Answer
$\left( -2,3,4 \right)$
Work Step by Step
According to Cramer’s rule
$x=\frac{{{D}_{x}}}{D}$, $y=\frac{{{D}_{y}}}{D}$, $z=\frac{{{D}_{z}}}{D}$.
$D$ is the determinant and all three denominators consist of x, y and z co-efficients.
$D=\left| \begin{matrix}
1 & -1 & 2 \\
2 & 3 & 1 \\
-1 & -1 & 3 \\
\end{matrix} \right|$
${{D}_{x}}$, is the determinant in the numerator for x obtained by replacing the x- co-efficient in D with the constants on the right sides of the equations.
${{D}_{x}}=\left| \begin{matrix}
3 & -1 & 2 \\
9 & 3 & 1 \\
11 & -1 & 3 \\
\end{matrix} \right|$
${{D}_{y}}$, is the determinant in the numerator for y obtained by replacing the y- co-efficient in D with the constants on the right sides of the equations.
${{D}_{y}}=\left| \begin{matrix}
1 & 3 & 2 \\
2 & 9 & 1 \\
-1 & 11 & 3 \\
\end{matrix} \right|$
${{D}_{z}}$, is the determinant in the numerator for z obtained by replacing the z- co-efficient in D with the constants on the right sides of the equations.
${{D}_{z}}=$ $\left| \begin{matrix}
1 & -1 & 3 \\
2 & 3 & 9 \\
-1 & -1 & 11 \\
\end{matrix} \right|$.
Calculate the four determinants.
$\begin{align}
& D=\left| \begin{matrix}
1 & -1 & 2 \\
2 & 3 & 1 \\
-1 & -1 & 3 \\
\end{matrix} \right| \\
& =1\left( 9-\left( -1 \right) \right)+1\left( 6-\left( -1 \right) \right)+2\left( -2-\left( -3 \right) \right) \\
& =19
\end{align}$
$\begin{align}
& {{D}_{x}}=\left| \begin{matrix}
3 & -1 & 2 \\
9 & 3 & 1 \\
11 & -1 & 3 \\
\end{matrix} \right| \\
& =3\left( 9+1 \right)+1\left( 27-11 \right)+2\left( -9-33 \right) \\
& =-38
\end{align}$
$\begin{align}
& {{D}_{y}}=\left| \begin{matrix}
1 & 3 & 2 \\
2 & 9 & 1 \\
-1 & 11 & 3 \\
\end{matrix} \right| \\
& =1\left( 27-11 \right)-3\left( 6+1 \right)+2\left( 22+9 \right) \\
& =57
\end{align}$
$\begin{align}
& {{D}_{z}}=\left| \begin{matrix}
1 & -1 & 3 \\
2 & 3 & 9 \\
-1 & -1 & 11 \\
\end{matrix} \right| \\
& =1\left( 33+9 \right)+1\left( 22+9 \right)+3\left( -2+3 \right) \\
& =76
\end{align}$
Substitute the given values
$\begin{align}
& x=\frac{{{D}_{x}}}{D} \\
& =\frac{-38}{19} \\
& =-2
\end{align}$
$\begin{align}
& y=\frac{{{D}_{y}}}{D} \\
& =\frac{57}{19} \\
& =3
\end{align}$
$\begin{align}
& z=\frac{{{D}_{z}}}{D} \\
& =\frac{76}{19} \\
& =4
\end{align}$
Therefore, $\left( x,y,z \right)=\left( -2,3,4 \right)$
