College Algebra, Chapter 7, Review Exercises, Section Review Exercises, Problem 8

Find the complete solution of the system
$
\left\{\begin{equation}
\begin{aligned}

x - y + z =& 2
\\
x + y + 3z =& 6
\\
2y + 3z =& 5

\end{aligned}
\end{equation} \right.
$
using Gaussian Elimination.

For this system we have

$\displaystyle \left[ \begin{array}{cccc}
1 & -1 & 1 & 2 \\
1 & 1 & 3 & 6 \\
0 & 2 & 3 & 5
\end{array} \right]$

$R_2 - R_1 \to R_2$

$\displaystyle \left[ \begin{array}{cccc}
1 & -1 & 1 & 2 \\
0 & 2 & 2 & 4 \\
0 & 2 & 3 & 5
\end{array} \right]$

$\displaystyle \frac{1}{2} R_2$

$\displaystyle \left[ \begin{array}{cccc}
1 & -1 & 1 & 2 \\
0 & 1 & 1 & 2 \\
0 & 2 & 3 & 5
\end{array} \right]$

$R_3 - 2R_2 \to R_3$

$\displaystyle \left[ \begin{array}{cccc}
1 & -1 & 1 & 2 \\
0 & 1 & 1 & 2 \\
0 & 0 & 1 & 1
\end{array} \right]$

Now we have equivalent matrix in row-echelon form and the corresponding system is


$
\left\{
\begin{equation}
\begin{aligned}

x - y + z =& 2
\\
y + z =& 2
\\
z =& 1

\end{aligned}
\end{equation}
\right.
$


Then we back-substitute $z = 1$ into the second equation and solve for $y$


$
\begin{equation}
\begin{aligned}

y + 1 =& 2
&& \text{Back-substitute } z= 1
\\
y =& 1
&& \text{Subtract } 1

\end{aligned}
\end{equation}
$


Now we back-substitute $y=1$ and $z=1$ into the first equation and solve for $x$


$
\begin{equation}
\begin{aligned}

x -1 + 1 =& 2
&& \text{Back-substitute } y = 1 \text{ and } z = 1
\\
x =& 2
&& \text{Simplify}

\end{aligned}
\end{equation}
$