Answer
8
Work Step by Step
First, use cofactor expansion (Theorem 4.1) along the second column because that is the row/column with the most zeros:
\[
\begin{vmatrix}
2 & 0 & 3 & -1\\1 & 0 & 2 & 2\\0 & -1 & 1 & 4\\2 & 0 & 1 & -3
\end{vmatrix}
=
-0 \cdot
\begin{vmatrix}
1 & 2 & 2\\0 & 1 & 4\\2 & 1 & -3
\end{vmatrix}
+0 \cdot
\begin{vmatrix}
2 & 3 & -1\\0 & 1 & 4\\2 & 1 & -3
\end{vmatrix}
-(-1) \cdot
\begin{vmatrix}
2 & 3 & -1\\1 & 2 & 2\\2 & 1 & -3
\end{vmatrix}
+0 \cdot
\begin{vmatrix}
2 & 3 & -1\\1 & 2 & 2\\0 & 1 & 4
\end{vmatrix}
\]
Then simplify the expansion:
$=0 + 0 + 1 \cdot
\begin{vmatrix}
2 & 3 & -1\\1 & 2 & 2\\2 & 1 & -3
\end{vmatrix}
+0$
Now expand along the first row:
=$2 \cdot
\begin{vmatrix}
2 & 2\\1 & -3
\end{vmatrix}
-3 \cdot
\begin{vmatrix}
1 & 2\\2 & -3
\end{vmatrix}
-1 \cdot
\begin{vmatrix}
1 & 2\\2 & 1
\end{vmatrix}$
Lastly, simplify the expansion:
$=2 \cdot (-6-2) - 3 \cdot (-3-4) - 1 \cdot (1 - 4 ))$
$=(2 \cdot -8)- (3 \cdot -7) - (1 \cdot-3)$
$=-16+21+3$
$=8$
