Although the row echelon form has the simplest shape, the individual entries can
still be further simplified. The following summarizes the simplest *matrix*
(as opposed to just *shape*) one can get by row operations.

A reduced row echelon form is characterized by two properties

- The matrix is a row echelon form
- The pivots are occupied by 1, and the entries above the pivots are occupied by 0

The following example justifies the definition.

Example Consider a row echelon form (`a` ≠ 0)

[ | * | # | # | c |
# | ] |

0 | * | # | b |
# | ||

0 | 0 | 0 | a |
# |

The operations (-`b`/`a`)[row 3] + [row 2], (-`c`/`a`)[row 3] + [row 1], and (-1/`a`)[row 3] give us

[ | * | # | # | 0 | # | ] |

0 | * | # | 0 | # | ||

0 | 0 | 0 | 1 | # |

Similar operations can further simplify [col 2] to become

[ | 1 | 0 | # | 0 | # | ] |

0 | 1 | # | 0 | # | ||

0 | 0 | 0 | 1 | # |

This matrix not only has the simplest shape, but also has the simplest entries.

Given a matrix, we may first compute a row echelon form. Next we use the nonzero numbers at the pivots to cancel the numbers above the pivots. Then we may divide the numbers at the pivots to make them equal to 1. The result is the reduced row echelon form.

We remark that different choices of row operations will produce the same reduced row echelon form at the end (for a rigorous proof, see here).

Example We simplify some row echelon forms given in a previous example. By (1/9)[row 1], the row echelon form

[ | 9 | ] |

becomes a reduced row echelon form.

[ | 1 | ] |

By (1/3)[row 1], the row echelon form

[ | 3 | 2 | ] |

becomes a reduced row echelon form.

[ | 1 | 2/3 | ] |

The row echelon form

[ | 0 | 0 | 0 | ] |

0 | 0 | 0 |

is already reduced. By (-1)[row 1], the row echelon form

[ | 0 | -1 | ] |

0 | 0 | ||

0 | 0 |

becomes a reduced row echelon form.

[ | 0 | 1 | ] |

0 | 0 | ||

0 | 0 |

The row echelon form

[ | 1 | 0 | 0 | ] |

0 | 1 | 0 | ||

0 | 0 | 1 |

is already reduced. To reduce the row echelon form

[ | 1 | 3 | 2 | 0 | 1 | 0 | ] |

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

0 | 0 | 0 | 2 | 1 | 0 | ||

0 | 0 | 0 | 0 | 0 | 1 |

we start from the rightmost pivot column. By (-1)[row 4] + [row 2], we get

[ | 1 | 3 | 2 | 0 | 1 | 0 | ] |

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

0 | 0 | 0 | 2 | 1 | 0 | ||

0 | 0 | 0 | 0 | 0 | 1 |

Then we use (1/2)[row 3] + [row 2] and (1/2)[row 3] to get

[ | 1 | 3 | 2 | 0 | 1 | 0 | ] |

0 | 2 | 1 | 0 | 1/2 | 0 | ||

0 | 0 | 0 | 1 | 1/2 | 0 | ||

0 | 0 | 0 | 0 | 0 | 1 |

Finally, we use (-3/2)[row 2] + [row 1] and (1/2)[Row 1] to get

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

0 | 1 | 1/2 | 0 | 1/4 | 0 | ||

0 | 0 | 0 | 1 | 1/2 | 0 | ||

0 | 0 | 0 | 0 | 0 | 1 |

This is a reduced row echelon form.

Example Let us compute the reduced row echelon form of the matrix

[ | 1 | -1 | 1 | 0 | 2 | ] |

2 | -2 | 0 | 2 | 2 | ||

-1 | 1 | 2 | -3 | 1 | ||

-2 | 2 | 1 | -3 | -1 |

By (-2)[row 1] + [row 2], [row 1] + [row 3], and 2[row 1] + [row 2], we get

[ | 1 | -1 | 1 | 0 | 2 | ] |

0 | 0 | -2 | 2 | -2 | ||

0 | 0 | 3 | -3 | 3 | ||

0 | 0 | 3 | -3 | 3 |

By (-1/2)[row 2], we get

[ | 1 | -1 | 1 | 0 | 2 | ] |

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

0 | 0 | 3 | -3 | 3 | ||

0 | 0 | 3 | -3 | 3 |

By (-3)[row 2] + [row 3] and (-3)[row 2] + [row 4], we get

[ | 1 | -1 | 1 | 0 | 2 | ] |

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

0 | 0 | 0 | 0 | 0 | ||

0 | 0 | 0 | 0 | 0 |

which is a row echelon form. By -[row 2] + [row 1], we get

[ | 1 | -1 | 0 | 1 | 1 | ] |

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

0 | 0 | 0 | 0 | 0 | ||

0 | 0 | 0 | 0 | 0 |

which is a reduced row echelon form.

It appears that reduced row echelon forms are more desirable than mere row echelon forms. However, the solution of many linear algebra problems only requires the row echelon form. For these problems, one may save much work by computing the row echelon form only.

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