Answer
$\det (A)=-4$
Work Step by Step
We have the cofactor expansion theorem for the 1st row:
$\det (A)=a_{11}C_{11}+a_{12}C_{12}+a_{13}C_{13}+a_{14}C_{14}$
with $C_{ij}=(-1)^{i+j}.M_{ij}$
So, we get:
$\det (A)=1C_{11}+0C_{12}+(-1)C_{13}+0C_{14}$
$\det (A)=1\begin{vmatrix}
1 & 0 & -1 \\
0 & -1 & 0 \\
1 & 0 & 1
\end{vmatrix}+0\begin{vmatrix}
0 & 0 & -1 \\
-1 & -1 & 0 \\
0 & 0 & 1
\end{vmatrix}+(-1)\begin{vmatrix}
0 & 1 & -1 \\
-1 & 0 & 0 \\
0 & 1 & 1
\end{vmatrix}+0\begin{vmatrix}
0 & 1 & 0 \\
-1 & 0 & -1 \\
0 & 1 & 0
\end{vmatrix}$
Plug in the given values:
$\det (A)=[1.(-1)+1.(-1)]+0-[-(-1).(1+1)+0$
$\det (A)=-2-2$
$\det (A)=-4$
