Answer
$\{(2,3,5)\}$.
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
x& -3y &+z&=&-2 \\
x& +2y & +0z&=&8\\
2x& -y &+0z &=&1
\end{matrix}\right.$
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.
$D=\begin{vmatrix}
1& -3 &1 \\
1& 2 &0 \\
2 &-1 &0
\end{vmatrix}=-5$
For determinant $D_x$ replace the $x−$ coefficients with the constants.
$D_x=\begin{vmatrix}
-2& -3 &1 \\
8& 2 &0 \\
1 &-1 &0
\end{vmatrix}=-10$
For determinant $D_y$ replace the $y−$ coefficients with the constants.
$D_y=\begin{vmatrix}
1& -2 &1 \\
1& 8 &0 \\
2 &1 &0
\end{vmatrix}=-15$
For determinant $D_z$ replace the $z−$ coefficients with the constants.
$D_z=\begin{vmatrix}
1& -3 &-2 \\
1& 2 &8 \\
2 &-1 &1
\end{vmatrix}=-25$
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
$x=\frac{D_z}{D}=\frac{-25}{-5}=5$
Hence, the solution set is $\{(x,y,z)\} =\{(2,3,5)\}$.
