Electronic – How to find the resistance between three terminals

It's actually quite straightforward to find the Thevenin impedance of this circuit.

The equivalent impedance looking into the port ab is defined by:

$$Z_{ab} \equiv \frac{V_{ab}}{I_{ab}} = \frac{V_{ab}}{i_{\Delta}} = Z_{th}$$

But you can write by inspection a simple KVL equation for \$V_{ab}\$ in terms of \$i_{\Delta}\$:

$$V_{ab} = i_{\Delta}(-jX_{1uF} + 10k\Omega) + 200 i_{\Delta} \cdot 100 \Omega) = i_{\Delta}(-jX_{1uF} + 30k)\Omega$$

Generally speaking, to find the Thevenin equivalent of a circuit with only dependent source(s), you must be sure to "activate" the dependent source(s) with a test source.

This is what was done above. We solved for the voltage across the port due to a test current source, \$I_{ab}\$ which, in this case, equals the controlling variable thus making this problem particularly easy to solve.

Consider \$N = 3\$. The resistance \$R_{12}\$ would be $$ R_{12} = X_{12} || (X_{13} + X_{23}) = \frac{X_{12} (X_{13} + X_{23})}{X_{12} + X_{23} + X_{13}} $$ This is a problem - your matrix multiplication can only make terms that look like $$ R_{ij} = a X_{12} + b X_{13} + c X_{23} $$ where \$a\$, \$b\$, and \$c\$ are constants, so you can't write the first equation in matrix form. That means the method you've suggested won't work - you'll need to do this without linear algebra.

There might be a method that skips this matrix multiplication (something closer to star-mesh transforms), but I'm not seeing it...