Answer
$(1,1,2)$
Work Step by Step
Start with the augmented matrix, and row reduce to
reduced row echelon form (Gauss-Jordan.)
$\left(\begin{array}{ccccc}
{1}&{-2}&{1}&{1}\\
{0}&{1}&{2}&{5}\\
{1}&{1}&{3}&{8}\end{array}\right)\rightarrow\left(\begin{array}{l}
.\\
.\\
R_{3}\leftarrow R_{3}-R_{1}
\end{array}\right)$
$\rightarrow\left(\begin{array}{cccc}
{1}&{-2}&{1}&{1}\\
{0}&{1}&{2}&{5}\\
{0}&{3}&{2}&{7}\end{array}\right)\rightarrow\left(\begin{array}{l}
R_{1}\leftarrow R_{1}+2R_{2}.\\
.\\
R_{3}\leftarrow R_{3}-3R_{2}.
\end{array}\right)$
$\rightarrow\left(\begin{array}{cccc}
{1}&{0}&{5}&{11}\\
{0}&{1}&{2} &{5}\\
{0}&{0}&{-4} &{-8}\end{array}\right)\rightarrow\left(\begin{array}{l}
.\\
.\\
R_{3}\leftarrow-\frac{1}{4}R_{3}
\end{array}\right)$
$\rightarrow\left(\begin{array}{cccc}
{1}&{0}&{5}&{11}\\
{0}&{1}&{2} &{5}\\
{0}&{0}&{1} &{2}\end{array}\right)\rightarrow\left(\begin{array}{l}
R_{1}\leftarrow R_{1}-5R_{3}.\\
R_{2}\leftarrow R_{2}-2R_{3}.\\
.
\end{array}\right)$
$\rightarrow\left(\begin{array}{cccc}
{1}&{0}&{0}&{1}\\
{0}&{1}&{0} &{1}\\
{0}&{0}&{1} &{2}\end{array}\right)$
The solution is $(1,1,2)$
