Hi

I have a equation $$Ax=b,$$ where matrix $A$ is invertible, $b$ is a constant vector, and $x$ is the unknown vector. To obtain $x$, it is obvious $x=A^{-1}b$. Alternatively, if $A$ is Hurwitz, one can build a differential equation $$\dot{x}=Ax-b$$ whose stable equilibrium point is $x=A^{-1}b$. My question is that if $A$ is an arbitrary invertible matrix, are we still able to build a similar differential equation as the above one to obtain $x$? Thanks!