Answer
$x=1/2, \ y=-1/2, \ z=1$
or $(1/2,-1/2,1)$
Work Step by Step
Writing the system in matrix form, $AX=B,$ the solution is $X=A^{-1}B$.
Here, $A=\left[\begin{array}{rrr}{1}&{-1}&{1}\\{0}&{-2}&{1}\\{-2}&{-3}&{0}\end{array}\right]$, as in exercise 37.
There, we found $A^{-1}=A^{-1}=\left[\begin{array}{rrr}
{3}&{-3}&{1}\\
{-2}&{2}&{-1}\\
{-4}&{5}&{-2}\end{array}\right]$
With $B=\left[\begin{array}{l}
2\\
2\\
1/2
\end{array}\right]$, the solution, $X=A^{-1}B$ equals
$X=\left[\begin{array}{l}
6-6+1/2\\
-4+4-1/2\\
-8+10-1
\end{array}\right]=\left[\begin{array}{l}
1/2\\
-1/2\\
1
\end{array}\right]$
$x=1/2, \ y=-1/2, \ z=1$
or $(1/2,-1/2,1)$
