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Thread: Solve for a matrix

  1. #1
    Junior Member
    Aug 2008

    Solve for a matrix

    Assuming that all matrices are nxn and invertible, solve for D in the equation.

    $\displaystyle ABC^TDBA^T = AB^T$

    How can I do it?
    Please advise me the direction to solve this.

    Thank you.
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  2. #2
    Master Of Puppets
    pickslides's Avatar
    Sep 2008
    I am a bit rusty on these ones but i'll give it a shot.

    We need to do some left multiplying on both sides

    $\displaystyle ABC^TDBA^T = AB^T$

    $\displaystyle A^{-1}ABC^TDBA^T = A^{-1}AB^T$


    $\displaystyle IBC^TDBA^T = IB^T$

    $\displaystyle BC^TDBA^T = B^T$

    try the same thing with B

    $\displaystyle B^{-1}BC^TDBA^T = B^{-1}B^T$


    $\displaystyle IC^TDBA^T = B^{-1}B^T$

    $\displaystyle C^TDBA^T = B^{-1}B^T$

    this is where it gets tricky I would try

    $\displaystyle C^{T^{-1}}C^TDBA^T = C^{T^{-1}}B^{-1}B^T$

    which gives

    $\displaystyle IDBA^T = C^{T^{-1}}B^{-1}B^T$

    $\displaystyle DBA^T = C^{T^{-1}}B^{-1}B^T$

    Now we need to do some right multiplying

    $\displaystyle DBA^TA^{T^{-1}} = C^{T^{-1}}B^{-1}B^TA^{T^{-1}}$

    $\displaystyle DBI = C^{T^{-1}}B^{-1}B^TA^{T^{-1}}$

    $\displaystyle DB = C^{T^{-1}}B^{-1}B^TA^{T^{-1}}$


    $\displaystyle DBB^{-1} = C^{T^{-1}}B^{-1}B^TA^{T^{-1}}B^{-1}$

    $\displaystyle DI = C^{T^{-1}}B^{-1}B^TA^{T^{-1}}B^{-1}$

    $\displaystyle D = C^{T^{-1}}B^{-1}B^TA^{T^{-1}}B^{-1}$

    I wouldn't take this as bible!

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