# Why must we turn the admittance into Y matrix, then turn the Y matrix into Z matrix before we calculate the fault current?

The one-line diagram is as below,and a three-phase fault has occurred in the bus 1,now find the fault current

\begin{align} Y=\begin{bmatrix}Y_{11}&Y_{12}\\Y_{21}&Y_{22}\end{bmatrix}=\begin{bmatrix}-j\frac{1}{0.15}-j\frac{1}{(0.1+0.1024+0.1)}&j\frac{1}{(0.1+0.1024+0.1)}\\j\frac{1}{(0.1+0.1024+0.1)}&-j\frac{1}{0.2}-j\frac{1}{(0.1+0.1024+0.1)}\end{bmatrix} \end{align}

\begin{align} Z=Y^{-1}=\begin{bmatrix} j0.1155 & j0.04598\\ j0.04598 & j0.13869\end{bmatrix} \end{align}

$$\I_f=\frac{V_f}{Z_{11}}=-j8.657 p.u.\$$

1.Why does the $$\Y_{11}\$$ just ignore the motor admittance,$$\\frac{-j1}{0.2}\$$ ?
2.$$\Y_{11}\$$ includes the admittance of $$\Tr_1\$$,$$\Tr_2\$$ and the generator itself,but why is $$\Y_{11}=-j\frac{1}{0.15}-j\frac{1}{(0.1+0.1024+0.1)}\$$,instead of $$\Y_{11}=-j\frac{1}{0.15}-j\frac{1}{0.1}-j\frac{1}{0.1024}-j\frac{1}{0.1}\$$ ?
3.we will calculate the fualt current $$\I_f=\frac{V_f}{Z}\$$,but $$\X\$$ is admittance,and admittance is equal to impedance when the resistance is zero,so why don't we just add all of the admittance directly ? I mean $$\Z=0.15+0.1+0.1024+0.1\$$,apparently it is wrong ,but why? why can't we use this method?