TITLE: Alternative proof for continuity of matrix inversion
QUESTION [5 upvotes]: I am to show that 
$\mathrm{inv}: \mathrm{GL}_{n \times n}(\mathbb{R}) \to \mathrm{GL}_{n \times n}(\mathbb{R}); A \mapsto A^{-1}$ 
is a continuous function. I have shown this by showing that the deteriminant is continuous, and that forming the adjunct is continuous, in which case one can apply the formula 
$A^{-1} = \frac{1}{\mathrm{det}(A)}\mathrm{adj}(A)$. 
This is all fine, but I would love to see a more intuitive proof that doesn’t simply rely on showing that some formula is composed of continuous functions. I tried finding some sort of relation between the operator norms of $A$ and $A^{-1}$ but couldn’t find anything that holds in the general case. For example, if something along the lines of 
$\min_{x \in \mathbb{S}^{n-1}}{(Ax)} \times \max_{x \in \mathbb{S}^{n-1}}{(A^{-1}x)} = 1$ 
were true, one could construct an epsilon-delta proof via that. However, the above expression while inutive, appears to be false (it seems to hold only if the chosen x happen to be eigenvectors unless my attempts to verify it in mathematica were somehow wrong). Does anyone know of a more intuitive proof of the statement than the one I outlined above?

REPLY [5 votes]: In finite dimension, also works the implicit function theorem. The matrix product $(X,Y)\longmapsto XY$ is bilinear and continuous, so $C^1$. The inverse is given by the equation $XA = I$ Check that
$$\frac{\partial(XA - I)}{\partial A}$$
is inversible and this proves that $A\longmapsto X = A^{-1}$ is $C^1$.