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Linear Algebra ABC Proof

  1. Feb 8, 2009 #1
    1. The problem statement, all variables and given/known data
    Prove that if A,B,and C are square matrices and ABC = I, then B is invertible and B^-1 = CA.

    3. The attempt at a solution

    [tex]ABC = I[/tex]
    [tex]CABC = CI[/tex]
    [tex]CABC = C[/tex]
    [tex]CABCA = CA[/tex]

    so we have these two things:
    [tex](CAB)CA = CA[/tex]
    [tex]CA(BCA) = CA[/tex]

    so I thought that since CA times CAB = CA then CAB = I, and same for BCA. But, that is only true if the matrix is invertible, and the problem doesn't say whether C and A are invertible. Any suggestions? Thanks.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Feb 8, 2009 #2


    Staff: Mentor

    If M and N are square matrices, and MN = I, then both M and N are invertible.

    You're given that ABC = I, so A(BC) = I, and (AB)C = I, which says that A, BC, AB, and C are all invertible, and that A^(-1) = BC, and so on.

    Your last equation is CA(BCA) = CA, which suggests to me that BCA = I, or that B(CA) = I. What's the relationship between B and CA?
  4. Feb 9, 2009 #3
    Is an if you multiply invertible matrices together does it always yield an invertible matrix?

    If so then CA(BCA) = CA implies that B(CA) = I, which means that B^-1=CA.
  5. Feb 9, 2009 #4
    Yes because det(AB) = det(A)*det(B) and if det(AB) is non-zero, then neither A nor B can have zero determinant, i.e., they must both be invertible. [if we're dealing with real or complex valued matrices at least]
  6. Feb 9, 2009 #5
    Yes we are. Alright then, thank you both.
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