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

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0this is solve for C my question was to solve for D

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Lol 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.

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I 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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0ok so once your albe to reduce to I you can just remove it for the equation
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