -3
\end{array}\right]$
