Mjollnir

Augmented Inverse

In the previous sections we saw how Gaussian Elimination could be successfully applied to the rows of an n-by-n square invertible matrix. We also saw how (AT)-1 = (A-1)T (or rather ((AT)-1)T = A-1). In this section we'll examine what happens when you use both methods simultaneously. For this section, keep in mind the property (BC)-1 = C-1B-1.

123 123
122-3100
2-102010
311-2001
1100
2010
3001

Add row 3 to row 2 and subtract 2 of row 3 from row 1.

123 123
100110-2
2010011
311-2001
1100
2010
3001

Subtract column 1 from column 2 and add 2 of column 1 to column 3.

123 123
100110-2
2010011
3100001
11-12
2010
3001

Swap columns 1 and 3 (or rows 1 and 3) to get the identity matrix.

123 123
110010-2
2010011
3001001
12-11
2010
3100

We could stop for a moment and think about what we've got now... neither the upper-right nor the lower-left is the inverse of the original matrix, but their product is:

123123123
12-1110-22-1-4
2010011=011
310000110-2

What we've done is effectively factored the original A matrix problem into two other matrices B and C, remembering that (B • C)-1 = C-1 • B-1. In this case, A = B • C. This refactoring is not unique. Note that one of the matrices is upper-triangular, and had we chosen our operations differently, we might have ended up with a lower- and an upper-triangular factorizations.

The augmented problem started:

A = B • CI
I

And it was reduced to:

IB-1
C-1

For our particular factorization, B is:

123
1102
201-1
3001

The product of B with its inverse is identity.

123123123
110210-2100
201-1011=010
3001001001

And C is:

123
1001
2010
311-2

The product of C with its inverse is identity.

123123123
10012-11100
2010010=010
311-2100001

We had to work backwards from the B-1 and C-1 to get the B and C matrices to demonstrate this, of course. And if we multiply B times C we get our original A matrix.

123123123
110200122-3
201-1010=-102
300111-211-2

And likewise, if we multiply C-1 times B-1 we get A-1.

123123123
12-1110-22-1-4
2010011=011
310000110-2

By using rows and columns, we've effectively factored the original matrix into two simpler matrices and inverted them, without the complexity of actually factoring the original matrix. Our use of column and row operations simplifies the inversion, and makes mathematical sense. But we're still only considering invertible matrices. Most problems in the real world are considerably more difficult.

For the next part we'll apply this method to the over-specified problem.