Remark The purpose of this exercise is to emphasis that one should not forget the original meaning of consistency.

Exercise Without any computation, determine the consistency of the systems.

1)

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

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

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

Answer The system has `x`_{1} = `x`_{2}
= `x`_{4} = 0, `x`_{3} = 1 as an obvious solution.
Therefore the system is consistent.

Note that in this exercise, we have used row operations to determine the system is consistent.

2)

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

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

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

- x_{1} |
- 2x_{2} |
= | b |

Answer As far as the coefficients on the left
side are concerned, the 2nd, 3rd, and the 4th equations are respectively twice
of, equal to, and negative of the 1st equation. Therefore the same should hold
for the right side in order for the system to be consistent. The condition is
`a` = 1, `b` = -1.

3)

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

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

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

Answer As far as the coefficients on the left side are concerned, the sum of the 1st and the 2nd equations is the 3rd one. The same should hold for the right side in order for the system to be consistent. Since 1 + 2 ≠ 0, the system is not consistent.