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Finding Inverse from Known Linear System

  1. Aug 1, 2012 #1
    Hello All:
    Suppose I have a completely known linear system: A*x=b. I know the matrix A, and an x and the associated RHS vector b (and it is non-trivial). Is there some tricky way to directly determine the inverse of A without performing an inversion by typical means (Gauss elimination, LU, etc) ?

    Thanks much.
  2. jcsd
  3. Aug 1, 2012 #2
  4. Aug 1, 2012 #3
    True. But my situation is that it is a large system. To use Cramer's rule would be expensive. I was wondering if because I have a known transformation (a known X and its corresponding B vector), that there may be something else that could be done.
    For example, if I write
    x = A^-1 * b
    Then perform a dyadic product with a known vector, d, (of my choosing):
    xd = A^-1 * bd
    (xd)((bd)^-1) = A^-1

    Something like this. However I know that you can't take an inverse of a dyad. This was my idea, but I just haven't used tensors in a long while and don't even know if what I am asking is possible.
  5. Aug 1, 2012 #4


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    hey Blue, try using the LaTeX feature here. it helps you express your mathematical thinking and it helps us read what you express accurately.

    do you simply want to solve for [itex] \mathbf{x} [/itex]? or do you want [itex] \mathbf{A}^{-1} [/itex]?
  6. Aug 1, 2012 #5
    Sorry. Yes I want [itex]\mathbf{A}^{-1}[/itex] (actually, particular entries in [itex]\mathbf{A}^{-1}[/itex]) . I have A and an x,b pair. My math from earlier (which I know you can't do unless the dyad is complete... but it was my original thinking):

    [itex]\mathbf{x}[/itex] = [itex]\mathbf{A}^{-1}[/itex][itex]\mathbf{b}[/itex]
    Make dyadic product with clever vector [itex]\mathbf{d}[/itex]:
    [itex]\mathbf{xd}[/itex] = [itex]\mathbf{A}^{-1}[/itex][itex]\mathbf{bd}[/itex]
    [itex]\mathbf{(xd)}[/itex][itex]\mathbf{(bd)}^{-1}[/itex] = [itex]\mathbf{A}^{-1}[/itex]

    My thought was that [itex]\mathbf{d}[/itex] could be chosen so that [itex]\mathbf{(bd)}^{-1}[/itex] might be simple. Again the tensor math here is not correct, I just wanted to throw out my thought.
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