Transition matrix M = $\left[\begin{matrix} A & 0 & 1/2 & 0 \\ B & 1 & 0 & 1 \\ C & 1 & 1/2 & 0 \\ \end{matrix}\right]$

Page Rank

$r = \beta M.r + (1- \beta) e/n$

$\beta = 0.8 = 4/5$

$(1 - \beta) = 0.2 = 1/5$

e/n = $\left[ \begin{matrix} 1/3 \\ 1/3 & \\ 1/3\\ \end{matrix}\right]$

Iteration 1:

$r^1 = \frac{4}{5} \left[\begin{matrix} 0 & 1/2 & 0 \\ 1 & 0 & 1 \\ 0 & 1/2 & 0 \\ \end{matrix}\right] \left[ \begin{matrix} 1/3 \\ 1/3 & \\ 1/3\\ \end{matrix}\right] + \frac{1}{5} \left[ \begin{matrix} 1/3 \\ 1/3 & \\ 1/3\\ \end{matrix}\right]$

= $\left[ \begin{matrix} 4/30 \\ 4/30 & \\ 4/30\\ \end{matrix}\right] + \left[ \begin{matrix} 1/15 \\ 1/15 & \\ 1/15 \end{matrix}\right]$

$r^1 = \left[ \begin{matrix} 6/30 \\ 10/30 & \\ 6/30 \\ \end{matrix}\right] = \left[ \begin{matrix} 3/15 \\ 5/15 & \\ 3/15 \\ \end{matrix}\right]$

Iteration 2:

$r " = \frac{4}{5} \left[\begin{matrix} 0 & 1/2 & 0 \\ 1 & 0 & 1 \\ 1 & 1/2 & 0 \\ \end{matrix}\right]$ $ \left[ \begin{matrix} 3/15 \\ 5/15 & \\ 3/15 \\ \end{matrix}\right] + \left[ \begin{matrix} 1/15 \\ 1/15 & \\ 1/15 \\ \end{matrix}\right]$

= $\left[ \begin{matrix} 2/15 \\ 8/25 & \\ 2/15 \\ \end{matrix}\right] + \left[ \begin{matrix} 1/15 \\ 1/15 & \\ 1/15 \\ \end{matrix}\right]$

= $\left[ \begin{matrix} 3/15 \\ 29/75 & \\ 3/15 \\ \end{matrix}\right]$

page rank after two iterations:

$r = \left[ \begin{matrix} 3/15 \\ 29/75 & \\ 3/15 \\ \end{matrix}\right]$