Editing Gaussian elimination (section)

=== Example of the algorithm ===
Suppose the goal is to find and describe the set of solutions to the following system of linear equations:
<math display="block">
\begin{alignat}{4}
  2x &{}+{}& y &{}-{}&  z &{}={}&   8 & \qquad (L_1) \\
 -3x &{}-{}& y &{}+{}& 2z &{}={}& -11 & \qquad (L_2) \\
 -2x &{}+{}& y &{}+{}& 2z &{}={}&  -3 & \qquad (L_3)
\end{alignat}
</math>

The table below is the row reduction process applied simultaneously to the system of equations and its associated [[augmented matrix]]. In practice, one does not usually deal with the systems in terms of equations, but instead makes use of the augmented matrix, which is more suitable for computer manipulations. The row reduction procedure may be summarized as follows: eliminate {{mvar|x}} from all equations below {{math|''L''<sub>1</sub>}}, and then eliminate {{mvar|y}} from all equations below {{math|''L''<sub>2</sub>}}. This will put the system into [[triangular form]]. Then, using back-substitution, each unknown can be solved for.

: {| class="wikitable"
|-
! System of equations !! Row operations !! Augmented matrix
|- align="center"
| <math>
\begin{alignat}{4}
  2x &{}+{}& y &{}-{}&  z &{}={}&   8 & \\
 -3x &{}-{}& y &{}+{}& 2z &{}={}& -11 & \\
 -2x &{}+{}& y &{}+{}& 2z &{}={}&  -3 &
\end{alignat}
</math>
|
| <math>
\left[\begin{array}{rrr|r}
  2 &  1 & -1 &   8 \\
 -3 & -1 &  2 & -11 \\
 -2 &  1 &  2 &  -3
\end{array}\right]
</math>
|- align="center"
| <math>
\begin{alignat}{4}
 2x &{}+{}&          y &{}-{}&          z &{}={}& 8 & \\
    &     & \tfrac12 y &{}+{}& \tfrac12 z &{}={}& 1 & \\
    &     &         2y &{}+{}&          z &{}={}& 5 &
\end{alignat}
</math>
| <math>
\begin{align}
 L_2 + \tfrac32 L_1 &\to L_2 \\
 L_3 +          L_1 &\to L_3
\end{align}
</math>
| <math>
\left[\begin{array}{rrr|r}
 2 &      1  &     -1  & 8 \\
 0 & \frac12 & \frac12 & 1 \\
 0 &      2  &      1  & 5
\end{array}\right]
</math>
|- align="center"
| <math>
\begin{alignat}{4}
 2x &{}+{}&          y &{}-{}&          z &{}={}& 8 & \\
    &     & \tfrac12 y &{}+{}& \tfrac12 z &{}={}& 1 & \\
    &     &            &     &         -z &{}={}& 1 &
\end{alignat}
</math>
| <math>
 L_3 + -4 L_2 \to L_3</math>
| <math>
\left[\begin{array}{rrr|r}
 2 &      1  &     -1  & 8 \\
 0 & \frac12 & \frac12 & 1 \\
 0 &      0  &     -1  & 1
\end{array}\right]
</math>
|-
|colspan=3 align="center"| The matrix is now in echelon form (also called triangular form)
|- align="center"
| <math>
\begin{alignat}{4}
 2x &{}+{}&          y &     &   &{}={}       7  & \\
    &     & \tfrac12 y &     &   &{}={} \tfrac32 & \\
    &     &            &{}-{}& z &{}={}       1  &
\end{alignat}
</math>
| <math>
\begin{align}
 L_1 -          L_3 &\to L_1\\
 L_2 + \tfrac12 L_3 &\to L_2 
 
\end{align}
</math>
| <math>
\left[\begin{array}{rrr|r}
 2 &      1  &  0 &      7  \\
 0 & \frac12 &  0 & \frac32 \\
 0 &      0  & -1 &      1
\end{array}\right]
</math>
|- align="center"
| <math>
\begin{alignat}{4}
 2x &{}+{}& y &\quad&   &{}={}&  7 & \\
    &     & y &\quad&   &{}={}&  3 & \\
    &     &   &\quad& z &{}={}& -1 &
\end{alignat}
</math>
| <math>
\begin{align}
 2 L_2 &\to L_2 \\
  -L_3 &\to L_3
\end{align}
</math>
| <math>
\left[\begin{array}{rrr|r}
 2 & 1 & 0 &  7 \\
 0 & 1 & 0 &  3 \\
 0 & 0 & 1 & -1
\end{array}\right]
</math>
|- align="center"
| <math>
\begin{alignat}{4}
 x &\quad&   &\quad&   &{}={}&  2 & \\
   &\quad& y &\quad&   &{}={}&  3 & \\
   &\quad&   &\quad& z &{}={}& -1 &
\end{alignat}
</math>
| <math>
\begin{align}
          L_1 - L_2 &\to L_1 \\
 \tfrac12 L_1       &\to L_1
\end{align}
</math>
| <math>
\left[\begin{array}{rrr|r}
 1 & 0 & 0 &  2 \\
 0 & 1 & 0 &  3 \\
 0 & 0 & 1 & -1
\end{array}\right]
</math>
|}

The second column describes which row operations have just been performed. So for the first step, the {{mvar|x}} is eliminated from {{math|''L''<sub>2</sub>}} by adding {{math|{{sfrac|3|2}}''L''<sub>1</sub>}} to {{math|''L''<sub>2</sub>}}. Next, {{mvar|x}} is eliminated from {{math|''L''<sub>3</sub>}} by adding {{math|''L''<sub>1</sub>}} to {{math|''L''<sub>3</sub>}}. These row operations are labelled in the table as
<math display="block">\begin{align}
 L_2 + \tfrac32 L_1 &\to L_2, \\
 L_3 +          L_1 &\to L_3.
\end{align}</math>

Once {{mvar|y}} is also eliminated from the third row, the result is a system of linear equations in triangular form, and so the first part of the algorithm is complete. From a computational point of view, it is faster to solve the variables in reverse order, a process known as back-substitution. One sees the solution is {{math|1=''z'' = −1}}, {{math|1=''y'' = 3}}, and {{math|1=''x'' = 2}}. So there is a unique solution to the original system of equations.

Instead of stopping once the matrix is in echelon form, one could continue until the matrix is in ''reduced'' row echelon form, as it is done in the table. The process of row reducing until the matrix is reduced is sometimes referred to as Gauss–Jordan elimination, to distinguish it from stopping after reaching echelon form.