I can find the equilibrium probabilities for a 2 x 2 matrix:

$\displaystyle (\pi_0, \pi_1) = (\pi_0 , 1 - \pi_0 )\[ \left( \begin{array}{cc}

0.2 & 0.8 \\

0.4 & 0.6 \end{array} \right)\]$

$\displaystyle \pi_0(0.2)+(1- \pi_0)0.4= \pi_0$

$\displaystyle \pi_0 = 0.4/1.2=1/3$ and $\displaystyle \pi_1 = 0.8/1.2=2/3$

But how do I find the probabilities for a 3 x 3 matrix?

Can somebody start me off on:

$\displaystyle \[ \left( \begin{array}{ccc}

0 & 1 & 0\\

0 & 0.4 & 0.6\\

0.4 & 0 & 0.6 \end{array} \right)\]$