A Markov Chain Flea Problem A Flea moves around the vertices of a triangle in the following manner: Whenever it is at vertex i it moves to its clockwise neighbor vertex with probability $p_i$ and to the counterclockwise neighbor with probability $q_{i}=1−p_{i}$, for $i=1,2,3$. 
(a) Find the proportion of time that the flea is at each of the vertices.
I know how to set up the system of equations and am having trouble solving it.
$$
\left\{
\begin{aligned} 
q_2 π_2+p_3 π_3&= π_1 \\ 
p_1 π_1+q_3 π_3&= π_2 \\ 
q_1 π_1+p_2 π_2&= π_3  \\
 π_1+ π_2+ π_3 &= 1
\end{aligned} 
\right. 
$$
I got $$\begin{cases}π_1 = π_2 (1+p_2)/(1+p1) \\\\ π_1 = π_2 (1-p_3p_2)/(p_3q_1 + p_1)\end{cases}$$
In which case the $π_2$ would cancel out.
 A: Using the fact that $p_i+q_i=1$ for $i=1,2,3$ you can solve the system of 4 equations in the three unknowns (which of course means that one equation will be linearly dependent and thus redundant) as follows
$$\begin{align*}&\begin{cases} 
q_2 π_2+p_3 π_3&=& π_1 \\ 
p_1 π_1+q_3 π_3&=& π_2 \\ 
q_1 π_1+p_2 π_2&=& π_3  \\
 π_1+ π_2+ π_3 &=& 1
\end{cases} \\\implies& \left\lbrace\begin{array}{rcrcrcl} 
-π_1&+&q_2 π_2&+&p_3 π_3&=& 0 \\ 
p_1 π_1&-&π_2&+&q_3 π_3&=& 0 \\ 
q_1 π_1&+&p_2 π_2&-&π_3&=& 0  \\
 π_1&+& π_2&+& π_3 &=& 1
\end{array}\right.\overset{R_2+R_3}{\underset{R_2+R_1}\implies} \left\lbrace\begin{array}{rcrcrcl} 
-q_1π_1&-&p_2 π_2&+&π_3&=& 0 \\ 
p_1 π_1&-&π_2&+&q_3 π_3&=& 0 \\ 
π_1&-&q_2π_2&-&p_3π_3&=& 0  \\
 π_1&+& π_2&+& π_3 &=& 1
\end{array}\right.\\\overset{R_2-R_3}\implies &\left\lbrace\begin{array}{rcrcrcl} 
-q_1π_1&-&p_2 π_2&+&π_3&=& 0 \\ 
p_1 π_1&-&π_2&+&q_3 π_3&=& 0 \\ 
-q_1π_1&-&p_2π_2&+&π_3&=& 0  \\
 π_1&+& π_2&+& π_3 &=& 1
\end{array}\right. \overset{\not R_3}{\underset{R_2+R_4-R_1}\implies} \left\lbrace\begin{array}{rcrcrcl} 
2π_1&+&p_2 π_2&+&q_3π_3&=& 1 \\ 
p_1 π_1&-&π_2&+&q_3 π_3&=& 0 \\ 
 π_1&+& π_2&+& π_3 &=& 1
\end{array}\right. \\\overset{R_1-R_2}{\underset{R_2+R_3}\implies}& \left\lbrace\begin{array}{rcrcrcl} 
(2-p_1)π_1&+&(1+p_2)π_2&&&=& 1 \\ 
(1+p_1)π_1&&&+&(1+q_3)π_3&=& 1 \\ 
 π_1&+& π_2&+& π_3 &=& 1
\end{array}\right. \\\implies& \left\lbrace\begin{array}{rcrcrcl} 
&π_2&=& \frac{1-(1+q_1)π_1}{1+p_2} \\ 
&π_3&=& \frac{1-(1+p_1)π_1}{1+q_3} \\ 
&π_1&=& 1-π_2-π_3
\end{array}\right.
\implies \left\lbrace\begin{array}{rcrcrcl} 
&π_2&=& \frac{1-p_3q_1}{2+p_1p_3+q_2q_1-p_3q_1} \\ 
&π_3&=& \frac{1-p_1q_2}{2+p_3p_1+q_3q_2-p_1q_2} \\ 
&π_1&=& \frac{1-p_2q_3}{2+p_1p_2+q_1q_3-p_2q_3}
\end{array}\right. \end{align*}$$
