$$\pmatrix{\pi_1 & \pi_2 & \pi_3} = \pmatrix{\pi_1 & \pi_2 & \pi_3}\pmatrix{\frac{1}{6} & \frac{4}{6} & \frac{1}{6} \\ \frac{5}{12} & \frac{2}{12} & \frac{5}{12} \\ \frac{1}{6} & \frac{4}{6} & \frac{1}{6} \\}$$
Answer: $$\pi_1=\frac{5}{18}$$ $$\pi_2=\frac{8}{18}$$ $$\pi_3=\frac{5}{18}$$
Could someone please explain how to find these variables.
The key terms are the Perron-Frobenius theorem and the stable state of a Markov chain. In practice, this vector $\pi$ is found as the eigenvector for the eigenvalue $1$, and is normalized so that the sum of its components is $1$.
Subtract $1$ along the diagonal to get
$$B=\pmatrix{-\frac{5}{6} & \frac{4}{6} & \frac{1}{6} \\ \frac{5}{12} & -\frac{10}{12} & \frac{5}{12} \\ \frac{1}{6} & \frac{4}{6} & -\frac{5}{6} }$$
Since we multiply by the matrix on the right, the desired vector is in the kernel of $B^T$. The Scilab command v=kernel(B')' outputs
v = 0.4682929 0.7492686 0.4682929
This is normalized to have length $1$, but we want the sum of components to be $1$. The command v/sum(v) does the job:
0.2777778 0.4444444 0.2777778
which is your vector. Of course, with sufficient patience one can do it by hand as well.
