Answer
-72
Work Step by Step
Given $$\begin{array}{c}{A=\left[\begin{array}{ccc}
{1} & {-1} & 2&4 \\
{3} & {1} & 2 &4 \\
-1& 1 &1& 3\\
2 &1 & 4 &2\end{array}\right] }\end{array} $$
So, we get
$$A=\begin{vmatrix} {1} & {-1} & 2&4 \\
3& 1 & 2 &4 \\
-1& 1 &1& 3\\
2 &1 & 4 &2
\end{vmatrix} \xrightarrow{R_{2}-3R_{1} ,R_3+R_{1},R_4-2R_1} \begin{vmatrix}
1 & -1 & 2&4 \\
0& 4 & -4 &-8 \\
0& 0&5& 6\\
0 &3 & 0&-6
\end{vmatrix} \xrightarrow{R_{4}-\frac{3}{4}R_{2}} \begin{vmatrix}
1 & -1 & 2&4 \\
0& 4 & -4 &-8 \\
0& 0&5& 6\\
0 &0 & 3&0
\end{vmatrix} \xrightarrow{R_{4}-\frac{3}{5}R_{3}} \begin{vmatrix}
1 & -1 & 2&4 \\
0& 4 & -4 &-8 \\
0& 0&5& 6\\
0 &0 & 0&-\frac{18}{5}
\end{vmatrix}$$
\begin{array}{l}{\text { Determinant of triangular matrix is the product of its diagonal elements, so: }} \\ {\text {det }A=1 \cdot 4 \cdot 5\cdot-\frac{18}{5}=-72}\end{array}