Why must metric tensor be invertible? The metric can be written as a matrix, but why must this matrix be invertible? At the points where the matrix is singular, why is the metric not defined?

This worried me at one time as well. The way I thought about it was by working at a fixed point and using the Gram-Schmidt process for inner products on the coordinate basis $\partial_1,...,\partial_n$ to produce an orthonormal basis $e_1,...,e_n$. It's a standard and easy fact that the matrices that represent these bilinear forms are related by $I=A^tgA$, where $I$ is the identity matrix (since $\{e_i\}$ is orthonormal), $g$ is the metric $g_{ij}$ with respect to the $\{\partial_i\}$ basis, and $A$ is the change of basis matrix taking the basis of partials to the orthonormal basis. Taking the determinant of both sides of this equation, and using the fact that $\det(A)=\det(A^t)\neq 0$, we see that $\det(g)\neq 0$ so $g$ is invertible.
