Answer
$\det(A)=-15$
Work Step by Step
We will apply Corollary 3.2.6. The determinant of the matrix of coefficients of the system is:
$\begin{vmatrix} -1 &3 & 3
\\ 4 & -6 & -3 \\
2 & -1 & 4
\end{vmatrix}=1\begin{vmatrix} -1 &3 & 3
\\ 0 & 6 & 9 \\
0 & 5 &10
\end{vmatrix}=6\begin{vmatrix} -1 &3 & 3
\\ 0 & 1& \frac{3}{2} \\
0 & 5 &10
\end{vmatrix}=6\begin{vmatrix} -1 &3 & 3
\\ 0 & 1& \frac{3}{2} \\
0 & 0 & \frac{5}{2}
\end{vmatrix}$
Use the Cofactor ExpansionTheorem along column 1:
$\det(A)=6.[a_{11}C_{11}+a_{21}C_{21}+a_{31}C_{31}]$
where $C_{ij}=(_1)^{i+j}.M_{ij}$
Hence, the the determinant is:
$\det(A)=6[-1.(-1)^{1+1}.\begin{vmatrix} 1 & \frac{3}{2}
\\ 0 & \frac{5}{2}
\end{vmatrix}+0.(-1)^{2+1}.\begin{vmatrix} 3 & 3
\\ 0 & \frac{5}{2}
\end{vmatrix}+0.(-1)^{3+1}.\begin{vmatrix} 3 & 3
\\ 1 & \frac{3}{2}
\end{vmatrix}]$
$\det (A)=6[-1.1.\frac{5}{2}]$
$\det (A)=6.\frac{-5}{2}$
$\det(A)=-15$
