Answer
4
Work Step by Step
First, use cofactor expansion (Theorem 4.1) along the third row because that is the row/column with the most zeros:
\[
\begin{vmatrix}
1 & -1 & 0 & 3\\2 & 5 & 2 & 6\\0 & 1 & 0 & 0\\1 & 4 & 2 & 1
\end{vmatrix}
=
0 \cdot
\begin{vmatrix}
-1 & 0 & 3\\5 & 2 & 6\\4 & 2 & 1
\end{vmatrix}
-1 \cdot
\begin{vmatrix}
1 & 0 & 3\\2 & 2 & 6\\1 & 2 & 1
\end{vmatrix}
+0 \cdot
\begin{vmatrix}
1 & -1 & 3\\2 & 5 & 6\\1 & 4 & 1
\end{vmatrix}
-0 \cdot
\begin{vmatrix}
1 & -1 & 0\\2 & 5 & 2\\1 & 4 & 2
\end{vmatrix}
\]
Then simplify the expansion:
$=0 -1 \cdot
\begin{vmatrix}
1 & 0 & 3\\2 & 2 & 6\\1 & 2 & 1
\end{vmatrix}
+0-0$
Now expand along the second column:
=$-1 \cdot (-0+2 \cdot
\begin{vmatrix}
1 & 3\\1 & 1
\end{vmatrix}
-2 \cdot
\begin{vmatrix}
1 & 3\\2 & 6
\end{vmatrix})$
Lastly, simplify the expansion:
$=-1 \cdot (2 \cdot (1-3)-2 \cdot(6-6))$
$=-(2 \cdot -2)$
$=4$
