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john123456789

  • 2 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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  1. vf321
    • 2 years ago
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    \[ABC^TDBA^T = AB^T\]\[(AB)^{-1}ABC^TDBA^T =(AB)^{-1} AB^T\]\[I_nC^TDBA^T =(AB)^{-1} AB^T\]\[C^TDBA^T (DBA^T)^{-1}=(AB)^{-1} AB^T(DBA^T)^{-1}\]\[C^T=((AB)^{-1} AB^T(DBA^T)^{-1})\]\[C=(B^{-1}A^{-1}AB^T(A^T)^{-1}B^{-1}D^{-1})^T\]\[C=(B^{-1}I_nB^T(A^T)^{-1}B^{-1}D^{-1})^T\]\[C=(D^T)^{-1}(B^T)^{-1}A^{-1}B(B^T)^{-1}\]

  2. john123456789
    • 2 years ago
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    this is solve for C my question was to solve for D

  3. vf321
    • 2 years ago
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    Lol whoops do you see what I did at least? Try to replicate it, I don't have time to re-do this. If no one answers this question next time I check (may be a couple days), then I'll answer it.

  4. john123456789
    • 2 years ago
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    yeah, but on the 3 line there IC^T...and on the four the I is gone why is that ?

  5. vf321
    • 2 years ago
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    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

  6. john123456789
    • 2 years ago
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    ok so once your albe to reduce to I you can just remove it for the equation

  7. vf321
    • 2 years ago
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    yes

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