Answer
$\{(-1,2,-3)\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
x+2y-z=6\\
2x-y+3z=-13\\
3x-2y+3z=-16
\end{cases}$
Compute $D,D_x,D_y,D_z$:
$D=\begin{vmatrix}1&2&-1\\2&-1&3\\3&-2&3\end{vmatrix}=1(-3+6)-2(6-9)+(-1)(-4+3)=10$
$D_x=\begin{vmatrix}6&2&-1\\-13&-1&3\\-16&-2&3\end{vmatrix}=6(-3+6)-2(-39+48)+(-1)(26-16)=-10$
$D_y=\begin{vmatrix}1&6&-1\\2&-13&3\\3&-16&3\end{vmatrix}=1(-39+48)-6(6-9)+(-1)(-32+39)=20$
$D_z=\begin{vmatrix}1&2&6\\2&-1&-13\\3&-2&-16\end{vmatrix}=1(16-26)-2(-32+39)+6(-4+3)=-30$
Determine $x$:
$x=\dfrac{D_x}{D}=\dfrac{-10}{10}=-1$
Determine $y$:
$y=\dfrac{D_y}{D}=\dfrac{20}{10}=2$
Determine $z$:
$z=\dfrac{D_z}{D}=\dfrac{-30}{10}=-3$
The solution set is:
$\{(-1,2,-3)\}$
