Answer
$\{(-2,3,4)\}$.
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
x& -y &+2z&=&3 \\
2 x& +3y & +z&=&9\\
-x& -y &+3z &=&11
\end{matrix}\right.$
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.
$D=\begin{vmatrix}
1& -1 &2 \\
2& 3 &1 \\
-1 &-1 &3
\end{vmatrix}=19$
For determinant $D_x$ replace the $x−$ coefficients with the constants.
$D_x=\begin{vmatrix}
3& -1 &2 \\
9& 3 &1 \\
11 &-1 &3
\end{vmatrix}=-38$
For determinant $D_y$ replace the $y−$ coefficients with the constants.
$D_y=\begin{vmatrix}
1& 3 &2 \\
2& 9 &1 \\
-1 &11 &3
\end{vmatrix}=57$
For determinant $D_z$ replace the $z−$ coefficients with the constants.
$D_z=\begin{vmatrix}
1& -1 &3 \\
2& 3 &9 \\
-1 &-1 &11
\end{vmatrix}=76$
By using Cramer's rule we have.
$x=\frac{D_x}{D}=\frac{-38}{19}=-2$
and
$y=\frac{D_y}{D}=\frac{57}{19}=3$
and
$x=\frac{D_z}{D}=\frac{76}{19}=4$
Hence, the solution set is $\{(x,y,z)\} =\{(-2,3,4)\}$.
