The markov chain with states {0,1,2,3,4} has transition matrix $$P=\begin{pmatrix} 0 & p & 0 & 0 & 1-p \\ 1-p & 0 & p & 0 & 0 \\ 0 & 1-p & 0 &p &0 \\ 0 & 0 &1-p &0 & p \\ p &0 &0 &1-p & 0 \end{pmatrix}$$ This chain is irreducible and finite which menas that it has a unique stationary distribution, since it's also aperiodic we have a limiting distribution.
Im trying to calculate this limiting distribution, but it I've been stuck with the equations, could somebody help me out?
I know that I should solve $$\begin{cases} \pi P= \pi \\ \sum_{k=0}^4 \pi_k =1 \end{cases} $$ Which gives $$\begin{cases} \pi_1(1-p)+\pi_4p=\pi_0 \\ \pi_0p+\pi_2(1-p)=\pi_1 \\ \pi_1p+\pi_3(1-p)=\pi_2 \\ \pi_2p+\pi_4(1-p)=\pi_3 \\ \pi_0(1-p)+\pi_3p=\pi_4 \\ \pi_0+\pi_1+\pi_2+\pi_3+\pi_4=1 \end{cases} $$ How do I solve this system of equations, it doesnt look to easy?