By the equivalence between systems of linear equations and augmented matrices, the three operations in the gaussian elimination are equivalent to three row operations.

`r`[row`i`] + [row`j`]- add a multiple of one row to another.
[ [row `i`]] ⇒ [ [row `i`]] : : [row `j`]`r`[row`i`] + [row`j`] - [row
`i`] ↔ [row`j`] - exchange rows.
[ [row `i`]] ⇒ [ [row `j`]] : : [row `j`][row `i`] `d`[row`i`],`d`≠ 0- multiply a nonzero number to a row.
[ : ] ⇒ [ : ] [row `i`]`d`[row`i`]: :

Example The operation (-3)[equation 2] + [equation 1] on the system

3x_{1} |
+ x_{2} |
- x_{3} |
= | 2 | ⇒ | 4x_{2} |
- 4x_{3} |
= | -4 | |

x_{1} |
- x_{2} |
+ x_{3} |
= | 2 | x_{1} |
- x_{2} |
+ x_{3} |
= | 2 | |

2x_{1} |
+ 2x_{2} |
+ x_{3} |
= | 6 | 2x_{1} |
+ 2x_{2} |
+ x_{3} |
= | 6 |

is equivalent to the operation (-3)[row 2] + [row 1] on the augmented matrix.

[ | 3 | 1 | -1 | 2 | ] ⇒ [ | 0 | 4 | -4 | -4 | ] |

1 | -1 | 1 | 2 | 1 | -1 | 1 | 2 | |||

2 | 2 | 1 | 6 | 2 | 2 | 1 | 6 |

The operation [equation 2] ↔ [equation 3] on the system

4x_{2} |
- 4x_{3} |
= | -4 | ⇒ | x_{1} |
- x_{2} |
+ x_{3} |
= | 2 | |

x_{1} |
- x_{2} |
+ x_{3} |
= | 2 | 4x_{2} |
- 4x_{3} |
= | -4 | ||

3x_{3} |
= | 6 | 3x_{3} |
= | 6 |

is equivalent to the operation [row 2] ↔ [row 3] on the augmented matrix.

[ | 0 | 4 | -4 | -4 | ] ⇒ [ | 1 | -1 | 1 | 2 | ] |

1 | -1 | 1 | 2 | 0 | 4 | -4 | -4 | |||

0 | 0 | 3 | 6 | 0 | 0 | 3 | 6 |

The operations (1/4)[equation 2] and (1/3)[equation 3] on the system

x_{1} |
- x_{2} |
+ x_{3} |
= | 2 | ⇒ | x_{1} |
- x_{2} |
+ x_{3} |
= | 2 |

4x_{2} |
- 4x_{3} |
= | -4 | x_{2} |
- x_{3} |
= | -1 | |||

3x_{3} |
= | 6 | x_{3} |
= | 2 |

is equivalent to the operation (1/4)[row 2] and (1/3)[row 3] on the augmented matrix.

[ | 1 | -1 | 1 | 2 | ] ⇒ [ | 1 | -1 | 1 | 2 | ] |

0 | 4 | -4 | -4 | 0 | 1 | -1 | -1 | |||

0 | 0 | 3 | 6 | 0 | 0 | 1 | 2 |

The goal of gaussian elimination is to eliminate as many variables as possible. Correspondingly, the goal of row operation is to create as many zeros as possible.

Example Consider the system (compare with this example).

3x_{1} |
+ x_{2} |
- x_{3} |
= | 2 |

x_{1} |
- x_{2} |
+ x_{3} |
= | 2 |

2x_{1} |
- x_{2} |
+ x_{3} |
= | 6 |

We simplify the system by doing row operations on the augmented matrix.

[ | 3 | 1 | -1 | 2 | ] |

1 | -1 | 1 | 2 | ||

2 | -1 | 1 | 6 |

Simplifying `x`_{1} means creating two zeros in the [col 1] (1st column).
By (-3)[row 2] + [row 1] and (-2)[row 2] + [row 3], we get

[ | 0 | 4 | -4 | -4 | ] |

1 | -1 | 1 | 2 | ||

0 | 1 | -1 | 2 |

By [row 1] ↔ [row 2] and [row 2] ↔ [row 3], we get

[ | 1 | -1 | 1 | 2 | ] |

0 | 1 | -1 | 2 | ||

0 | 4 | -4 | -4 |

Then (-4)[row 2] + [row 3] gives us

[ | 1 | -1 | 1 | 2 | ] |

0 | 1 | -1 | 2 | ||

0 | 0 | 0 | -12 |

This is the simplest shape we can get by row operations. The corresponding simple system is

x_{1} |
- x_{2} |
+ x_{3} |
= | 2 |

x_{2} |
- x_{3} |
= | 2 | |

0 | = | -12 |

Since the 3rd equation 0 = -12 is a contradiction, the original system has no solution.

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