Mean first passage time of a Markov Chain If I have this matrix:$$ P=
        \begin{pmatrix}
        0.2 & 0.3 & 0.5 \\
        0.5 & 0.1 & 0.4 \\
        0.3 & 0.3 & 0.4 \\
        \end{pmatrix}
$$
How do I find the mean first passage time $m_0,_2$ and $m_2,_0$?
I know that $m_{i,j}$ = $1 +$ $\sum_{k≠j} P_{i,k}m_{k,j}$ so, is this how you would go about this:
$m_2,_0$ = $1 + P_2,_1m_1,_0 + P_2,_3m_3,_0$ = $1 + (0.5)$ $m_1,_0$? + $(0.4)$ $m_3,_0$? Then, where do I go from here? 
 A: Here's how to do this by solving for all the unknown mean first-passage times element wise.
\begin{align}
\left\{\begin{array}{}
\left.\begin{array}{}
m_{10}=1+P_{11\,}m_{10}+P_{12\,}m_{20},\\
m_{20}=1+P_{21\,}m_{10}+P_{22\,}m_{20},
\end{array}\,\right\}\rightarrow\mbox{ to State }0\\
\left.\begin{array}{}
m_{01}=1+P_{00\,}m_{01}+P_{02\,}m_{21},\\
m_{21}=1+P_{20\,}m_{01}+P_{22\,}m_{21},\\
\end{array}\,\right\}\rightarrow\mbox{ to State }1\\
\left.\begin{array}{}
m_{02}=1+P_{00\,}m_{02}+P_{01\,}m_{12},\\
m_{12}=1+P_{10\,}m_{02}+P_{11\,}m_{12}.
\end{array}\,\right\}\rightarrow\mbox{ to State }2
\end{array}\right.
\end{align}
The equations with different destinations (second index) are uncoupled and can be solved separately. So to find $m_{02}$ you solve the $2$ coupled equations for $m_{02},m_{12}$, and to find $m_{20}$ you solve the $2$ coupled equations for $m_{10},m_{20}$. I'll let you work out the numbers. This method is equivalent to the absorbing state method, in which you set the destination as an absorbing state.
