Hi, I am trying to solve a rigging problem which essentially boils down to solving this equation:

AXBX’ = C

where

X’ = inverse of X

A, B, C are 4x4 matrices with known values

I would like to find X. Is it possible? Any help would be appreciated!

Hi, I am trying to solve a rigging problem which essentially boils down to solving this equation:

AXBX’ = C

where

X’ = inverse of X

A, B, C are 4x4 matrices with known values

I would like to find X. Is it possible? Any help would be appreciated!

If you start simplifying the equation you get this:

XX’ = C / AB

Going even further, you get:

1 = C / AB

Because anything multiplied by its inverse is 1 (or in this case the identity matrix). So, in the original equation, X has an infinite number of possible values.

Now all that is assuming my basic algebra isn’t worse than I think it is:)

What “rigging problem” are you trying to solve? It might be easier to help you if the equation was in a better context.

It’s been a good few years since I touched “real maths” ™, but as far as I know the solution is quite a lot more complicated. Matrix multiplication can’t just be swapped around and stuff.

AXBX’=C

-> A’AXBX’X=A’CX

->XB=A’CX

0=A’CX-XB

So with Q = A’C and R = -B, you’ve got QX+XR = 0

So you’re now effectively down to solving an equation of the form “AX+XB=0”, which is a lot easier to Google, and whose solution is pretty complicated for 2x2 and 3x3 matrices. I didn’t see a 4x4 one, but a generalised paper can be found here: http://aero-comlab.stanford.edu/Papers/jameson_007.pdf, and I figure that’d get you the rest of the way. Again, sorry, it’s been ages, so by all accounts I may have made mistakes too, but at least I can see why you got stuck.

(You may well end up with X=0 or some other trivial solution anyway, but it’ll at least be for the right reasons! :P)