Answer
$\left( -5,-2,7 \right)$
Work Step by Step
According to Cramer’s rule
$x=\frac{{{D}_{x}}}{D}$, $y=\frac{{{D}_{y}}}{D}$, $z=\frac{{{D}_{z}}}{D}$.
As
$\begin{align}
& D=\left| \begin{matrix}
1 & 1 & 1 \\
2 & -1 & 1 \\
-1 & 3 & -1 \\
\end{matrix} \right| \\
& {{D}_{x}}=\left| \begin{matrix}
0 & 1 & 1 \\
-1 & -1 & 1 \\
-8 & 3 & -1 \\
\end{matrix} \right| \\
& {{D}_{y}}=\left| \begin{matrix}
1 & 0 & 1 \\
2 & -1 & 1 \\
-1 & -8 & -1 \\
\end{matrix} \right| \\
& {{D}_{z}}=\left| \begin{matrix}
1 & 1 & 0 \\
2 & -1 & -1 \\
-1 & 3 & -8 \\
\end{matrix} \right|
\end{align}$
Calculate the four determinants.
$\begin{align}
& D=\left| \begin{matrix}
1 & 1 & 1 \\
2 & -1 & 1 \\
-1 & 3 & -1 \\
\end{matrix} \right| \\
& =1\left( 1-3 \right)-1\left( -2-\left( -1 \right) \right)+1\left( 6-1 \right) \\
& =4
\end{align}$
$\begin{align}
& {{D}_{x}}=\left| \begin{matrix}
0 & 1 & 1 \\
-1 & -1 & 1 \\
-8 & 3 & -1 \\
\end{matrix} \right| \\
& =0\left( 1-3 \right)+1\left( -1-\left( 3 \right) \right)-8\left( 1-\left( -1 \right) \right) \\
& =-20
\end{align}$
$\begin{align}
& {{D}_{y}}=\left| \begin{matrix}
1 & 0 & 1 \\
2 & -1 & 1 \\
-1 & -8 & -1 \\
\end{matrix} \right| \\
& =1\left( 1-\left( -8 \right) \right)+1\left( -16-1 \right) \\
& =-8
\end{align}$
$\begin{align}
& {{D}_{z}}=\left| \begin{matrix}
1 & 1 & 0 \\
2 & -1 & -1 \\
-1 & 3 & -8 \\
\end{matrix} \right| \\
& =1\left( 8-\left( -3 \right) \right)-2\left( -8 \right) \\
& =28
\end{align}$
Substitute the given values
$\begin{align}
& x=\frac{{{D}_{x}}}{D} \\
& =\frac{-20}{4} \\
& =-5
\end{align}$
$\begin{align}
& y=\frac{{{D}_{y}}}{D} \\
& =\frac{-8}{4} \\
& =-2
\end{align}$
$\begin{align}
& z=\frac{{{D}_{z}}}{D} \\
& =\frac{28}{4} \\
& =7
\end{align}$
Therefore, $\left( x,y,z \right)=\left( -5,-2,7 \right)$
