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Invertible matrix proof

  1. Nov 3, 2008 #1
    1. The problem statement, all variables and given/known data
    Prove: If A and B are both nxn matrices, A is invertible and AB=BA, then A-1B=BA-1


    2. Relevant equations
    (AB)-1=B-1A-1


    3. The attempt at a solution
    A-1B=BA-1
    A-1BB-1=BA-1B-1
    A-1BB-1=B(A-1B-1)
    A-1(BB-1)=B(BA)-1
    B-1A-1(I)=B-1B(BA)-1
    (AB)-1=I(BA)-1
    (AB)-1[tex]\neq[/tex]A-1B-1

    Therefore A-1B does not equal BA-1

    I've been struggling with proofs and was wondering if this was correct?
     
  2. jcsd
  3. Nov 3, 2008 #2

    gabbagabbahey

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    First, you aren't told that [itex]B^{-1}[/itex] exists, so any proof that relied on the use of [itex]B^{-1}[/itex] is out of the question.

    Second, even if [itex]B^{-1}[/itex] did exist then [itex](AB)^{-1}=(BA)^{-1} \Rightarrow AB=BA[/itex] which was already given in the question and so your attempt to find a contradiction has failed.

    Third try starting with one of the conditions you're given: [itex]AB=BA[/itex] you're also told that [itex]A^{-1}[/itex] exists, so why not try multiplying both sides of the equation [itex]AB=BA[/itex] by [itex]A^{-1}[/itex] from either side and see what you get...
     
  4. Nov 3, 2008 #3
    I took your advice and came up with this:

    AB=BA
    A-1AB=A-1BA
    IB=A-1BA
    B=A-1BA
    BA-1=A-1BAA-1
    BA-1=A-1B(AA-1)
    BA-1=A-1BI
    BA-1=A-1B

    Read from right to left, this proves the statement. Is this right?
     
  5. Nov 3, 2008 #4

    Dick

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    It sure does.
     
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