Matrix question

let's i assume it's not homework. Tho, i might be mistaken.
A = B * C * D * E
let Q = B * C
then
A = Q * D * E
A * E^-1 = Q * D
Q^-1 * A * E^-1 = D
that is,

D = (B * C)^-1 * A * E^-1

also, that identity might be useful for you
(B * C)^-1 = C^-1 * B^-1

Dmytry is way too smart for us human beings. I don't really know what I'm talking about, but I want to see if I can understand this..

First, you have to invert E and multiply A with it. Temp = A * inverted(E). In that order because E is after D in the original equation, so it must be after A to put it on the opposite side of =.

Then you invert B * C and multiply that with Temp. D = inverted( B * C ) * Temp. Also in that order, because the B and C are behind the D.

Basically you multiply everything on the right side of the requested matrix, invert it, and multiply with the other side of = on the right. Then multiply everything on the left side, invert it, multiply it on the left.

So you could have A * B * C * D == E * F * G * H * I;
G = inverted(E * F) * A * B * C * D * inverted(H * I);

Is that right?
Sorry to pollute your thread.

Perfect, works a treat. Thanks.

P.S. It wasn't homework

Idea is simple: in
A = B * C * D * E
I rightmultiply™ both sides of equation with E^-1 and get
A * E^-1 = B * C * D * E * E^-1 = B * C * D * (E * E^-1) = B * C * D
as E * E^-1 = identity, and can be removed.
Then, I leftmultiply™ both sides with (B * C)^-1 and get
(B * C)^-1 * A * E^-1 = (B * C)^-1 * B * C * D = D

so we just found D. It's quite simple, almost same as with numbers (idea is that if you do same thing with both sides of equation, equation is still true), just need to be accurate with left/right multiply.