Answer
$x=-2$
$y=3$
$z=4$
Work Step by Step
First we have to compute the determinants $D,D_x,D_y,D_z$:
$D=\begin{vmatrix}1&-1&2\\2&3&1\\-1&-1&3\end{vmatrix}=1(9+1)-(-1)(6+1)+2(-2+3)=19$
$D_x=\begin{vmatrix}3&-1&2\\9&3&1\\11&-1&3\end{vmatrix}=3(9+1)-(-1)(27-11)+2(-9-33)=-38$
$D_y=\begin{vmatrix}1&3&2\\2&9&1\\-1&11&3\end{vmatrix}=1(27-11)-3(6+1)+2(22+9)=57$
$D_z=\begin{vmatrix}1&-1&3\\2&3&9\\-1&-1&11\end{vmatrix}=1(33+9)-(-1)(22+9)+3(-2+3)=76$
We use Cramer's Rule to determine the solutions of the system:
$x=\dfrac{D_x}{D}=\dfrac{-38}{19}=-2$
$y=\dfrac{D_y}{D}=\dfrac{57}{19}=3$
$z=\dfrac{D_z}{D}=\dfrac{76}{19}=4$
The solution is:
$x=-2$
$y=3$
$z=4$
