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cjohnston8Best ResponseYou've already chosen the best response.0
When taking the transverse, is it done after the adjoint of gBC and Fc? \[F ^{B} = Ad _{g ^{B}_{C}}^{T} * F ^{C}\]
 one year ago

Chipper10Best ResponseYou've already chosen the best response.0
It's actually transpose. You transpose the adjoint matrix in this formula (before multiplying by F^X). If you didn't do that, you would end up with a 1x3 vector, which is not what you want.
 one year ago

cjohnston8Best ResponseYou've already chosen the best response.0
I meant transpose, sorry. But how would that work? I thought adjoining \[g ^{B}_{C}\] with \[F ^{C}\] would look like this \[g ^{B}_{C} * F ^{C} * g ^{C}_{B}\] so wouldn't it only be possible to transpose after doing that multiplication?
 one year ago

Chipper10Best ResponseYou've already chosen the best response.0
In last week's homework, there was a formula for the SE2 adjoint matrix. It's just a matrix multiplication. You can transpose the adjoint matrix first.
 one year ago
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