Thread: Gauss-Jordan elimination

Normally have no problem with these types of problems to find the inverse of a matrix. However I'm not sure how to start with this one, usually the top left value is rather conveniently 1.

Find where .

Do I multiply by -1 and then proceed as normal?

Thanks for your help

Not sure which method you have been using for finding A^{-1}, but if you think about the augmented 3x6 matrix with the last 3 columns as the Identity matrix (3x3 matrix with 1's down the diagonal), then doing GJ elimination on this matrix should yield a 3x6 matrix with the first 3 columns being the identity matrix and the last 3 columns now being A^{-1}. So to change {-1,-2,2} to {1,2,-2} you would multiply the whole first row {-1,-2,2,1,0,0} by -1 and get {1,2,-2,-1,0,0}. You can proceed through rows 2 and 3 the same way.

Originally Posted by Bgrasty

Not sure which method you have been using for finding A^{-1}, but if you think about the augmented 3x6 matrix with the last 3 columns as the Identity matrix (3x3 matrix with 1's down the diagonal), then doing GJ elimination on this matrix should yield a 3x6 matrix with the first 3 columns being the identity matrix and the last 3 columns now being A^{-1}. So to change {-1,-2,2} to {1,2,-2} you would multiply the whole first row {-1,-2,2,1,0,0} by -1 and get {1,2,-2,-1,0,0}. You can proceed through rows 2 and 3 the same way.

Thanks for the reply.

So I'd just multiply the first row by -1, and do nothing to the second and third ones?

After you multiply the first row by -1, add 2 times the first row to the second row and 2 times the first row to the 3rd row. This leaves the first row as the only one with a number in the first column.

Originally Posted by kaylakutie

After you multiply the first row by -1, add 2 times the first row to the second row and 2 times the first row to the 3rd row. This leaves the first row as the only one with a number in the first column.

Thankyou.