Relating the Error in Matrix Inversion to the Determinant 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.
 A: 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 $\alpha\neq 1$.
