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