Is it possible for a metric to have an undefined term (and still be invertible/non-singular at that point) and if so what would this correspond to physically?
how could we have an undefined term when the overall determinant of the matrix is a constant?
This is actually something that happens a lot, since not all the coordinate system you can choose to describe locally a physical system are globally well defined.
The maybe simplest but most relevant case is the Schwarzschild solution, which describe a spherically symmetric and static vacuum solution of the Einstein field equations. In unit G_N = 1, it can be written in a diagonal form as:
ds^2 = -f(r)dt^2 + f(r)^-1 dr^2 + r^2d\Omega^2,
where \Omega is the solid angle and f(r) = (1-2M/r).
Clearly the determinant is not divergent for any positive r, since in it appears the product f(r)*f(r)^-1. It is also not invertible everywhere, since in r = 0 the determinant is 0. On the other hand, there is a coordinate singularity in r = 2m, which has many interesting physical conseguences, even if its not a real singularity.
To prove that the singularity is only of the coordinate system it is enough to find a coordinate transformation in which the singularity is not present. For the case of Schwarzschild these are, for example, the Kruscal coordinates.
However in most cases it is possible to evaluate the nature of a singularity by studying scalars built on from its curvature. One of the most common and relatively simple to compute is the Kretschmann scalar, which is the contraction of the Rienmann tensor with itself. For the Schwarzschild example, it is not divergent at r=2m, which then is surely a coordinate singularity, and divergent at 0 which instead it is a true singularity.
About the physical interpretation of a coordinate singularity, it strongly depends on the class of observer you are trying to describe with those coordinates. For the case of Schwarzschild, the fact that a metric component blows at r = 2M implies a peculiar behavior for observers at the spatial infinity (r \rightarrow \infty). Locally speaking, an observer A crossing the sphere r=2M will not experience nothing particular, i.e. standard physics will apply for any experiment he will try to perform locally. On the other hand, an observer B at spatial infinity will never see the outcome of any of those experiments because it would require an infinite amount of time for him to see A actually crossing the r = 2M radius.
This example is to explain that the law of physics are still the same at a coordinate singularity, but it is not possible to describe within the same set of coordinate physics on the global space-time.
Edit:
For the Schwarzschild example, it is not divergent at r=2m, which then is surely a coordinate singularity, and divergent at 0 which instead it is a singularity.
As pointed out by @AndrewSteane the above sentence is wrong, in particular the word surely, since a non-diverging curvature invariant does not imply that a singularity is not true. Only the opposite applies, a diverging curvature invariant implies the existence of a true singularity. For the Scharzschild case, however, the existence of a coordinate system (the kruskal one) in which the singularity r=2m does not show implies that indeed it is a curvature singularity, and thats why its Kretschmann scalar does not diverge.
