Sampled data often provides more equations than there are unknowns. Let's look now at the over-specified problem.
We'll use the same original equations but drop out the
`z`

values.

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

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

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

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

1 | 1 | 0 | ||||

2 | 0 | 1 |

We'll apply the same row and column operations to get at the "inverse". Begin with subtracting one of column
`1`

from column
`2`

.

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

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

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

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

1 | 1 | -1 | ||||

2 | 0 | 1 |

Add row
`3`

to row
`2`

and subtract 2 of row
`3`

from row
`1`

.

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

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

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

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

1 | 1 | -1 | ||||

2 | 0 | 1 |

Now swap rows
`1`

and
`3`

.

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

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

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

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

1 | 1 | -1 | ||||

2 | 0 | 1 |

The third row is identically
`0`

, as expected, so we'll multiply the lower left matrix and the top two rows of the upper right.

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

1 | 1 | -1 | • | 0 | 0 | 1 | = | 0 | -1 | 0 | ||||||

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

This is now our
```
A
```

. We'll confirm this by multiplying it times the
^{-1}
`A`

matrix.

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

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

2 | 0 | 1 | 1 | • | -1 | 0 | = | 0 | 1 | ||||||

3 | 1 | 1 |

With the
```
A
```

being the identity matrix, there's not much point in trying to find a null-space. There's not much point in
calculating the
^{-1} • A
```
(A
```

or the
^{-1} • A) • A^{-1}
```
A • (A
```

because anything times identity or identity times anything is itself. However, it is worth noting that
^{-1} • A)
```
A • A
```

is not identity.
^{-1}

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

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

2 | -1 | 0 | • | 0 | 1 | 1 | = | 0 | 1 | 0 | ||||||

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

I may come back to this if I can find the time.

For the next part we'll do the same using row and column operations on the under-specified problem.