[Bf-committers] Matrix "differencing" procedure
Sun, 1 Aug 2004 12:16:06 -0700 (PDT)
--- Chris Want <firstname.lastname@example.org> wrote:
> 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
> 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
That's what I get for not double checking after
changing stuff (at first, I was doing AR = B -> iAAR =
iAB and decided to switch that around).
That's not nitpicking, it's called being precise.
Lets rewrite that as a whole with correct
RA = B
(RA)iA = BiA
R(AiA) = BiA
R(I) = BiA
R = BiA
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