Answer
$\det (A)=5$
Work Step by Step
We have the cofactor expansion theorem is:
$\det (A)=a_{11}C_{11}+a_{12}C_{12}$
with $C_{ij}=(-1)^{i+j}.M_{ij}$
to evaluate the given determinant of row 1:
$\det (A)=1C_{11}+(-2)C_{12}$
$\det (A)=1(-1)^{1+1}.M_{11}+(-2)(-1)^{1+2}.M_{12}$
Plug in the given values:
$\det (A)=1(-1)^{1+2}.3+(-2)(-1)^{1+2}.1$
$\det (A)=3+2$
$\det (A)=5$
