6
$\begingroup$

At first I thought it should be LinearSolve, which however turns out aiming at x, given m and b in m.x == b.

$\endgroup$

1 Answer 1

5
$\begingroup$

Yes, but the solution is not unique. In some sense, the optimal solution is the following rank-one matrix:

m = KroneckerProduct[b, x/x.x]
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.