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For any two elements A and B that form AB, neither A nor B have to be

  1. Nov 4, 2009 #1
    For any two elements A and B that form AB, neither A nor B have to be invertible for (AB)-1 to exist, right?
     
  2. jcsd
  3. Nov 4, 2009 #2
    Re: Inverses

    Both have to be invertible. Inv(AB) = inv(B)inv(A).
     
  4. Nov 4, 2009 #3

    Mark44

    Staff: Mentor

    Re: Inverses

    Right. For example..
    [tex] \left[
    \begin{array}{c c c}
    1&0&0\\
    0&1&0
    \end{array}
    \right]
    \left[
    \begin{array}{c c}
    1&0\\
    0&1\\
    0&0
    \end{array}
    \right]~=~\left[
    \begin{array}{c c}
    1&0\\
    0&1\\
    \end{array}
    \right]
    [/tex]

    The matrix on the right is clearly invertible, while the two matrices in the product aren't event square, let alone invertible.
     
  5. Nov 4, 2009 #4

    Mark44

    Staff: Mentor

    Re: Inverses

    See my counterexample.
     
  6. Nov 4, 2009 #5
    Re: Inverses

    Thanks for the reply. So that would mean that Inv(AB) = inv(B)inv(A) iff inv(B) and inv(A) exist (whereas Inv(AB) may exist without a defined inv(A) or inv(B)), right?
     
  7. Nov 4, 2009 #6
    Re: Inverses

    Looks correct to me. The condition that I am familiar with is:

    If A and B are invertible matrices of the same size, then AB is invertible and (AB)-1 = B-1A-1.

    This is easy to prove by showing that (AB)(B-1A-1) = A(BB-1)A-1 = AIA-1 = AA-1 = I.

    As far as the other way of your "iff", if (AB)-1 = B-1A-1 is given then it seems to me that the existence of B-1 and A-1 would directly follow since they are used in the initial condition.
     
  8. Nov 5, 2009 #7
    Re: Inverses

    I see. Thanks again for the replies. I have an additional question:

    For a matrix C that is not a square matrix, there is no defined inverse; however, it is possible that there is a left inverse A and a right inverse B, in which A =/= B, for the matrix C, right?
     
  9. Nov 5, 2009 #8

    Mark44

    Staff: Mentor

    Re: Inverses

    Yes. See the wikipedia article here, under the section titled Matrices.
     
  10. Nov 5, 2009 #9
    Re: Inverses

    Thanks for the link.

    So in the case of linear systems Ax = b, I suppose it wouldn't always be possible to use the left inverse of A to isolate x as a general method since the left inverse of A does not necessarily exist. How would one isolate x in these linear systems (other than parametrization of x)?
     
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