Answer
$\det(A)=2908$
Work Step by Step
$\det(A)=\begin{vmatrix} -3 & 2 & 0 & -1 & -4
\\ 1&3 &1&3&5 \\
0& 2 & 1 &-3&-1 \\
2 & 0 & 4 & -2 &-3 \\
-1 & 2 & 6 & 0 & -1
\end{vmatrix}=-\begin{vmatrix} 0 & 2 & 1 & -3 & -1
\\ 1&3 &1&3&5 \\
-3& 2 & 0 &-1&-4 \\
2 & 0 & 4 & -2 &-3 \\
-1 & 2 & 6 & 0 & -1
\end{vmatrix}=-\begin{vmatrix} 0 & 2 & 1 & -3 & -1
\\ 1&3 &1&3&5 \\
0& 11 & 3 &8&11 \\
0 & -6 & 2 & -8 &-13 \\
0 & 5& 7 & 3 & 4
\end{vmatrix}=-2\begin{vmatrix} 0 & 1 & \frac{1}{2} & -\frac{3}{2} & -\frac{1}{2}
\\ 1&3 &1&3&5 \\
0& 11& 3 &8&11 \\
0 & -6 & 2 & -8 &-13 \\
0 & 5 & 7 & 3 &4
\end{vmatrix}=-2\begin{vmatrix} 0 & 1 & \frac{1}{2} & -\frac{3}{2} & -\frac{1}{2}
\\ 1&3 &1&3&5 \\
0& 0& \frac{-5}{2} &\frac{-17}{2}&\frac{11}{2}\\
0 & 0 & 5& -17 &-16 \\
0 & 0 & \frac{9}{2} &\frac{21}{2} &\frac{13}{2}
\end{vmatrix}=5\begin{vmatrix} 0 & 1 & \frac{1}{2} & -\frac{3}{2} & -\frac{1}{2}
\\ 1&3 &1&3&5 \\
0& 0& 1 &\frac{17}{5}&\frac{-11}{5} \\
0 & 0 & 5 & -17 &-16 \\
0 & 0 & \frac{9}{2} &\frac{21}{2} &\frac{13}{2}
\end{vmatrix}=-170\begin{vmatrix} 0 & 1 & \frac{1}{2} & -\frac{3}{2} & -\frac{1}{2}
\\ 1&3 &1&3&5 \\
0& 0& 1 &\frac{17}{5}&\frac{-11}{5} \\
0 & 0 & 0 & 1 &\frac{5}{34} \\
0 & 0 &0 &\frac{-24}{5} &\frac{82}{5}
\end{vmatrix}=-170\begin{vmatrix} 0 & 1 & \frac{1}{2} & -\frac{3}{2} & -\frac{1}{2}
\\ 1&3 &1&3&5 \\
0& 0& 1 &\frac{17}{5}&\frac{-11}{5} \\
0 & 0 & 0 & 1 &\frac{5}{34} \\
0 & 0 &0 &0 &\frac{1454}{85}
\end{vmatrix}=170\begin{vmatrix} 1 & 3 &1 & 3 & 5
\\ 0&1 &\frac{1}{2}&\frac{-3}{2}&\frac{-1}{2} \\
0& 0& 1 &\frac{17}{5}&\frac{-11}{5} \\
0 & 0 & 0 & 1 &\frac{5}{34} \\
0 & 0 &0 &0 &\frac{2908}{170}
\end{vmatrix}$
We will apply Corollary 3.2.6. The determinant of the matrix of coefficients of the system is:
$\det(A)=a_{11}C_{11}+a_{21}C_{21}+a_{31}C_{31}$
where $C_{ij}=(_1)^{i+j}.M_{ij}$
$\det (A)=170.[1(-1)^{1+1}.1.1.\frac{2908}{170}]$
$\det (A)=170.\frac{2908}{170}$
$\det(A)=2908$
