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:
$\det (A)=-1C_{11}+0C_{21}+2C_{31}$
$\det (A)=-1\begin{vmatrix}
1 & 4 \\
-1 & 3
\end{vmatrix}+0\begin{vmatrix}
2 & 3 \\
-1 & 3
\end{vmatrix}+2\begin{vmatrix}
2 & 3 \\
1 & 4
\end{vmatrix}$
Plug in the given values:
$\det (A)=-1.7+0+2.5$
$\det (A)=-7+10$
$\det (A)=3$
