Answer
$\det(A)=-170$
Work Step by Step
$\det(A)=2\begin{vmatrix}1 & 0 & \frac{-1}{2} & \frac{3}{2} & 0
\\ 0&3 &0&1&2 \\
0& 1 & 3 &0&4 \\
1 & 0 & 1 & -1 &0 \\
3 & 0 & 2 & 0 & 5
\end{vmatrix}=2\begin{vmatrix}1 & 0 & \frac{-1}{2} & \frac{3}{2} & 0
\\ 0&3 &0&1&2 \\
0& 1 & 3 &0&4 \\
0 & 0 & \frac{3}{2} & \frac{-5}{2} &0 \\
0& 0 & \frac{7}{2} & \frac{-9}{2} & 5
\end{vmatrix}=-2\begin{vmatrix}1 & 0 & \frac{-1}{2} & \frac{3}{2} & 0
\\ 0&1 &3&0&4 \\
0& 3 & 0 &1&2 \\
0 & 0 & \frac{3}{2}&\frac{-5}{2} &0 \\
0& 0 & \frac{7}{2} & \frac{-9}{2} & 5
\end{vmatrix}=-2\begin{vmatrix}1 & 0 & \frac{-1}{2} & \frac{3}{2} & 0
\\ 0&1 &3&0&4 \\
0& 0 & -9 &1&-10 \\
0 & 0 & \frac{3}{2}&\frac{-5}{2} &0 \\
0& 0 & \frac{7}{2} & \frac{-9}{2} & 5
\end{vmatrix}=18\begin{vmatrix}1 & 0 & \frac{-1}{2} & \frac{3}{2} & 0
\\ 0&1 &3&0&4 \\
0& 0 & 1 &\frac{-1}{9}&\frac{10}{9} \\
0 & 0 & \frac{3}{2}&\frac{-5}{2} &0 \\
0& 0 & \frac{7}{2} & \frac{-9}{2} & 5
\end{vmatrix}=18\begin{vmatrix}1 & 0 & \frac{-1}{2} & \frac{3}{2} & 0
\\ 0&1 &3&0&4 \\
0& 0 & 1 &\frac{-1}{9}&\frac{10}{9} \\
0 & 0 & 0&\frac{-14}{6}&\frac{-5}{3} \\
0& 0 & 0& \frac{74}{18} & \frac{10}{9}
\end{vmatrix}=-42\begin{vmatrix}1 & 0 & \frac{-1}{2} & \frac{3}{2} & 0
\\ 0&1 &3&0&4 \\
0& 0 & 1 &\frac{-1}{9}&\frac{10}{9} \\
0 & 0 & 0&1&\frac{5}{7} \\
0& 0 & 0& \frac{74}{18} & \frac{10}{9}
\end{vmatrix}=-42\begin{vmatrix}1 & 0 & \frac{-1}{2} & \frac{3}{2} & 0
\\ 0&1 &3&0&4 \\
0& 0 & 1 &\frac{-1}{9}&\frac{10}{9} \\
0 & 0 & 0&1&\frac{15}{14} \\
0& 0 & 0& 0 & \frac{255}{63}
\end{vmatrix}$
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 $C_{ij}=(_1)^{i+j}.M_{ij}$
$\det (A)=-42.[1(-1)^{1+1}.\frac{255}{63}]$
$\det (A)=-42.\frac{255}{63}$
$\det(A)=-170$
