Split Jacobian result into Matrix factors

Hello,

I'm using the symbolic toolbos to calculate the following Jacobian:

jacobian(A4,phi_)

where

A4 = - TLB*R*TLB2*w

with

The result of jacobian(A4,phi_) calculated by Matlab is

[wz*(r2 + psi*r5 - r8*theta) - wy*(r3 + psi*r6 - r9*theta),   wy*(r8 + phi*r9 - psi*r7) - wz*(r1 - r9 + phi*r8 + psi*r4 - 2*r7*theta) + wx*(r3 + r7 + psi*r6 + psi*r8 - 2*r9*theta), - wz*(r6 - phi*r5 + r4*theta) - wy*(r5 - r1 + phi*r6 - 2*psi*r4 + r7*theta) - wx*(r2 + r4 + 2*psi*r5 - r6*theta - r8*theta)]
[   - wx*(r7 + psi*r8 - r9*theta) - wz*(r9 - r5 - 2*phi*r8 + psi*r2 + r7*theta) - wy*(r6 + r8 + 2*phi*r9 - psi*r3 - psi*r7),wx*(r6 + phi*r9 - psi*r3) - wz*(r4 + phi*r7 - psi*r1),       wz*(r3 - phi*r2 + r1*theta) - wx*(r5 - r1 + phi*r8 - 2*psi*r2 + r3*theta) + wy*(r2 + r4 + phi*r3 + phi*r7 - 2*psi*r1)]
[ wx*(r4 + psi*r5 - r6*theta) - wy*(r9 - r5 - 2*phi*r6 + psi*r4 + r3*theta) + wz*(r6 + r8 - 2*phi*r5 + r2*theta + r4*theta), - wy*(r2 + phi*r3 - psi*r1) - wx*(r1 - r9 + phi*r6 + psi*r2 - 2*r3*theta) - wz*(r3 + r7 - phi*r2 - phi*r4 + 2*r1*theta), wy*(r7 - phi*r4 + r1*theta) - wx*(r8 - phi*r5 + r2*theta)]

which as you can see isn't very nice looking.

My question is if I can somehow turn this result into a product of my original matrices (if even mathematically possible),

for example ans = TLB*TLB2*R^T*inv(TLB*TLB2) or something like that.

Hope someone can help me here, thanks.

[David Goodmanson]

Hi Simon,

if you want a result in symbolic variables you are not going to do significantly better than what you have. If you are looking for straight numerical results then you can get epressions that shorter than the result above.

[Simon Detmer]

So no possibility to simplify the result while keeping the symbolic variables/matrices? shame...

Thank you for your answer though.

[David Goodmanson]

With fifteen independent variables sprinkled around by matrix multiplication, you can't expect the results to be sleek. And if you do have a better looking alternative expression such as the one you suggest (assuming it were true) then of course the end result has to be exactly the same as the one you have. So a better starting point doesn't help.