College Algebra, Chapter 7, Review Exercises, Section Review Exercises, Problem 16

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

x-y =& 3
\\
2x + y =& 6
\\
x - 2y =& 9

\end{aligned}
\end{equation}
\right.
$
using Gauss-Jordan Elimination.

We transform the system into reduced row-echelon form

$\displaystyle \left[
\begin{array}{ccc}
1 & -1 & 3 \\
2 & 1 & 6 \\
1 & -2 & 9
\end{array}
\right]$

$R_2 - 2 R_1 \to R_2$

$\displaystyle \left[
\begin{array}{ccc}
1 & -1 & 3 \\
0 & 3 & 0 \\
1 & -2 & 9
\end{array}
\right]$

$R_3 - R_1 \to R_3$

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

$\displaystyle \frac{1}{3} R_2$

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

$R_3 + R_2 \to R_3$

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

$\displaystyle \frac{1}{6} R_3$

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

$R_1 - 3 R_3 \to R_1$

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

$R_1 + R_2 \to R_1$

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

This is in reduced row echelon form. If we translate the last row back into equation, we get $0x + 0y = 1$, or $0 = 1$, which is false. This that the system has no solution or it is inconsistent.