Answer
$\{(2,3,1)\}$.
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
x& +0y &+2z&=&4 \\
0x& +2y & -z&=&5\\
2x& +3y &+0z &=&13
\end{matrix}\right.$
The formula to determine the determinant is
$D=\begin{vmatrix}
a& b &c \\
d& e &f \\
g &h &i
\end{vmatrix}=a\begin{vmatrix}
e& f \\
h&i
\end{vmatrix}-b\begin{vmatrix}
d& f \\
g&i
\end{vmatrix}+c\begin{vmatrix}
d& e \\
g&h
\end{vmatrix}$
Determinant $D$ consists of the $x,y$ and $z$ coefficients.
$D=\begin{vmatrix}
1& 0 &2 \\
0& 2 &-1 \\
2 &3 &0
\end{vmatrix}=-5$
For determinant $D_x$ replace the $x−$ coefficients with the constants.
$D_x=\begin{vmatrix}
4& 0 &2 \\
5& 2 &-1 \\
13 &3 &0
\end{vmatrix}=-10$
For determinant $D_y$ replace the $y−$ coefficients with the constants.
$D_y=\begin{vmatrix}
1& 4 &2 \\
0& 5 &-1 \\
2 &13 &0
\end{vmatrix}=-15$
For determinant $D_z$ replace the $z−$ coefficients with the constants.
$D=\begin{vmatrix}
1& 0 &4 \\
0& 2 &5 \\
2 &3 &13
\end{vmatrix}=-5$
By using Cramer's rule we have.
$x=\frac{D_x}{D}=\frac{-10}{-5}=2$
and
$y=\frac{D_y}{D}=\frac{-15}{-5}=3$
and
$x=\frac{D_z}{D}=\frac{-5}{-5}=1$
Hence, the solution set is $\{(x,y,z)\} =\{(2,3,1)\}$.
