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If $M$ and $\tilde M$ are invertible square matrices which are almost the same (you get to pick the norm) $$\tilde M-M<\epsilon$$ Then can we say that their inverses are almost the same (of course)? Here's the catch: can we have the upper-bound on the error be a function of the determinant? $$\tilde M^{-1}-M^{-1}<f(\det M,\epsilon)$$ Of course, you get to choose the norms.

Dave
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The perturbation theory of inverses is based on condition numbers. If $M$ and $\tilde{M}$ are nonsingular such that $\|\tilde{M}-M\|\leq\epsilon\|M\|$ then $$ \frac{\|\tilde{M}^{-1}-M^{-1}\|}{\|\tilde{M}^{-1}\|}\leq \epsilon\,\kappa(M), $$ where $\kappa(M)=\|M\|\|M^{-1}\|$ is the condition number of $M$. If we have $\epsilon\,\kappa(M)<1$, then also $$ \frac{\|\tilde{M}^{-1}-M^{-1}\|}{\|M^{-1}\|}\leq\frac{\epsilon\,\kappa(M)}{1-\epsilon\,\kappa(M)}. $$ Since the bounds are sharp, a relevant question would be:

Is there a relation between the condition number and determinant?

Consider the matrix $\infty$-norm $\|\cdot\|_{\infty}$ and the associated condition number $\kappa_{\infty}(X)=\|X\|_{\infty}\|X^{-1}\|_{\infty}$. The following two examples show that there's obviously no direct link between the determinant and the condition number:

1) Let $$ X=\begin{bmatrix} 1 & -1 & -1 & \cdots & -1 \\ & 1 & -1 & \cdots & -1 \\ & & \ddots & \cdots & \vdots \\ & & & 1 & -1 \\ & & & & 1 \end{bmatrix}\in\mathbb{R}^{n\times n}. $$ Then $$ \|X\|_\infty=n, \quad \|X^{-1}\|_{\infty}=2^{n-1}, \quad \kappa_{\infty}(X)=n2^{n-1}, \quad \det(X)=1. $$ So $\kappa_{\infty}(X)$ can be very large even for small $n$ while $\det(X)$ remains equal to 1 independently of $n$.

2) Let $$ X=\alpha I\in\mathbb{R}^{n\times n}, \quad \alpha\in\mathbb{R}\setminus\{0\}, $$ where $I$ is the identity matrix. Then $$ \|X\|_{\infty}=|\alpha|, \quad \|X^{-1}\|_{\infty}=\frac{1}{|\alpha|}, \quad \kappa_{\infty}(X)=1, \quad \det(X)=|\alpha|^n. $$ So while the matrix $X$ is "ideally" conditioned for any feasible $\alpha$ and $n$, its determinant can be made arbitrarily large or small if $\alpha\neq 1$.