Letting $\Phi$ be a square matrix with continuous, real-valued entries. Assume $\Phi(t)$ is invertible for all $t$ in some interval $\mathcal{l}$.
Show that if $\Phi$ is differentiable, then so is the inverted matrix $\Phi^{-1}$.
p.s. the question comes, I think, from the same book/assignment as this one, but I have no trouble showing the relation given there, I just do not know how to show that $\Phi^{-1}$ is differentiable.
On
Remark that the set of invertible matrices $Gl(n,\mathbb{R})$ is an open subset of $\mathbb{R}^{n^2}$. the map $g(A)=A^{-1}$ is a differentiable map since it is a rational map of the entries.
If $f:I\rightarrow Gl(n,\mathbb{R})$ is differentiable, so is $g\circ f$ since it is a composition of differentiable maps.
Recall that the inverse of a matrix $A$ can be written as
$(\det A)^{-1}\text{adj} A$
where adj$A$ is the adjugate of $A$. Since the determinant is a polynomial of the entries of $A$, it is differentiable if the entries of $A$ are, and therefor its inverse is aswell.
The adjugate of a matrix is has entries which are also polynomial functions of the entries of $A$, and therefore, all the entries of adj$A$ are differentiable aswell, giving that the inverse is differentiable.