Over Specified

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-1 . We'll confirm this by multiplying it times the 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-1 • 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 (A-1 • A) • A-1 or the A • (A-1 • A) because anything times identity or identity times anything is itself. However, it is worth noting that A • A-1 is not identity.

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.