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Transpose of matrix

  1. Jul 19, 2006 #1
    [tex](ABC)^T, A,B,C[/tex] are all symmetric, then why isn't [tex](ABC)^T = CBA[/tex]? If you consider that [tex](ABC)^T = (C^T)(B^T)(A^T)[/tex] and in symmetrix cases, then [tex]C^T = C[/tex] and so on...?

    (Latex edit by HallsofIvy)
     
    Last edited by a moderator: Jul 28, 2006
  2. jcsd
  3. Jul 19, 2006 #2

    matt grime

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    Who says that (ABC)^T is not CBA when all three are symmetric?
     
  4. Jul 19, 2006 #3
    The solutions manual to Gilbert Strang Linear Algebra...
     
  5. Jul 19, 2006 #4

    matt grime

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    Right, why don't you post the full question and the full answer from this book? I mean, is the question:

    Q. if A,B, and C are symmetric does (ABC)^T = CBA?
    A. No.
     
  6. Jul 19, 2006 #5

    matt grime

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    Right, why don't you post the full question and the full answer from this book? I mean, is the question:

    Q. if A,B, and C are symmetric does (ABC)^T = CBA?
    A. No.
     
  7. Jul 19, 2006 #6
    Yes, that is the case
     
  8. Jul 19, 2006 #7

    matt grime

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    Then the asnwer book is wrong, if that is the precise statement of the question.
     
  9. Jul 27, 2006 #8

    Bob

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    ABC [tex]\neq[/tex]CBA
     
    Last edited: Jul 28, 2006
  10. Jul 28, 2006 #9

    matt grime

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    And no one is claiming that they are equal.
     
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