G-S one iteration, generator at bus 3

Bus 1 is the slack bus ($1.05\angle 0^\circ$), bus 2 a PQ (load) bus with $P_2 = -3$ pu, $Q_2 = -1.5$ pu, and bus 3 a PV bus with $P_3 = 2$ pu, $|V_3| = 1.03$ pu. Initial estimates: $V_2^{(0)} = 1.0\angle 0^\circ$; for the PV bus the specified magnitude is kept, $V_3^{(0)} = 1.03\angle 0^\circ$. Bus 2 is solved first, then bus 3 with the new $V_2$.

Y-bus

$$\begin{aligned} y_{12} &= 1/j0.06 = -j16.6667 \\ y_{13} &= 1/j0.05 = -j20.0000 \\ y_{23} &= 1/j0.08 = -j12.5000 \end{aligned}$$ $$Y_{bus} = \begin{bmatrix} -j36.6667 & j16.6667 & j20.0000 \\ j16.6667 & -j29.1667 & j12.5000 \\ j20.0000 & j12.5000 & -j32.5000 \end{bmatrix}\ \text{pu}$$

First Gauss-Seidel iteration

Iteration 1

$$\begin{aligned} V_2^{(1)} &= \frac{1}{Y_{22}}\left[\frac{P_2 - jQ_2}{V_2^{*}} - Y_{21}V_1 - Y_{23}V_3\right] \\ \frac{P_2 - jQ_2}{V_2^{*}} &= \frac{-3.0000 + j1.5000}{1.0000} = -3.0000 + j1.5000 \\ \textstyle\sum_{k\ne 2} Y_{2k}V_k &= j30.3750 \\ V_2^{(1)} &= \frac{(-3.0000 + j1.5000) - (j30.3750)}{-j29.1667} = 0.9900 - j0.1029 \\ &= 0.9953\angle -5.9315^\circ\ \text{pu} \end{aligned}$$ $$\begin{aligned} \sum_{k=1}^{3} Y_{3k}V_k &= 1.2857 - j0.1000 \\ Q_3^{(1)} &= -\text{Im}\{V_3^{*}\textstyle\sum_k Y_{3k}V_k\} = 0.1030\ \text{pu} \end{aligned}$$ $$\begin{aligned} V_3^{(1)} &= \frac{1}{Y_{33}}\left[\frac{P_3 - jQ_3}{V_3^{*}} - Y_{31}V_1 - Y_{32}V_2\right] \\ \frac{P_3 - jQ_3}{V_3^{*}} &= \frac{2.0000 - j0.1030}{1.0300} = 1.9417 - j0.1000 \\ \textstyle\sum_{k\ne 3} Y_{3k}V_k &= 1.2857 + j33.3750 \\ V_3^{(1)} &= \frac{(1.9417 - j0.1000) - (1.2857 + j33.3750)}{-j32.5000} = 1.0300 + j0.0202 \\ \delta_3 &= 1.1227^\circ \Rightarrow V_3^{(1)} = 1.0300\angle 1.1227^\circ = 1.0298 + j0.0202\ \text{(magnitude reset to }1.03\text{)} \end{aligned}$$

Result after one iteration

Answer: $V_2 = 0.9953\angle -5.9315^\circ$ pu, $V_3 = 1.0300\angle 1.1227^\circ$ pu (with $Q_3 = 0.1030$ pu).