|Mar27-10, 12:30 PM||#1|
Finding the inverse of an m*n matrix
Prove that if A is an mXn matrix, there is an invertible matrix C such that CA is in reduced row-echelon form. I think I know how to get the inverse of a square or an nXn matrix B, i.e., each elementary row operation carried out on B is also carried out on an identity matrix I. [B|I] to give [I|B^-1]. But I have no idea how to do the same to an mXn matrix A in order to find C. In fact are all mXn matrices invertible? I doubt it. I am studing this on my own, so please give a hint or two to get started. Thanks.
|Mar27-10, 02:51 PM||#2|
No, an "m by n matrix" (without m= n) does not have a true "inverse" and you don't need one. Instead think about how you would row-reduce A. In order that CA exist and C be invertible, C must be an "n by n" square matrix. Every row operation corresponds to an "elementary" matrix- the same row operation applied to the n by n identity matrix. C will be the product of the elementary matrices corresponding to the row operations required to row-reduce A.
|Mar27-10, 05:08 PM||#3|
I tried what you said first on a 2 by 3 matrix A, and I found the matrix C without any problem. I need to extend it now to a the general m by n matrix A. Thanks for your help.
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