Answer
$x=3$
$y=-1$
$z=2$
Work Step by Step
First we have to compute the determinants $D,D_x,D_y,D_z$:
$D=\begin{vmatrix}1&1&1\\1&-2&1\\1&3&2\end{vmatrix}=1(-4-3)-1(2-1)+1(3+2)=-3$
$D_x=\begin{vmatrix}4&1&1\\7&-2&1\\4&3&2\end{vmatrix}=4(-4-3)-1(14-4)+1(21+8)=-9$
$D_y=\begin{vmatrix}1&4&1\\1&7&1\\1&4&2\end{vmatrix}=1(14-4)-4(2-1)+1(4-7)=3$
$D_z=\begin{vmatrix}1&1&4\\1&-2&7\\1&3&4\end{vmatrix}=1(-8-21)-1(4-7)+4(3+2)=-6$
We use Cramer's Rule to determine the solutions of the system:
$x=\dfrac{D_x}{D}=\dfrac{-9}{-3}=3$
$y=\dfrac{D_y}{D}=\dfrac{3}{-3}=-1$
$z=\dfrac{D_z}{D}=\dfrac{-6}{-3}=2$
The solution is:
$x=3$
$y=-1$
$z=2$
