Finding Inverse from Known Linear System

[Blue2Sky]

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.

 

[antibrane]

You could use Cramer's rule.

 

[Blue2Sky]

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
Then:
(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.
Thanks

 

[rbj]

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 $\mathbf{x}$? or do you want $\mathbf{A}^{-1}$?

 

[Blue2Sky]

Sorry. Yes I want $\mathbf{A}^{-1}$ (actually, particular entries in $\mathbf{A}^{-1}$) . 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):

$\mathbf{x}$ = $\mathbf{A}^{-1}$$\mathbf{b}$
Make dyadic product with clever vector $\mathbf{d}$:
$\mathbf{xd}$ = $\mathbf{A}^{-1}$$\mathbf{bd}$
then,
$\mathbf{(xd)}$$\mathbf{(bd)}^{-1}$ = $\mathbf{A}^{-1}$

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