Solving Backwards Chapman-Kolmogorov Questions

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A flea hops on the vertices A, B and C of a triangle. Each hop takes it from one vertex to the next and the times between successive hops are independent random variables, each with an exponential distribution with mean $\frac{1}{\lambda}$. Each hop is equally likely to be in the clockwise direction or in the anticlockwise direction. Find the probability that the flea is at vertex A at a given time t > 0, starting from A at time t = 0.

I have to solve this using Backwards Kolmogorov equations but it's just so messy that I don't have an intuition on what equations to find since there are $9$ of them...
I want $P_{AA}(t)$ and currently I have $$P'_{AA}(t) = \lambda(P_{BA}(t) + P_{CA}(t) - 2P_{AA}(t))\\ P'_{BA}(t) = \lambda(P_{AA}(t) + P_{CA}(t) - 2P_{BA}(t))\\ P'_{CA}(t) = \lambda(P_{AA}(t) + P_{BA}(t) - 2P_{CA}(t)). $$

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The ODE system observes the following equivalent matrix form $$ \frac{\rm d}{{\rm d}t}\left( \begin{array}{c} P_{AA}\\ P_{BA}\\ P_{CA} \end{array} \right)=-\lambda\left( \begin{array}{ccc} 2&-1&-1\\ -1&2&-1\\ -1&-1&2 \end{array} \right)\left( \begin{array}{c} P_{AA}\\ P_{BA}\\ P_{CA} \end{array} \right), $$ or $$ \frac{{\rm d}\mathbf{p}}{{\rm d}t}=-\lambda Q\mathbf{p} $$ for short. Here the symmetric matrix $Q$ yields the following orthogonal diagonalization $$ Q=U\Lambda U^{\top}, $$ where $$ U=\left( \begin{array}{ccc} -\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{6}}&\frac{1}{\sqrt{3}}\\ 0&\frac{\sqrt{2}}{\sqrt{3}}&\frac{1}{\sqrt{3}}\\ \frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{6}}&\frac{1}{\sqrt{3}} \end{array} \right)\quad\text{and}\quad\Lambda=\left( \begin{array}{ccc} 3&&\\ &3&\\ &&0 \end{array} \right). $$ Therefore, if $$ \mathbf{p}(0)=\mathbf{p}_0, $$ the solution reads $$ \mathbf{p}(t)=e^{-\lambda Qt}\mathbf{p}_0=Ue^{-\lambda\Lambda t}U^{\top}\mathbf{p}_0=U\left( \begin{array}{ccc} e^{-3\lambda t}&&\\ &e^{-3\lambda t}&\\ &&1 \end{array} \right)U^{\top}\mathbf{p}_0. $$