We are currently looking into using a symmetric closeness definition for numpy.isclose. To judge the impact this move would have on other libraries we ran scipy's test suite against a modified numpy.

While doing this a lot of tests emitted this warning

because it relies on the asymmetric definition.

I've tracked it down to #9081 and specifically #9081 (comment) (cc @Kai-Striega @mdhaber). If I understand the intention correctly it should check if the pivval is significantly greater than the tolerance. If that is the case

if abs(pivval) <= tol * 1e4:
    ...

would be a better choice, because

1. it communicates the intent better, and
2. is safer since np.isclose basically checks abs(pivval - tol) <= ... which might lead to unwanted behavior in an edge case.