Take a linear system of the form $Ax=b$. Usually, for obvious reasons, we want to find $x$ given $A$ and $b$. As you all know, the solutions may not be unique, or exist, and many algorithms have been proposed to find them effectively for different $A$.
Are there any methods to generate $A$ given $x$ and $b$? I realise that, of course, a) this might not be a useful problem to consider, and b) $A$ is of course not unique. Apologies if c) I'm missing an obvious trick to generate a trivial $A$.
Here's some background about my problem: I'm solving a large linear system with an LSQR algorithm. Here, $A$ is about $4000\times2000$, not sparse, and ill conditioned -- $\text{cond}(A)$ is about $10^{10}$. It transpires that my problem is "semi convergent", in the sense that this large condition number means that any attempt to solve it eventually fails due to effects of numerical precision. The literature states that people take enough iterations of LSQR to decrease the residual a lot, and then give up as further iterations serve to make the solution worse; commonly the number of iterations, $n$, used is only 4-10; for my problem, all hell breaks loose when $n>10$ and $n=4$ is, amazingly, pretty good.
I've had success building some physics into my model such that every evaluation of $A$ is very slightly (but deterministically) different, with changes that are physically motivated. The solution is quantitatively and qualitatively "much better", and is still good for $n>200$. I am certain that the effect of my modifications to $A$ are essentially preconditioning it with something like a Fourier matrix multiplied by a filter that kills high frequency noise. I'd like to know what the condition number of my 'equivalent' $A$ is, and moreover just how different a given $A$ is from the problem I started with.
Apologies if this should be on another SE instead, but my fundamental question is one of linear algebra.
EDIT: For my problem, $x$ and $b$ are just vectors $\in\mathbb{C}$, and $A$ is a complex rectangular matrix. I'd of course be interested in general answers, however.