Вот ещё немножко контекста для capacitance-only-tie-sets и inductance-only-cut-sets.
Since the presence of capacitance-only-tie-sets is solely
responsible for the non-vanishing of the matrix

, it is clear that if
one connects a small linear resistance in series with one of the capacitances
in each capacitance-only-tie-set, the matrix

becomes zero.
Again, the small resistance can be chosen small enough so as to simulate
the lead resistance.
Since the presence of inductance-only-cut-sets is solefy responsible
for the non-vanishing of the matrix

it is clear that if one connects
a large linear resistance in parallel with one of the inductances in each
inductance-only-cut-set, the matrix

becomes zero. Again, the large
resistance can be chosen large enough so as to simulate the insulator resistance.