1. Mar 17, 2014

### Sledge

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

2. Mar 17, 2014

### D H

Staff Emeritus
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.

3. Feb 17, 2017

### seba

what would be the uses of matrix division?