Answer
$\{(0,4,2)\}$.
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
3x& +0y &+2z&=&4 \\
5x& -y & +0z&=&-4\\
0x& +4y &+3z &=&22
\end{matrix}\right.$
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.
$D=\begin{vmatrix}
3& 0 &2 \\
5& -1 &0 \\
0 &4 &3
\end{vmatrix}=31$
For determinant $D_x$ replace the $x−$ coefficients with the constants.
$D_x=\begin{vmatrix}
4& 0 &2 \\
-4& -1 &0 \\
22 &4 &3
\end{vmatrix}=0$
For determinant $D_y$ replace the $y−$ coefficients with the constants.
$D_y=\begin{vmatrix}
3& 4 &2 \\
5& -4 &0 \\
0 &22 &3
\end{vmatrix}=124$
For determinant $D_z$ replace the $z−$ coefficients with the constants.
$D_z=\begin{vmatrix}
3& 0 &4 \\
5& -1 &-4 \\
0 &4 &22
\end{vmatrix}=62$
By using Cramer's rule we have.
$x=\frac{D_x}{D}=\frac{0}{31}=0$
and
$y=\frac{D_y}{D}=\frac{124}{31}=4$
and
$x=\frac{D_z}{D}=\frac{62}{31}=2$
Hence, the solution set is $\{(x,y,z)\} =\{(0,4,2)\}$.
