1. Sledge

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Is it possible to compute matrix (A/B) without first finding the inverse of matrix B but ending with EITHER { A * (Inverse of B) } OR { (Inverse of B * A }....i think i discovered the trick

Staff: Mentor

Another way to write ##X=A/B \equiv AB^{-1}## is ##XB=A##. This has a unique solution X if B is not singular. You can solve for X in XB=A using Gaussian elimination.

Another way to write ##X=B \backslash A \equiv B^{-1}A## is ##BX=A##. This, too, has a a unique solution X if B is not singular. You can solve for X in BX=A using Gaussian elimination.

What if B is singular? The standard approach is to use the pseudo-inverse, and now you have but no choice to compute that inverse, typically via singular value decomposition.