How to find Thevenin voltage of two-port network from admittance matrix
To calculate the Thevenin voltage the port 2 has to be opened, thus the current is zero: \$ i_2=0 \$.
You get: \$ 0.5⋅u_1+2⋅10^{-4}⋅u_2=0 \to u_2= -2500⋅u_1 \to u_2=10mV*2500=25V\$.
To calculate the Thevenin resistance, do the same but with \$ u_2=0 \to i_2=0.5u_1 \to i_2=5mA\$.
Then, since the voltage drop is all on the Thevenin resistance
$$ R_{th}=\frac{u_2}{i_2} \to \frac{25}{5m}=5k\Omega$$
Notice it is the \$(y_{22})^{-1}\$.