Answer
$x=2$
$y=3$
$z=5$
Work Step by Step
First we have to compute the determinants $D,D_x,D_y,D_z$:
$D=\begin{vmatrix}1&-3&1\\1&2&0\\2&-1&0\end{vmatrix}=1(0+0)-(-3)(0-0)+1(-1-4)=-5$
$D_x=\begin{vmatrix}-2&-3&1\\8&2&0\\1&-1&0\end{vmatrix}=(-2)(0+0)-(-3)(0-0)+1(-8-2)=-10$
$D_y=\begin{vmatrix}1&-2&1\\1&8&0\\2&1&0\end{vmatrix}=1(0-0)-(-2)(0-0)+1(1-16)=-15$
$D_z=\begin{vmatrix}1&-3&-2\\1&2&8\\2&-1&1\end{vmatrix}=1(2+8)-(-3)(1-16)+(-2)(-1-4)=-25$
We use Cramer's Rule to determine the solutions of the system:
$x=\dfrac{D_x}{D}=\dfrac{-10}{-5}=2$
$y=\dfrac{D_y}{D}=\dfrac{-15}{-5}=3$
$z=\dfrac{D_z}{D}=\dfrac{-25}{-5}=5$
The solution is:
$x=2$
$y=3$
$z=5$
