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Linear algebra question

  1. Mar 15, 2013 #1
    1. The problem statement, all variables and given/known data

    Assuming that all matrices are [itex]n\times n[/itex] and invertible, solve for [itex]D[/itex].

    [itex]C^{T}B^{-1}A^{2}BAC^{-1}DA^{-2}B^{T}C^{-2}=C^{T}[/itex]

    3. The attempt at a solution

    I tried to group all like terms and simplify. I'm pretty sure this is not allowed but I'm not really sure how to approach this question. Thanks a lot any help appreciated!

    [itex]C^{T}C^{-1}C^{-1}C^{-1}B^{-1}BB^{T}AAAA^{-1}A^{-1}D=C^{T}[/itex]

    [itex]C^{T}C^{-1}C^{-1}C^{-1}IB^{T}IIAD=C^{T}[/itex]

    [itex]((C^T)^{-1}C^{T})C^{-1}C^{-1}C^{-1}B^{T}AD=C^{T}(C^T)^{-1}[/itex]

    [itex]I(C^{-1}C^{-1}C^{-1}CCC)(B^{T}(B^T)^{-1})(AA^{-1})D=ICCC(B^T)^{-1}A^{-1}[/itex]

    [itex]D=C^{3}(B^T)^{-1}A^{-1}[/itex]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 15, 2013 #2

    haruspex

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    No, you can't change the order like that. What you can do is multiply both sides by the same thing at the same end. I.e. You can go from Y = Z to XY = XZ or to YX = ZX. Start with multiplying both sides on the left by C-T. Use the fact that you can change the order when the two matrices being multiplied are inverses of each other: X-1X = I = XX-1
     
  4. Mar 15, 2013 #3

    Dick

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    Don't assume the matrices commute. You can't interchange the order like you did. Aside from the fact this problem is needlessly complex, just use patience and cancel each matrix M by M^(-1) on the appropriate side.
     
  5. Mar 15, 2013 #4
    Thanks guys, but what happens when I get down to D after I multiply each term preceding it by its inverse?
     
  6. Mar 15, 2013 #5
    Ahhhh, multiply on the right!
     
  7. Mar 15, 2013 #6

    Dick

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    Right! No pun intended.
     
  8. Mar 15, 2013 #7

    Lol thanks
     
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