Answer
$\{(-3,2,1)\}$.
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
2x& +y &&=&-4\\
& y & -2z&=&0\\
3x& &-2z &=&-11
\end{matrix}\right.$
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.
$D=\begin{vmatrix}
2& 1 &0 \\
0& 1 &-2 \\
3 &0 &-2
\end{vmatrix}=-10$
For determinant $D_x$ replace the $x−$ coefficients with the constants.
$D_x=\begin{vmatrix}
-4& 1 &0 \\
0& 1 &-2 \\
-11&0 &-2
\end{vmatrix}=30$
For determinant $D_y$ replace the $y−$ coefficients with the constants.
$D_y=\begin{vmatrix}
2& -4 &0 \\
0& 0 &-2 \\
3 &-11 &-2
\end{vmatrix}=-20$
For determinant $D_z$ replace the $z−$ coefficients with the constants.
$D_z=\begin{vmatrix}
2& 1 &-4 \\
0& 1 &0 \\
3 &0 &-11
\end{vmatrix}=-10$
By using Cramer's rule we have.
$x=\frac{D_x}{D}=\frac{30}{-10}=-3$
and
$y=\frac{D_y}{D}=\frac{-20}{-10}=2$
and
$z=\frac{D_z}{D}=\frac{-10}{-10}=1$
Hence, the solution set is $\{(x,y,z)\} =\{(-3,2,1)\}$.
