Mjollnir

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.

123123
122-3100
2-102010
3000001
1100
2010
3001

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.

12312
122-310
2-10201
1100
2010
3001

Add 2 of row 2 to row 1.

12312
102112
2-10201
1100
2010
3001

Add 2 of column 1 to column 3.

12312
102112
2-10001
1102
2010
3001

Subtract 2 of column 3 from column 2.

12312
100112
2-10001
11-42
2010
30-21

Multiply -1 through row 2.

12312
100112
21000-1
11-42
2010
30-21

Swap column 3 into the first column.

12312
110012
20100-1
121-4
2001
310-2

To get our inverse A-1 we multiply the lower left times the upper right.

121212
1211223
2000-1=00
31012

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.

12123123
12322-3140
200-102=000
312021

This "inverse" times the original matrix would give us the identity matrix I if the original problem wasn't underspecified. This (A-1 • A) contains some additional information we need to consider; it contains the null-space of the original problem.

Multiply the original A matrix times this (A-1 • A) and we should get the original A matrix returned.

123123123
122-314022-3
2-102000=-102
3021

This gives us (A • A-1 • A) = A. The "inverse" matrix is obviously functioning as an inverse of sorts. Let's multiply the (A-1 • A) times the "inverse". It should give us the "inverse" again.

1231212
11402323
200000=00
30211212

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:

123123
1100120
20100-10
3000001
121-4
2001
310-2

The product of the column and row parts is:

123123123
121-412023-4
20010-10=001
310-200112-2

This matrix (in this case) has the null-space as the third column. The (A-1 • A) is:

123123123
123-422-3140
2001-102=000
312-2000021

This is exactly the same (A-1 • 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 (A-1 • A) • A-1 - the result isn't quite the A-1 matrix we calculated here, but us identically zero for the third column and the same as the previous A-1 with a zero column.

123123123
114023-4230
2000001=000
302112-2120

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