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Question: find solutions to the normal equations (A^T).A.x=(A^T).b

Question:find solutions to the normal equations (A^T).A.x=(A^T).b

GPY 80 Maple , MaplePrimes
matrix transpose inverse
February 09 2015
0

where A^T is the transpose of A and it's given that (A^T).A is not invertible.

I am stuck as to how to arrive at the solution for x in this case. I initially thought I could multiply both sides by the inverse of A^T reducing it to Ax=b but that was obviously wrong since A^T is itself not invertible(it is singular).
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