Answer
$x=-2,y=3,z=5$
or $(-2,3,5)$
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 $A^{-1}=\left[\begin{array}{rrr}
{3}&{-3}&{1}\\
{-2}&{2}&{-1}\\
{-4}&{5}&{-2}\end{array}\right]$
With $B=\left[\begin{array}{l}
0\\
-1\\
-5
\end{array}\right]$, the solution, $X=A^{-1}B$ equals
$X=\left[\begin{array}{l}
0+3-5\\
0-2+5\\
0-5+10
\end{array}\right]=\left[\begin{array}{l}
-2\\
3\\
5
\end{array}\right]$
$x=-2,y=3,z=5$
or $(-2,3,5)$
