Relationships between Parameters

Example 2: Obtain the $ y $ parameters of the op amp circuit in Fig. 1. Show that the circuit has no $ z $ parameters. Solution:
Since no current can enter the input terminals of the op amp, $ \mathbf{I}_{1}=0 $, which can be expressed in terms of $ \mathbf{V}_{1} $ and $ \mathbf{V}_{2} $ as $$\mathbf{I}_{1}=0 \mathbf{V}_{1}+0 \mathbf{V}_{2}$$ Where it gives $$\mathbf{y}_{11}=0=\mathbf{y}_{12}$$ Also, $$\mathbf{V}_{2}=R_{3} \mathbf{I}_{2}+\mathbf{I}_{o}\left(R_{1}+R_{2}\right)$$ where $ \mathbf{I}_{o} $ is the current through $ R_{1} $ and $ R_{2} $. But $ \mathbf{I}_{o}=\mathbf{V}_{1} / R_{1} $. Hence, $$\mathbf{V}_{2}=R_{3} \mathbf{I}_{2}+\frac{\mathbf{V}_{1}\left(R_{1}+R_{2}\right)}{R_{1}}$$ which can be written as $$\mathbf{I}_{2}=-\frac{\left(R_{1}+R_{2}\right)}{R_{1} R_{3}} \mathbf{V}_{1}+\frac{\mathbf{V}_{2}}{R_{3}}$$ Comparing this with admittance equation shows that $$\mathbf{y}_{21}=-\frac{\left(R_{1}+R_{2}\right)}{R_{1} R_{3}}, \quad \mathbf{y}_{22}=\frac{1}{R_{3}}$$ The determinant of the $ [\mathbf{y}] $ matrix is $$\Delta_{y}=\mathbf{y}_{11} \mathbf{y}_{22}-\mathbf{y}_{12} \mathbf{y}_{21}=0$$ Since $ \Delta_{y}=0 $, the $ [\mathbf{y}] $ matrix has no inverse; therefore, the $ [\mathbf{z}] $ matrix does not exist according to Eq. (A). $$[y] = [z]^{-1} \tag{A}$$ Note that the circuit is not reciprocal because of the active element.