Answer
-103
Work Step by Step
Given $$\begin{array}{c}{A=\left[\begin{array}{ccc}
{3} & {7} & 1 \\
{5} & {9} & -6\\
2 & 1 & 3\end{array}\right] }\end{array} $$
So, we get
$$A=\begin{vmatrix}
3 & 7 & 1\\
5 & 9 &-6\\
2 &1 &3
\end{vmatrix} \xrightarrow{R_{2}-\frac{5}{3}R_{1} ,R_3-\frac{2}{3}R_{1}} \begin{vmatrix}
3 & 7 & 1\\
0 & -\frac{8}{3} & -\frac{23}{3}\\
0&-\frac{11}{3}&\frac{7}{3}
\end{vmatrix} \xrightarrow{R_{3}-\frac{11}{8}R_{2}} \begin{vmatrix}
3 & 7 & 1\\
0 & -\frac{8}{3} & -\frac{23}{3}\\
0&0&\frac{309}{24}
\end{vmatrix}$$
\begin{array}{l}{\text { Determinant of triangular matrix is the product of its diagonal elements, so: }} \\ {\text {det }A=-2 \cdot(-\frac{8}{3})\cdot \frac{309}{24}=-103}\end{array}
