College Algebra, Chapter 7, Review Exercises, Section Review Exercises, Problem 14

Find the complete solution of the system
$
\left\{
\begin{array}{ccccc}
x & -y & & = & 1 \\
x & +y & +2z & = & 3 \\
x & -3y & -2z & = & -1
\end{array}
\right.
$
using Gauss-Jordan Elimination.

We transform the system into reduced row-echelon form

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

$R_2 - R_1 \to R_2$

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

$R_3 - R_1 \to R_3$

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

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

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

$R_3 + 2 R_2 \to R_3$

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

$R_1 + R_2 \to R_1$

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

This is in reduced row echelon form. Since the last row represents the equation $0=0$, we may discard it. So the last matrix corresponds to the system


$
\left\{
\begin{array}{ccccc}
x & & +z & = & 2 \\
& y & +z & = & 1
\end{array}
\right.
$


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


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$