Answer
$\det (A)=9$
Work Step by Step
We have the cofactor expansion theorem for the 1st row:
$\det (A)=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}$
with $C_{ij}=(-1)^{i+j}.M_{ij}$
So, we get:
$\det (A)=1C_{11}+0C_{12}+(-2)C_{13}$
$\det (A)=1\begin{vmatrix}
1 & -1 \\
2 & 5
\end{vmatrix}+0\begin{vmatrix}
3 & -1 \\
7 & 5
\end{vmatrix}+(-2)\begin{vmatrix}
3 & 1 \\
7 & 2
\end{vmatrix}$
Plug in the given values:
$\det (A)=1.7+0+(-2)(-1)$
$\det (A)=7+2$
$\det (A)=9$
