Answer
$\det A=21$
Work Step by Step
Given $$\begin{array}{c}{A=\left[\begin{array}{ccc}
2 & {1} & 3&5 \\
3 & {0} & 1&2 \\
4& 1 &4& 3\\
5 &2 &5 &3\end{array}\right] }\end{array} $$
So, we get
$$A=\begin{vmatrix} 2 & {1} & 3&5 \\
3 & {0} & 1&2 \\
4& 1 &4& 3\\
5 &2 &5 &3
\end{vmatrix} \xrightarrow{R_2-\frac{3}{2}R_1,R_3-2R_{1},R_4-\frac{5}{2}R_1} \begin{vmatrix}
2 & {1} & 3&5 \\
0 &\frac{3}{2}&\frac{-7}{2}&\frac{-11}{2} \\
0& \frac{-1}{2} &-2& -7\\
0 &\frac{-1}{2} &-\frac{5}{2} &\frac{-14}{2}
\end{vmatrix} \xrightarrow{R_{3}-\frac{2}{3}R_{2},R_4-\frac{1}{3}R_2,R_4+4R_3 }\begin{vmatrix}
2 & {1} & 3&5 \\
0 &\frac{3}{2}&\frac{-7}{2}&\frac{-11}{2} \\
0& 0 &\frac{1}{3}& -\frac{10}{3}\\
0 &0&0 &-21
\end{vmatrix} $$
\begin{array}{l}{\text { Determinant of triangular matrix is the product of its diagonal elements, so: }} \\ {\text {det }A=2 \cdot -\frac{3}{2}\cdot \frac{1}{3}\cdot (-21)=21}\end{array}
