Derivation of inverse matrix

Question

My question is about how can reach to the formula $$\frac{dA^{-1}}{dt}=-A^{-1}\frac{dA}{dt}A^{-1}$$ when $A$ is a matrix. It is very similar to $$\frac{dx^{-1}}{dt}=-x^{-2}\frac{dx}{dt}$$And I know we can write $$A^{-1}=A^{-1}I\\A^{-1}=A^{-1}AA^{-1}$$ then it seems to be $$\frac{dA^{-1}}{dt}=\frac{dA^{-1}}{dt}AA^{-1} +A^{-1}\frac{dA}{dt}A^{-1}+A^{-1}A\frac{dA^{-1}}{dt}$$ in this part: how can I simplify right hand side? to obtain $\frac{dA^{-1}}{dt}=-A^{-1}\frac{dA}{dt}A^{-1}$
Any hint will be appreciated.

Answer 1

from your bottom line: $$\frac{dA^{-1}}{dt}=\frac{dA^{-1}}{dt}I+A^{-1}\frac{dA}{dt}A^{-1}+I\frac{dA^{-1}}{dt} \\=2I\frac{dA^{-1}}{dt}+A^{-1}\frac{dA}{dt}A^{-1}$$ which rearranges to: $$-\frac{dA^{-1}}{dt}=A^{-1}\frac{dA}{dt}A^{-1}$$ which is what you were looking for :)

Answer 2

Start with the identity $$ \def\A{A^{-1}} I \;=\; \A A $$ Differentiate each side $\left({\rm using}\;dA\;{\rm to\ denote}\;{\large\frac{dA}{dt}}\right)$ $$ 0 \;=\; d\A A \;+\; \A\,dA $$ Solve for $d\A$ $$ d\A = - \A\;dA\;\A $$