Hint: Whenever confronted with an ordinary differential equation of the form x′(t) = ax(t)+b(t), it might be beneficial to consider the function y(t) = $e^{−at}x(t)$.
$$Q = \left[ \begin{matrix} 0 & 2\lambda & 0 \\ \lambda & 0 & \lambda \\ 0 & 2\lambda & 0 \\ \end{matrix} \right] $$
So the backward Kolmogorov's equation I got is: $$P'_{12}(t) = 2\lambda P_{22}(t) -2\lambda P_{12}(t) $$ $$P'_{21}(t) = \lambda P_{11}(t) + \lambda P_{31}(t) - 2\lambda P_{21}(t) $$
I am lost about using the symmetry of Q to solve these equations.
Edit: I was asked to provide more information regarding what are $P_{ij}$ and how it is related to Q.
So $P_{ij}(t)$ = $P(X(t) =j| X(0) = i)$ be the probability that the continuous time Markov chain X(t), having started in sate i, will be in state j at time t.
The differential equation I got (as above) is by using Kolmogorov's Backward Equations $P'_{ij}(t)$ = $\sum_{k\neq i}^m q_{ik}p_{kj}(t) - v_{i}p_{ij}(t)$. Where $v_{i}$ is the transition rate out of state i and $q_{ij}$ is the transition rate from i to j ($v_{i}$$p_{ij}$).
I believe after solving the differential equation the answer is suppose to be like $p_{12}$ = $e^{t(-2 \lambda)}$. But I am not sure how to get this.
This solution doesn't really use the symmetries of $Q$, but it may be useful to have the answer.
We are solving the set of equations given by $$ P'(t) = QP(t) - 2\lambda P(t) = (Q-2\lambda I)P(t)\equiv AP(t) $$ where $I$ is the identity matrix, and I have defined $A\equiv Q-2\lambda I$. This is a linear differential equation with the formal solution $$P(t) = e^{At}P(0)= e^{At}I = e^{At}$$ where I have used the fact that the transition matrix at time zero is the identitiy matrix $I$.
So how do we exponentiate the matrix $At$? We can do this by decomposing $A$ as $$A = \left( \begin{array}{ccc} -2 \lambda & 2 \lambda & 0 \\ \lambda & -2 \lambda & \lambda \\ 0 & 2 \lambda & -2 \lambda \\ \end{array} \right) = VDV^{-1} = \frac14\left( \begin{array}{ccc} 1 & -1 & 1 \\ -1 & 0 & 1 \\ 1 & 1 & 1 \\ \end{array} \right)\left( \begin{array}{ccc} -4 \lambda & 0 & 0 \\ 0 & -2 \lambda & 0 \\ 0 & 0 & 0 \\ \end{array} \right)\left( \begin{array}{ccc} 1 & -2 & 1 \\ -2 & 0 & 2 \\ 1 & 2 & 1 \\ \end{array} \right)$$ where $V$ is matrix whose columns are eigenvectors of $A$ and the $D$ is a diagonal matrix with eigenvalues of $At$ as the diagonal entries. Now exponential of $A$ is given by $$P(t) = e^{At} = Ve^{Dt}V^{-1} = \frac14\left( \begin{array}{ccc} 1 & -1 & 1 \\ -1 & 0 & 1 \\ 1 & 1 & 1 \\ \end{array} \right)\left( \begin{array}{ccc} e^{-4 \lambda t} & 0 & 0 \\ 0 & e^{-2 \lambda t} & 0 \\ 0 & 0 & 1 \\ \end{array} \right)\left( \begin{array}{ccc} 1 & -2 & 1 \\ -2 & 0 & 2 \\ 1 & 2 & 1 \\ \end{array} \right) = \frac14\left( \begin{array}{ccc} e^{-4 \lambda t}+2 e^{-2 \lambda t}+1 & 2-2 e^{-4 \lambda t} & e^{-4 \lambda t}-2 e^{-2 \lambda t}+1 \\ 1-e^{-4 \lambda t} & 2 e^{-4 \lambda t}+2 & 1-e^{-4 \lambda t} \\ e^{-4 \lambda t}-2 e^{-2 \lambda t}+1 & 2-2 e^{-4 \lambda t} & e^{-4 \lambda t}+2 e^{-2 \lambda t}+1 \\ \end{array} \right)$$