Answer
$\det (A)=-4$
Work Step by Step
We have the cofactor expansion theorem for the 1st column:
$\det (A)=a_{11}C_{31}+a_{21}C_{21}+a_{31}C_{31}$
with $C_{ij}=(-1)^{i+j}.M_{ij}$
So, we get:
$\det (A)=-4C_{31}+7C_{32}-6C_{33}$
$\det (A)=-4\begin{vmatrix}
-3 & 2 \\
6 & 2
\end{vmatrix}+7\begin{vmatrix}
2 & -1 \\
-3 & 2
\end{vmatrix}-6\begin{vmatrix}
2 & -1 \\
6 & 2
\end{vmatrix}$
Plug in the given values:
$\det (A)=-4.(-18)+7.(-10)-6.1$
$\det (A)=72-70-6$
$\det (A)=-4$
