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Homework Help: Linear algebra determinant proof

  1. Feb 13, 2013 #1
    1. The problem statement, all variables and given/known data

    Let A and B be nxn matrices. Prove that if AB=I, then BA=I





    Equating 1 and two together I get det(B)*det(A)=det(A)*det(B)

    Thus AB=I, then BA=I

    Is this correct?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 13, 2013 #2
    What you showed is that the ##det(AB) = det(BA)## but that doesn't really imply that ##AB = BA##. If you want to show that ##AB = BA##, you need to use the properties of inverses and facts about them.

    What you did is a start, though. Since ##det(AB) = det(A)det(B) = 1##, that means neither ##det(A)## or ##det(B)## are 0. Ie, A and B are non-singular and have inverses. Now all you need to do is show that ##A## and ##B## are inverses of each other.
    Last edited: Feb 13, 2013
  4. Feb 13, 2013 #3
    Could I say that if B=A-1 then


  5. Feb 13, 2013 #4
    Sure, that is correct if [itex]B=A^{-1}[/itex]. The problem is that you don't know that [itex]B=A^{-1}[/itex]. You need to prove that.
  6. Feb 13, 2013 #5
    Would it be enough if I maybe do something like this

    assume A-1=B

    then A*B=I == A*A-1=I

    Which would make it true?

    Because if B=A-1 then A*B=I would be true
  7. Feb 13, 2013 #6


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    No, you can't start with an assumption, get a true result, and then claim that proves the assumption is true. It's possible to start with a false assumption and get a true result.
  8. Feb 13, 2013 #7
    What you're saying makes sense but i'm a bit unsure on how to use a false assumption in this case. For instance if I assume that B~=A-1 what can I do with that?
  9. Feb 13, 2013 #8


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    The point is you can't assume B=A-1. That's what you're trying to prove.
  10. Feb 13, 2013 #9
    That's where i'm stuck, How do I prove the assumption I made without using another assumption? Can you help me get started in the direction of proving B=A-1 so I dont run in circles.
  11. Feb 13, 2013 #10
    It's very simple. Start with your "If" statement and somehow isolate ##B##. Remember that what you are given is assumed to be true.
  12. Feb 13, 2013 #11


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    Let's back up for a second. You know what AB=I because that's a given. At this point, you don't know anything about A and B other than the fact that their product AB is the identity matrix. In particular, you don't know if A has an inverse or B has an inverse.

    You then took the determinant of both sides and got to det(A)det(B)=1. Now det(A)det(B)=1 implies that det(A)≠0. See what Karnage1993 said up in post 2. det(A)≠0 tells you A does indeed have an inverse. So now you know that AB=I and A-1 exists. Can you use this knowledge to solve for B?
  13. Feb 13, 2013 #12
    So the if statement is if AB=I

    then A-1*AB=A-1I

    Which is equal to B=A-1
  14. Feb 13, 2013 #13
    Yes, that's correct, but you can only multiply both sides by ##A^{-1}## once you know for a fact that ##A## does indeed have an inverse, which we already went over above.
  15. Feb 13, 2013 #14
    So I would have to state it like this

    Since det(A)det(b)=det(I)=1
    det(AB) is nonsingular meaning it has an inverse, and then I can show the proof for B=A^-1
  16. Feb 13, 2013 #15


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    Not exactly. Try again. Your logic is a bit muddied still.
  17. Feb 13, 2013 #16
    Looking at the book it says that a matrix A is termed invertable if there is a matrix B such that A*B=I.

    So if i'm quoting the definition in the book technically i'm not assuming anything, just explaining the if statement. Right?
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