Answer
$(x,y,z) =(1,3,-2)$
Work Step by Step
The given system of equations is
$\left\{\begin{matrix}
x& +y &-z&=&6\\
3x& -2y & +z&=&-5\\
x& +3y &-2z &=&14
\end{matrix}\right.$
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.
$D=\begin{vmatrix}
1& 1 &-1 \\
3& -2 &1 \\
1 &3 &-2
\end{vmatrix}=-3$
For determinant $D_x$ replace the $x−$ coefficients with the constants.
$D_x=\begin{vmatrix}
6& 1 &-1 \\
-5& -2 &1 \\
14 &3 &-2
\end{vmatrix}=-3$
For determinant $D_y$ replace the $y−$ coefficients with the constants.
$D_y=\begin{vmatrix}
1& 6 &-1 \\
3& -5 &1 \\
1 &14 &-2
\end{vmatrix}=-9$
For determinant $D_z$ replace the $z−$ coefficients with the constants.
$D_z=\begin{vmatrix}
1& 1 &6 \\
3& -2 &-5 \\
1 &3 &14
\end{vmatrix}=6$
By using Cramer's rule we have.
$x=\dfrac{D_x}{D}=\dfrac{-3}{-3}=1$
and
$y=\dfrac{D_y}{D}=\dfrac{-9}{-3}=3$
and
$z=\dfrac{D_z}{D}=\dfrac{6}{-3}=-2$
Hence, the solution set is $(x,y,z) =(1,3,-2)$.
