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A*B*(C^T) *D*B*(A^T) = A*(B^T) Assuming that all matrices are n × n invertible, solve for D
 one year ago
 one year ago
A*B*(C^T) *D*B*(A^T) = A*(B^T) Assuming that all matrices are n × n invertible, solve for D
 one year ago
 one year ago

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vf321Best ResponseYou've already chosen the best response.0
\[ABC^TDBA^T = AB^T\]\[(AB)^{1}ABC^TDBA^T =(AB)^{1} AB^T\]\[I_nC^TDBA^T =(AB)^{1} AB^T\]\[C^TDBA^T (DBA^T)^{1}=(AB)^{1} AB^T(DBA^T)^{1}\]\[C^T=((AB)^{1} AB^T(DBA^T)^{1})\]\[C=(B^{1}A^{1}AB^T(A^T)^{1}B^{1}D^{1})^T\]\[C=(B^{1}I_nB^T(A^T)^{1}B^{1}D^{1})^T\]\[C=(D^T)^{1}(B^T)^{1}A^{1}B(B^T)^{1}\]
 one year ago

john123456789Best ResponseYou've already chosen the best response.0
this is solve for C my question was to solve for D
 one year ago

vf321Best ResponseYou've already chosen the best response.0
Lol whoops do you see what I did at least? Try to replicate it, I don't have time to redo this. If no one answers this question next time I check (may be a couple days), then I'll answer it.
 one year ago

john123456789Best ResponseYou've already chosen the best response.0
yeah, but on the 3 line there IC^T...and on the four the I is gone why is that ?
 one year ago

vf321Best ResponseYou've already chosen the best response.0
I is the identity matrix, any matrix Q like so makes I: \[QQ^{1}=I\]In the spot you're looking at Q=AB. Also, I follows the following property: \[AI=IA=A\] for any A
 one year ago

john123456789Best ResponseYou've already chosen the best response.0
ok so once your albe to reduce to I you can just remove it for the equation
 one year ago
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