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