The map $A\mapsto A^{-1}$ is $C^{\infty}$?

Question

The Inverse Function Theorem states that for a continuously differentiable $f:\Bbb R^n \to \Bbb R^n$ with non-vanishing Jacobian is a homeomorphism around a small neighborhood, and that $$ (f^{-1})'(y)=[f'(f^{-1}(y))]^{-1}. $$ I have seen a stronger claim that if $f$ is $C^{\infty}$ then $f^{-1}$ is also one (i.e. $f$ is a diffeomorphism in that small neighborhood). I observed that we have the relation $$ (f^{-1})' = \Bbb I \circ f' \circ f^{-1} $$ where $\Bbb I:GL(\Bbb R^n)\to GL(\Bbb R^n)$ is the mapping $A\mapsto A^{-1}$. By chain rule and induction, it appears that we only need to show that $\Bbb I$ is $C^{\infty}$. My questions are:

1. What does the chain rule look like for composition with function of class $GL(\Bbb R^n)\to GL(\Bbb R^n)$? Is it the same as the usual $f:\Bbb R^n\to\Bbb R^m$ and $g:\Bbb R^m\to\Bbb R^k$?

2. How does one show that $\Bbb I$ is $C^{\infty}$?

Answer 1

1. The chain rule works the same in this context.
2. This function is a rational function, and therefore a $C^\infty$ function. This is so because $A^{-1}=\frac1{\det A}\operatorname{adj}(A)$, where $\operatorname{adj}(A)$ is the adjugate matrix of $A$.