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If I have a square invertible matrix and I take its determinant, and I find that $\det(A) \approx 0$, does this imply that the matrix is poorly conditioned?
Is the converse also true? Does an ill-conditioned matrix have a nearly zero determinant?
Here is something I tried in Octave:
a = rand(4,4);
det(a) %0.008
cond(a)%125
a(:,4) = 1*a(:,1) + 2*a(:,2) = 0.000000001*ones(4,1);
det(a)%1.8E-11
cond(a)%3.46E10
It's the largeness of the condition number $\kappa(\mathbf A)$ that measures the nearness to singularity, not the tininess of the determinant.
For instance, the diagonal matrix $10^{-50} \mathbf I$ has tiny determinant, but is well-conditioned.
On the flip side, consider the following family of square upper triangular matrices, due to Alexander Ostrowski (and also studied by Jim Wilkinson):
$$\mathbf U=\begin{pmatrix}1&2&\cdots&2\\&1&\ddots&\vdots\\&&\ddots&2\\&&&1\end{pmatrix}$$
The determinant of the $n\times n$ matrix $\mathbf U$ is always $1$, but the ratio of the largest to the smallest singular value (i.e. the 2-norm condition number $\kappa_2(\mathbf U)=\dfrac{\sigma_1}{\sigma_n}$) was shown by Ostrowski to be equal to $\cot^2\dfrac{\pi}{4n}$, which can be seen to increase for increasing $n$.
As $\det(kA)=k^n\det A$, the determinant can be made arbitrarily large or small by simple rescaling (which doesn't change the condition number). Especially in high dimensions, even scaling by an innocent factor of 2 changes the determinant by a huge amount.
Thus never use the determinant to assess condition or closeness to singularity.
On the other hand, for almost all well-posed numerical problems, the condition is closely related to the distance to singularity, in the sense of the smallest relative perturbation needed to make the problem ill-posed. In particular, this holds for linear systems.
