Answer
$(x,y,z) =\left(-3,\frac{1}{2},1\right)$
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
x& +2y &-z&=&-3\\
2x& -4y & +z&=&-7\\
-2x& +2y &-3z &=&4
\end{matrix}\right.$
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.
$D_x=\begin{vmatrix}
-3& 2 &-1 \\
-7& -4 &1 \\
4&2 &-3
\end{vmatrix}=-66$
For determinant $D_y$ replace the $y−$ coefficients with the constants.
$D_y=\begin{vmatrix}
1& -3 &-1 \\
2& -7 &1 \\
-2&4&-3
\end{vmatrix}=11$
For determinant $D_z$ replace the $z−$ coefficients with the constants.
$D_z=\begin{vmatrix}
1& 2 &-3 \\
2& -4 &-7 \\
-2&2 &4
\end{vmatrix}=22$
By using Cramer's rule we have.
$x=\dfrac{D_x}{D}=\dfrac{-66}{22}=-3$
and
$y=\dfrac{D_y}{D}=\dfrac{11}{22}=\dfrac{1}{2}$
and
$z=\dfrac{D_z}{D}=\dfrac{22}{22}=1$
Hence, the solution set is $(x,y,z) =\left(-3,\frac{1}{2},1\right)$.
