Answer
$\{(3,-1,2)\}$.
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
x& +y &+z&=&4 \\
x& -2y & +z&=&7\\
x& +3y &+2z &=&4
\end{matrix}\right.$
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.
$D=\begin{vmatrix}
1& 1 &1 \\
1& -2 &1 \\
1 &3 &2
\end{vmatrix}=-3$
For determinant $D_x$ replace the $x−$ coefficients with the constants.
$D_x=\begin{vmatrix}
4& 1 &1 \\
7& -2 &1 \\
4 &3 &2
\end{vmatrix}=-9$
For determinant $D_y$ replace the $y−$ coefficients with the constants.
$D_y=\begin{vmatrix}
1& 4 &1 \\
1& 7 &1 \\
1 &4 &2
\end{vmatrix}=3$
For determinant $D_z$ replace the $z−$ coefficients with the constants.
$D_z=\begin{vmatrix}
1& 1 &4 \\
1& -2 &7 \\
1 &3 &4
\end{vmatrix}=-6$
By using Cramer's rule we have.
$x=\frac{D_x}{D}=\frac{-9}{-3}=3$
and
$y=\frac{D_y}{D}=\frac{3}{-3}=-1$
and
$x=\frac{D_z}{D}=\frac{-6}{-3}=2$
Hence, the solution set is $\{(x,y,z)\} =\{(3,-1,2)\}$.
