# Calculate the inverse of a matrix

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!

• If A is a simple scalar matrix (1x1) of 1, this equation will not have a stable equilibrium, so I would say no – bobuhito Apr 21 '13 at 21:00
• If A=1, we can still do it by building $\dot{x}$=b-Ax, right? – Wei Apr 21 '13 at 21:12
• Wei: It will be unstable. – Misha Apr 21 '13 at 21:32

Your question requires the fixed point to be a node of some variety, so the real parts of the eigenvalues of $A$ must be the same sign.