## Under Specified

In previous sections we saw how augmenting a matrix for row and column operations effectively factors a matrix into two parts. Maybe we didn't see much difference in speed, but it does open up some possibilities. In this section we'll apply the same method to the under-specified problem - using our first two equations and ignoring the third.

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

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

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

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

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

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

3 | 0 | 0 | 1 |

The third equation is set to zero as in the above, but the zero row is unnecessary, as it has no effect on the solution
and the following is preferred. The identity matrix is square, so the one on the right drops to `2 x 2`

.

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

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

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

1 | 1 | 0 | 0 | |||

2 | 0 | 1 | 0 | |||

3 | 0 | 0 | 1 |

Add 2 of row `2`

to row `1`

.

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

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

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

1 | 1 | 0 | 0 | |||

2 | 0 | 1 | 0 | |||

3 | 0 | 0 | 1 |

Add 2 of column `1`

to column `3`

.

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

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

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

1 | 1 | 0 | 2 | |||

2 | 0 | 1 | 0 | |||

3 | 0 | 0 | 1 |

Subtract 2 of column `3`

from column `2`

.

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

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

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

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

2 | 0 | 1 | 0 | |||

3 | 0 | -2 | 1 |

Multiply `-1`

through row `2`

.

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

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

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

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

2 | 0 | 1 | 0 | |||

3 | 0 | -2 | 1 |

Swap column `3`

into the first column.

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

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

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

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

2 | 0 | 0 | 1 | |||

3 | 1 | 0 | -2 |

To get our inverse `A`

we multiply the lower left times the upper right.
^{-1}

1 | 2 | 1 | 2 | 1 | 2 | |||||||||

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

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

3 | 1 | 0 | 1 | 2 |

If we multiply the "inverse" times the original matrix, we expect the result to be the identity matrix, but what we get is more interesting than that.

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

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

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

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

This "inverse" times the original matrix would give us the identity matrix `I`

if the original problem wasn't underspecified.
This (`A`

) contains some additional information we need to consider; it contains the null-space of the
original problem.
^{-1} • A

Multiply the original `A`

matrix times this `(A`

and we should get
the original ^{-1} • A)`A`

matrix returned.

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

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

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

3 | 0 | 2 | 1 |

This gives us `(A • A`

.
The "inverse" matrix is obviously functioning as an inverse of sorts. Let's multiply the ^{-1} • A) = A`(A`

times the "inverse". It should give us the "inverse" again.
^{-1} • A)

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

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

2 | 0 | 0 | 0 | • | 0 | 0 | = | 0 | 0 | ||||||

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

This gives us `(A`

.
^{-1} • A • A^{-1}) = A^{-1}

Had we kept the full matrix at the start of this page and done the reduction, we'd have the following:

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

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

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

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

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

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

3 | 1 | 0 | -2 |

The product of the column and row parts is:

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

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

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

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

This matrix (in this case) has the null-space as the third column. The `(A`

is:
^{-1} • A)

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

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

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

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

This is exactly the same `(A`

that we saw from the method which ignored the third row
(the null-space component has already dropped out).
We can see this if we calculate ^{-1} • A)`(A`

- the result isn't quite the
^{-1} • A) • A^{-1}`A`

matrix we calculated here, but us identically zero for the third column and the same as the previous
^{-1}`A`

with a zero column.
^{-1}

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

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

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

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

For the next part we'll look at the null-space from this problem.