For a fixed matrix ` A`, the consistency of the system

Example We have studied the following system before

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

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

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

with a special choice on the right side. By applying the same row operations as before, the augmented matrix

[ | 3 | 1 | -1 | b_{1} |
] |

1 | -1 | 1 | b_{2} |
||

2 | 2 | 1 | b_{3} |

becomes

[ | 1 | -1 | 1 | b_{2} |
] |

0 | 4 | -4 | b_{1} - 3b_{2} |
||

0 | 0 | 3 | - b_{1} + b_{2}
+ b_{3} |

*Due to the fact* that there is no zero row [0 0 0] in
the darker part (which is a row echelon form of the coefficient matrix), there is no row of
the form [0 0 0 ≠0] for all choice of `b`_{1}, `b`_{2},
`b`_{3}. Therefore the system is consistent for any choice
of the right side.

Example In an earlier example, we already know that the system

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

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

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

has no solution for the special choice of
`b`_{1} = 2, `b`_{2} = 2, `b`_{3} = 6.
Thus the system is *not consistent for some right side* (i.e., it is not true that the system is
consistent for any right side).

We study the problem again by looking at the coefficient matrix only. The same row operations as before change the coefficient matrix

[ | 3 | 1 | -1 | ] |

1 | -1 | 1 | ||

2 | -1 | 1 |

to a row echelon form.

[ | 1 | -1 | 1 | ] |

0 | 1 | -1 | ||

0 | 0 | 0 |

The same row operations must also change the augmented matrix

[ | 3 | 1 | -1 | b_{1} |
] |

1 | -1 | 1 | b_{2} |
||

2 | -1 | 1 | b_{3} |

to

[ | 1 | -1 | 1 | b'_{1} |
] |

0 | 1 | -1 | b'_{2} |
||

0 | 0 | 0 | b'_{3} |

where `b`'_{1}, `b`'_{2}, `b`'_{3}
are some combinations of `b`_{1}, `b`_{2}, `b`_{3}.
Now we can choose some `b`_{1}, `b`_{2}, `b`_{3}
so that the last combination `b`'_{3} ≠ 0. Then the system
has no solution for this choice. Thus we conclude that the system is *not
always consistent*. Again we emphasis the *key reason* for this conclusion
is that the row echelon form of the coefficient matrix contains a zero row [0 0 0].

You may also check out the always consistent problem for this example. From all these examples, it is easy to conclude the criterion for the existence of solutions for any right side.

The system ** Ax** =

⇔ There is no zero row in the row echelon form of

⇔ All rows of

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