Exercise Compute the (reduced) row echelon form and list the pivot columns.

1)

[ | 0 | ] |

Answer The matrix is a reduced row echelon form, with no pivot columns.

2)

[ | 0 | 1 | ] |

Answer The matrix is a reduced row echelon form, with [col 2] pivot.

3)

[ | 1 | 2 | ] |

Answer The matrix is a reduced row echelon form, with [col 1] pivot.

4)

[ | 2 | 1 | ] |

Answer The matrix is a row echelon form, but not reduced. By (1/2)[row 1], we get the reduced row echelon form.

[ | 1 | 1/2 | ] |

[col 1] is pivot.

5)

[ | 0 | ] |

1 |

Answer The matrix is not a row echelon form. By exchanging the two rows, we get a row echelon form

[ | 1 | ] |

0 |

which is in fact reduced. [col 1] is pivot.

6)

[ | 1 | ] |

2 |

Answer The matrix is not a row echelon form. By (-2)[row 1] + [row 2], we get a row echelon form

[ | 1 | ] |

0 |

which is in fact reduced. [col 1] is pivot.

7)

[ | 2 | ] |

1 |

Answer The matrix is not a row echelon form. By (-1/2)[row 1] + [row 2], we get a row echelon form

[ | 2 | ] |

0 |

which is not yet reduced. By further doing (1/2)[row 1], we get the reduced row echelon form.

[ | 1 | ] |

0 |

[col 1] is pivot.

8)

[ | 0 | -1 | ] |

0 | 0 | ||

0 | 0 |

Answer The matrix is a row echelon form, but not reduced. By (-1)[row 1], we get the reduced row echelon form.

[ | 0 | 1 | ] |

0 | 0 | ||

0 | 0 |

[col 2] is pivot.

9)

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

0 | 0 | 1 | 0 | ||

0 | 1 | 0 | 0 | ||

1 | 0 | 0 | 0 |

Answer The matrix is not a row echelon form. By [row 1] ↔ [row 4] and [row 2] ↔ [row 3], we get the reduced row echelon form.

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

0 | 1 | 0 | 0 | ||

0 | 0 | 1 | 0 | ||

0 | 0 | 0 | 1 |

All columns are pivot.

10)

[ | 1 | 2 | ] |

0 | 0 | ||

0 | 0 | ||

0 | 0 |

Answer The matrix is a reduced row echelon form, with [col 1] pivot.

11)

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

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

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

0 | 0 | 0 | 0 | 0 |

Answer The matrix is a reduced row echelon form, with [col 2], [col 3], [col 4] pivot.

12)

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

-1 | 1 | 0 | 0 | ||

0 | -1 | 1 | 0 | ||

0 | 0 | -1 | 1 |

Answer The matrix is not a row echelon form. By [row 1] + [row 2], [row 2] + [row 3], [row 3] + [row 4], we get the reduced row echelon form.

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

0 | 1 | 0 | 0 | ||

0 | 0 | 1 | 0 | ||

0 | 0 | 0 | 1 |

All columns are pivot.

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

2 | 1 | 1 | 0 | ||

0 | 3 | 1 | 0 | ||

3 | 0 | 1 | 0 |

Answer The matrix is not a row echelon form. By (-2)[row 1] + [row 2] and (-3)[row 1] + [row 4], we have

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

0 | -3 | -1 | -2 | ||

0 | 3 | 1 | 0 | ||

0 | -6 | -2 | -3 |

Then by [row 3] + [row 2], 2[row 3] + [row 4], and [row 2] ↔ [row 3], we get

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

0 | 3 | 1 | 0 | ||

0 | 0 | 0 | -2 | ||

0 | 0 | 0 | 1 |

Finally, by 2[row 4] + [row 3] and [row 3] ↔ [row 4], we get a row echelon form.

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

0 | 3 | 1 | 0 | ||

0 | 0 | 0 | 1 | ||

0 | 0 | 0 | 0 |

[col 1], [col 2], [col 4] are pivot.

To compute the reduced row echelon form, we use (-1)[row 3] + [row 1] to make [col 4] into the right form.

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

0 | 3 | 1 | 0 | ||

0 | 0 | 0 | 1 | ||

0 | 0 | 0 | 0 |

By (-2/3)[row 2] + [row 1] and (1/3)[row 2], we make [col 2] into the right form.

[ | 1 | 0 | 1/3 | 0 | ] |

0 | 1 | 1/3 | 0 | ||

0 | 0 | 0 | 1 | ||

0 | 0 | 0 | 0 |

This is the reduced row echelon form.

14)

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

1 | 4 | -3 | -2 | ||

1 | -1 | 2 | 3 |

Answer The matrix is not a row echelon form. By (-1)[row 1] + [row 2] and (-1)[row 1] + [row 3], we have

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

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

0 | -1 | 1 | 1 |

By 4[row 3] + [row 2], [row 2] ↔ [row 3], we have

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

0 | -1 | 1 | 1 | ||

0 | 0 | 0 | 0 |

This is a row echelon form. [col 1] and [col 2] are pivot.

By (-1)[row 2], we get the reduced row echelon form.

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

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

0 | 0 | 0 | 0 |

15)

[ | 2 | -3 | 6 | 2 | 5 | ] |

-2 | 3 | -3 | -3 | -4 | ||

4 | -6 | 9 | 5 | 9 | ||

-2 | 3 | 3 | -4 | 1 |

Answer The matrix is not a row echelon form. By [row 1] + [row 2], (-2)[row 1] + [row 3], [row 1] + [row 4], we have

[ | 2 | -3 | 6 | 2 | 5 | ] |

0 | 0 | 3 | -1 | 1 | ||

0 | 0 | -3 | 1 | -1 | ||

0 | 0 | 9 | -2 | 6 |

By [row 2] + [row 3], (-3)[row 2] + [row 4], [row 3] ↔ [row 4], we have

[ | 2 | -3 | 6 | 2 | 5 | ] |

0 | 0 | 3 | -1 | 1 | ||

0 | 0 | 0 | 1 | 3 | ||

0 | 0 | 0 | 0 | 0 |

This is a row echelon form. [col 1], [col 3], [col 4] are pivot.

To get the reduced row echelon form, we do [row 3] + [row 2], (-2)[row 3] + [row 1] to get

[ | 2 | -3 | 6 | 0 | -1 | ] |

0 | 0 | 3 | 0 | 4 | ||

0 | 0 | 0 | 1 | 3 | ||

0 | 0 | 0 | 0 | 0 |

Then by (-2)[row 2] + [row 1], (1/3)[row 2], we have

[ | 2 | -3 | 0 | 0 | -9 | ] |

0 | 0 | 1 | 0 | 4/3 | ||

0 | 0 | 0 | 1 | 3 | ||

0 | 0 | 0 | 0 | 0 |

Finally, (1/2)[row 1] gives us the reduced row echelon form.

[ | 1 | -3/2 | 0 | 0 | -9/2 | ] |

0 | 0 | 1 | 0 | 4/3 | ||

0 | 0 | 0 | 1 | 3 | ||

0 | 0 | 0 | 0 | 0 |

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

-1 | 3 | 0 | 2 | -2 | ||

0 | 1 | 1 | 3 | 4 | ||

1 | 2 | 5 | 13 | 5 |

Answer The matrix is not a row echelon form. By [row 1] + [row 2], (-1)[row 1] + [row 4], we have

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

0 | 1 | 1 | 3 | 0 | ||

0 | 1 | 1 | 3 | 4 | ||

0 | 4 | 4 | 12 | 3 |

By (-1)[row 2] + [row 3], (-4)[row 2] + [row 4], (-3/4)[row 3] + [row 4], we get a row echelon form.

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

0 | 1 | 1 | 3 | 0 | ||

0 | 0 | 0 | 0 | 4 | ||

0 | 0 | 0 | 0 | 0 |

[col 1], [col 2], [col 5] are pivot.

To get the reduced row echelon form, we do (-1/2)[row 3] + [row 1], (1/4)[row 3] to get

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

0 | 1 | 1 | 3 | 0 | ||

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

0 | 0 | 0 | 0 | 0 |

Then 2[row 2] + [row 1] give us the reduced row echelon form.

[ | 1 | 0 | 3 | 7 | 0 | ] |

0 | 1 | 1 | 3 | 0 | ||

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

0 | 0 | 0 | 0 | 0 |

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

-2 | 3 | 1 | 2 | ||

1 | 0 | 1 | 5 | ||

1 | 2 | 3 | 13 | ||

2 | -2 | 4 | 5 |

Answer The matrix is not a row echelon form. By 2[row 1] + [row 2], (-1)[row 1] + [row 3], (-1)[row 1] + [row 4],(-2)[row 1] + [row 5], we have

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

0 | 1 | 1 | 4 | ||

0 | 1 | 1 | 4 | ||

0 | 3 | 3 | 12 | ||

0 | 0 | 4 | 3 |

By (-1)[row 2] + [row 3], (-3)[row 2] + [row 4], [row 5], [row 3] ↔ [row 5], we get a row echelon form.

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

0 | 1 | 1 | 4 | ||

0 | 0 | 4 | 3 | ||

0 | 0 | 0 | 0 | ||

0 | 0 | 0 | 0 |

[col 1], [col 2], [col 3] are pivot.

We may further use (-1/4)[row 3] + [row 2], [row 2] + [row 1], 1/4[row 3] to get the reduced row echelon form.

[ | 1 | 0 | 0 | 17/4 | ] |

0 | 1 | 0 | 13/4 | ||

0 | 0 | 1 | 3/4 | ||

0 | 0 | 0 | 0 | ||

0 | 0 | 0 | 0 |

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

0 | 3 | 2 | 1 | -1 | 0 | ||

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

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

4 | 1 | 3 | 2 | 2 | 1 |

Answer The matrix is not a row echelon form. By [row 3] + [row 4], [row 1] + [row 4], (-4)[row 1] + [row 5], we have

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

0 | 3 | 2 | 1 | -1 | 0 | ||

0 | 3 | 5 | -2 | 2 | 1 | ||

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

0 | -7 | -9 | 2 | -2 | -3 |

By 3[row 4] + [row 2], 3[row 4] + [row 3], (-7)[row 4] + [row 5], [row 2] ↔ [row 4], we get

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

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

0 | 0 | 5 | -5 | 5 | 1 | ||

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

0 | 0 | -9 | 9 | -9 | -3 |

By (-5/2)[row 4] + [row 3], (9/2)[row 4] + [row 5], (1/2)[row 4], [row 3] ↔ [row 4], we get

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

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

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

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

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

By 3[row 4] + [row 5], we get a row echelon form.

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

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

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

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

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

[col 1], [col 2], [col 3] [col 6] are pivot.

We may use (-1)[row 4] + [row 1], (-3)[row 3] + [row 1], 2[row 2] + [row 1], (-1)[row 2] to get the reduced row echelon form.

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

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

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

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

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