Answer
$\det (A)=0$
Work Step by Step
We have the cofactor expansion theorem for the 1st row:
$\det (A)=a_{11}C_{11}+a_{21}C_{21}+a_{31}C_{31}+a_{41}C_{41}$
with $C_{ij}=(-1)^{i+j}.M_{ij}$
So, we get:
$\det (A)=2C_{11}+1C_{21}+0C_{31}+1C_{41}$
$\det (A)=2\begin{vmatrix}
4 & -2 & 3 \\
2 & -1 & 0 \\
3 & -2 & 4
\end{vmatrix}+1\begin{vmatrix}
-1 & 3 & 1 \\
2 & -1 & 0 \\
0 & 0 & 1
\end{vmatrix}+(-1)\begin{vmatrix}
-1 & 3 & 1 \\
4 & -2 & 3 \\
3 & -2 & 4
\end{vmatrix}+0\begin{vmatrix}
-1 & 3 & 1 \\
4 & -2 & 3 \\
2 & -1 & 0
\end{vmatrix}$
Plug in the given values:
$\det (A)=2[4.(-4)-2.(8+6)+3.3]+(-1)[1(-4+3)+4(1-6)]+0+(-1)[1(-4+4)-3(1-6)]$
$\det (A)=-6+21+0-15$
$\det (A)=0$
