Answer
$x=2$
$y=3$
$z=1$
Work Step by Step
First we have to compute the determinants $D,D_x,D_y,D_z$:
$D=\begin{vmatrix}1&0&2\\0&2&-1\\2&3&0\end{vmatrix}=1(0+3)-0(0+2)+2(0-4)=-5$
$D_x=\begin{vmatrix}4&0&2\\5&2&-1\\13&3&0\end{vmatrix}=4(0+3)-0(0+13)+2(15-26)=-10$
$D_y=\begin{vmatrix}1&4&2\\0&5&-1\\2&13&0\end{vmatrix}=1(0+13)-4(0+2)+2(0-10)=-15$
$D_z=\begin{vmatrix}1&0&4\\0&2&5\\2&3&13\end{vmatrix}=1(26-15)-0(0-10)+4(0-4)=-5$
We use Cramer's Rule to determine the solutions of the system:
$x=\dfrac{D_x}{D}=\dfrac{-10}{-5}=2$
$y=\dfrac{D_y}{D}=\dfrac{-15}{-5}=3$
$z=\dfrac{D_z}{D}=\dfrac{-5}{-5}=1$
The solution is:
$x=2$
$y=3$
$z=1$
