College Algebra, Chapter 7, Review Exercises, Section Review Exercises, Problem 10

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

x-y+z =& 2
\\
x+y+3z =& 6
\\
3x-y+5z =& 10

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

For this system we have

$\displaystyle \left[ \begin{array}{cccc}
1 & -1 & 1 & 2 \\
1 & 1 & 3 & 6 \\
3 & -1 & 5 & 10
\end{array} \right]$

$R_2 - R_1 \to R_2$

$\displaystyle \left[ \begin{array}{cccc}
1 & -1 & 1 & 2 \\
0 & 2 & 2 & 4 \\
3 & -1 & 5 & 10
\end{array} \right]$

$\displaystyle R_3 - 3 R_1 \to R_3$

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

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

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

$R_3 - 2 R_2 \to R_3$

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


The corresponding system is


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

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

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


Now we solve for the leading variables $x$ and $y$ in terms of the non leading variable $z$.


$
\begin{equation}
\begin{aligned}

x =& y - z + 2 = 2 - z - z + 2 = 4 - 2z
\\
y =& 2 - z

\end{aligned}
\end{equation}
$


To obtain the complete solution; we let $t$ represent any real number and we express $x, y$ and $z$ in terms of $t$:


$
\begin{equation}
\begin{aligned}

x =& 4 - 2t
\\
y=& 2-t
\\
z =& t

\end{aligned}
\end{equation}
$