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