# Gauss-Jordan elimination

• Mar 16th 2010, 11:42 AM
craig
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 $\displaystyle A^{-1}$ where $\displaystyle A = \left( \begin{array}{ccc} -1 & -2 & 2 \\ -2 & -1 & 2 \\ -2 & -2 & 3 \\ \end{array} \right)$.

Do I multiply $\displaystyle A$ by -1 and then proceed as normal?

• Mar 16th 2010, 12:39 PM
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.
• Mar 17th 2010, 12:49 AM
craig
Quote:

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.