Answer
$x=1, \ y=-1, \ z=1$
or $(1, \ -1, \ 1)$
Work Step by Step
Writing the system in matrix form, $AX=B,$ the solution is $X=A^{-1}B$.
Here, $A=\left[\begin{array}{rrr}{3}&{3}&{1}\\{1}&{2}&{1}\\{2}&{-1}&{1}\end{array}\right]$, as in exercise $40$.
There, we found $A^{-1}=\left[\begin{array}{rrr}
{\displaystyle \frac{3}{7}}&{-\displaystyle \frac{4}{7}}&{\displaystyle \frac{1}{7}}\\
{\displaystyle \frac{1}{7}}&{\displaystyle \frac{1}{7}}&{-\displaystyle \frac{2}{7}}\\
{-\displaystyle \frac{5}{7}}&{\displaystyle \frac{9}{7}}&{\displaystyle \frac{3}{7}}\end{array}\right]$
With $B=\left[\begin{array}{l}
1\\
0\\
4
\end{array}\right]$, the solution, $X=A^{-1}B$ equals
$X=\left[\begin{array}{l}
\frac{1}{7}(3+0+4)\\
\frac{1}{7}(1+0-8)\\
\frac{1}{7}(-5+0+12)
\end{array}\right]=\left[\begin{array}{l}
1\\
-1\\
1
\end{array}\right]$
$x=1, \ y=-1, \ z=1$
or $(1, \ -1, \ 1)$
