Mjollnir

Column Operations

In the previous section we used Gaussian Elimination to find the inverse of a matrix A. In this section we'll bring back the idea from the curious identity (AT)-1 = (A-1)T and we'll augment the transposed matrix A. Well, we have a choice here.. we can transpose the matrix, perform the inverse, and transpose the result back to get the inverse, or we can augment the matrix with columns instead and save the transpose steps.

This is a choice between:

123 123
12-11100
2201010
3-32-2001

and this:

123
122-3
2-102
311-2
1100
2010
3001

The first one uses less 'real estate' on the page, and it's clear that it's the transposed matrix that we're inverting, but I prefer the second form because it saves doing the transpose and is closer to what I'm proposing further on. It should be obvious that these two are equivalent, as far as our operations are concerned. So let's begin by subtracting column 2 from column 1 and adding 2 of column 2 to column 3.

123
1021
2-102
3010
1100
2-112
3001

Add 2 of column 1 to column 3.

123
1021
2-100
3010
1102
2-110
3001

Subtract 2 of column 3 from column 2.

123
1001
2-100
3010
11-42
2-110
30-21

Multiply -1 through column 1.

123
1001
2100
3010
1-1-42
2110
30-21

And for the final step, swap columns to make column 3 the first, column 1 the second, and column 2 the third.

123
1100
2010
3001
12-1-4
2011
310-2

The lower matrix is now the inverse of the original A matrix, and we did it without having to do the transposes. Had we performed these same operations on the rows of the identity-augmented-transpose matrix instead of doing these on the columns of the non-transposed matrix, we'd have to apply the transpose to our solution to get back the inverse. Doing colum operations just removes our need to do the two transposes.

For the next part we'll see how we can use this to factor the original matrix into two parts and solve these separately.