[Bf-committers] Matrix "differencing" procedure

[Chris Want]

Martin Poirier wrote:
> RA = B
> 
> You're looking for R.
> 
> RiAA = BiA (where iA is inverse of A)

Blindly stuffing iA in the middle like that leads to
a really bad illustration, since it seems to imply that
matrix multiplication is commutitive (i.e.,
B*C = C*B, which is not generally true).
If this were a linear algebra class you would
definitely lose marks.

The better illustration would be RAiA = BiA,
or more clearly, as (RA)iA = (B)iA.
i.e., what happened was that both sides were
multiplied on the right by iA.

Chris <-- nitpicker, but it's important to get it straight