Step 1: Expand the determinant along the third column.
The sign pattern for cofactors in a $3 \times 3$ determinant is:
\[
\begin{bmatrix}
+ & - & +
- & + & -
+ & - & +
\end{bmatrix}
\]
The elements of the third column are:
\[
4,\quad 0,\quad 0
\]
Hence,
\[
|A|
=
4
\begin{vmatrix}
-1 & 3
1 & 2
\end{vmatrix}
-0
\begin{vmatrix}
1 & 2
1 & 2
\end{vmatrix}
+0
\begin{vmatrix}
1 & 2
-1 & 3
\end{vmatrix}
\]
Since the last two terms are multiplied by zero, they vanish completely.
Therefore,
\[
|A|
=
4
\begin{vmatrix}
-1 & 3
1 & 2
\end{vmatrix}
\]
Step 2: Evaluate the remaining $2 \times 2$ determinant.
For a matrix
\[
\begin{bmatrix}
a & b
c & d
\end{bmatrix},
\]
the determinant is given by
\[
ad-bc.
\]
Applying this formula:
\[
\begin{vmatrix}
-1 & 3
1 & 2
\end{vmatrix}
=
(-1)(B)-(C)(A)
\]
\[
=-2-3
\]
\[
=-5
\]
Thus,
\[
\begin{vmatrix}
-1 & 3
1 & 2
\end{vmatrix}
=-5.
\]
Step 3: Substitute the value into the expansion.
Using the result obtained above,
\[
|A|
=
4(-5)
\]
\[
|A|
=
-20
\]
Step 4: Verify the result.
Since only one non-zero cofactor term contributed to the expansion, the computation is straightforward and free from sign ambiguity.
Thus, the determinant of matrix $A$ is
\[
{-20}
\]
Hence, the correct answer is:
\[
{(D)\ -20}
\]