Exercise Which of the following systems have solutions?

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

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

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

Answer The augmented matrix is

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

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

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

By -2[row 2] + [row 1] and [row 1] ↔ [row 2], we get

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

0 | -2 | 4 | 1 | 3 | ||

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

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

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

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

0 | 0 | 0 | 3 | -1 |

Since there is no row of the form [0 0 0 0 ≠0], the system is consistent.

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

2x_{1} |
+ x_{2} |
+ x_{3} |
= | 0 |

3x_{2} |
+ x_{3} |
= | 0 | |

3x_{1} |
+ x_{3} |
= | 0 |

Answer The augmented matrix

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

2 | 1 | 1 | 0 | ||

0 | 3 | 1 | 0 | ||

3 | 0 | 1 | 0 |

appeared in this exercise, in which we have computed its row echelon form.

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

0 | 3 | 1 | 0 | ||

0 | 0 | 0 | 1 | ||

0 | 0 | 0 | 0 |

Since [row 3] is of the form [0 0 0 ≠0], the system has no solution.

x_{1} |
- x_{2} |
+ x_{3} |
+ 2x_{5} |
= | 1 | |

2x_{1} |
- 2x_{2} |
+ 2x_{4} |
+ 2x_{5} |
= | 0 | |

- x_{1} |
+ x_{2} |
+ 2x_{3} |
- 3x_{4} |
+ x_{5} |
= | 0 |

- 2x_{1} |
+ 2x_{2} |
+ x_{3} |
- 3x_{4} |
- x_{5} |
= | 0 |

x_{1} |
- x_{2} |
+ x_{3} |
+ 2x_{5} |
= | 3 | |

2x_{1} |
- 2x_{2} |
+ 2x_{4} |
+ 2x_{5} |
= | 2 | |

- x_{1} |
+ x_{2} |
+ 2x_{3} |
- 3x_{4} |
+ x_{5} |
= | 3 |

- 2x_{1} |
+ 2x_{2} |
+ x_{3} |
- 3x_{4} |
- x_{5} |
= | 0 |

Answer The two systems have the same coefficient
matrices. We compute both systems at the same time. Specifically, for
** Ax** =

[] = [A b c |
1 | -1 | 1 | 0 | 2 | 1 | 3 | ] |

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

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

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

Now we carry out row operations, keeping in mind that the last two columns are the right sides. By -2[row 1] + [row 2], [row 1] + [row 3], 2[row 1] + [row 4], and (1/2)[row 2], we get

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

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

0 | 0 | 3 | -3 | 3 | 1 | 6 | ||

0 | 0 | 3 | -3 | 3 | 2 | 6 |

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

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

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

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

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

Note that the coefficient matrix part has already become a row echelon form.
If we delete the last column, we see that the augmented matrix [** A b**]
has been simplified to

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

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

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

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

Because [row 3] is of the form [0 0 0 0 0 ≠0], the 3rd system has no
solution. Similarly, the augmented matrix [** A c**] has been simplified
to

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

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

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

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

From this we conclude that the 4th system has solutions.