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Inverse question

  1. Dec 11, 2007 #1
    For matrices:
    is (AB)[tex]^{-1}[/tex]=
    A[tex]^{-1}[/tex]B[tex]^{-1}[/tex]
    or
    B[tex]^{-1}[/tex]A[tex]^{-1}[/tex]
     
    Last edited: Dec 11, 2007
  2. jcsd
  3. Dec 11, 2007 #2
    Are you talking about linear operators, matrices, members of a group, or what?
     
  4. Dec 11, 2007 #3

    robphy

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    Since [tex]I=(AB)^{-1}(AB)=(AB)^{-1}AB[/tex]
    .. you can finish this off.
     
  5. Dec 11, 2007 #4

    HallsofIvy

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    In other words do it! What is [itex](A^{-1}B^{-1})(AB)[/itex]? What is [itex](B^{-1}A^{-1})(AB)[/itex]?
     
  6. Dec 11, 2007 #5
    so the answer is B[tex]^{-1}[/tex]A[tex]^{-1}[/tex]
     
  7. Dec 11, 2007 #6
    what im trying to get to is this:
    is there a general property.
    for example:
    is(BA)[tex]^{2}[/tex]
    equal to:
    B[tex]^{2}[/tex]A[tex]^{2}[/tex]
    or
    A[tex]^{2}[/tex]B[tex]^{2}[/tex]
     
  8. Dec 11, 2007 #7

    morphism

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    In general, no.
     
  9. Dec 11, 2007 #8
    so what would the the answer for (BA)^2
     
  10. Dec 11, 2007 #9

    morphism

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    (BA)^2 = BABA

    So if A and B are invertible...
     
  11. Dec 11, 2007 #10
    so its B^2A^2
     
  12. Dec 11, 2007 #11

    morphism

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    What is B^2A^2???
     
  13. Dec 11, 2007 #12

    rock.freak667

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    well normally to find a general way(an easy) way to find say A^99194

    you can just represent A in the form PDP[itex]^{-1}[/itex] where D is the diagonalizable matrix. But I do not think you have reached that far in your course yet. If you have done eigenvalues and eigenvectors then you will understand.
     
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