Answer
$\det (A)=3$
Work Step by Step
We have the cofactor expansion theorem for the 1st row:
$\det (A)=a_{11}C_{11}+a_{21}C_{21}+a_{31}C_{31}$
with $C_{ij}=(-1)^{i+j}.M_{ij}$
So, we get:
$\det (A)=2C_{11}+5C_{21}+3C_{31}$
$\det (A)=2\begin{vmatrix}
2 & 1 \\
-3 & 7
\end{vmatrix}+5\begin{vmatrix}
-1 & 3 \\
-3 & 7
\end{vmatrix}+3\begin{vmatrix}
-1 & 3 \\
2 & 1
\end{vmatrix}$
Plug in the given values:
$\det (A)=2.17+5(-2)+3.(-7)$
$\det (A)=34-10-21$
$\det (A)=3$
