The network admittance matrix of a four-bus power system is presented in the following. Determine the updated network admittance matrix if the second and the third buses are short-circuited:

\[
\left[\mathrm{Y}_{\text {Bus }}ight]=j\left[\begin{array}{cccc}
-5 & 4 & 3 & 2 \\
4 & -10 & 2 & 1 \\
3 & 2 & -10 & 4 \\
2 & 1 & 4 & -20
\end{array}ight] \text { p.u. }
\]

1) \(j\left[\begin{array}{ccc}-5 & 7 & 2 \\ 7 & -16 & 5 \\ 2 & 5 & -20\end{array}ight]\) p.u.

2) \(j\left[\begin{array}{ccc}-10 & 7 & 2 \\ 7 & -10 & 5 \\ 2 & 5 & -10\end{array}ight]\) p.u.

3) \(j\left[\begin{array}{ccc}-16 & -5 & 6 \\ -5 & -10 & 5 \\ 6 & 5 & -20\end{array}ight]\) p.u.

4) \(j\left[\begin{array}{ccc}0 & 10 & 20 \\ 10 & -10 & 6 \\ 20 & 6 & -20\end{array}ight]\) p.u.