What is the stationary distribution of this Markov chain? 
A Markov chain is shown in the figure above.
I am writing the transition matrix as:
$$P = \begin{bmatrix}
0.6 & 0.4 & 0 & 0\\
0 & 0 & 0.5 & 0.5\\
0 & 0 & 0.25 & 0.75\\
0.7 & 0.3 & 0 & 0
\end{bmatrix}$$
I using given formula to find stationary distribution:
$$AP=A$$
where $A=[w_1 \; w_2\; w_3\; w_4]$ is the stationary distribution. The answer for stationary distribution is given as:
$$w_1=\frac{4}{12} \quad w_2=\frac{3}{12} \quad w_3=\frac{3}{12} \quad w_4=\frac{2}{12}$$
which I can easily check the formula $$AP=A$$ is not satisfying. Where am I going wrong?
 A: I'm not convinced that the answer claimed is correct.
Your matrix indicates that the chain is regular, so there should be no issue approximating the stationary distribution by taking $P^n$ for large $n$. For instance, WolframAlpha gives
$$P^{500} \approx \begin{bmatrix}
0.396226 & 0.226415 & 0.150943 & 0.226415 \\
0.396226 & 0.226415 & 0.150943 & 0.226415 \\
0.396226 & 0.226415 & 0.150943 & 0.226415 \\
0.396226 & 0.226415 & 0.150943 & 0.226415
\end{bmatrix}$$
suggesting the stationary distribution is
$$A \approx \begin{bmatrix}
0.396226 & 0.226415 & 0.150943 & 0.226415 
\end{bmatrix}$$
and indeed, $AP \approx A$, suggesting this would be the correct stationary distribution.
A: You should be able to get the exact answers.
In the steady state, one more iteration won't make any difference, so
$\displaylines{(3/5)w_1 + (7/10)w_4 = w_1,\\(2/5)w_1+ (3/5)w_4= w_2,\\(1/2)w_2+ (1/4)w_3 = w_3,\\ (1/2)w_2+(3/4)w_3 = w_4\\w_1+w_2+w_3+w_4=1}$
Solving the above system of linear equations, I get
$w_1 = 21/53, w_2 = 12/53, w_3 = 8/53, w_4 = 12/53$
In order to get the exact answer,
you need to  use fractions rather than decimals.
