Question
Consider a Markov process $\{X(t)\}_{t \geq 0}$ with state space $$S = \{0, 1, 2, 3\}$$ and generator matrix $$Q = \begin{pmatrix} -q_0 & 2 & 0 & 0\\ 2 & -q_1 & 4 & 0\\ 0 & 4 & -q_2 & 2\\ 0 & 0 & 1 & -q_3 \end{pmatrix}.$$ Find the stationary distribution $\pi_Q$ associated with $Q$.
My working
I first fully determine the generator matrix to be $$Q = \begin{pmatrix} -2 & 2 & 0 & 0\\ 2 & -6 & 4 & 0\\ 0 & 4 & -6 & 2\\ 0 & 0 & 1 & -1 \end{pmatrix},$$ using the fact that the rows must sum to $0$.
Now, I also know that $\pi_Q$ must satisfy $$\pi_QQ = 0,$$ yielding the following system of linear equations: \begin{align} -2\pi_0 + 2\pi_1 & = 0, \tag{1}\\ 2\pi_0 - 6\pi_1 + 4\pi_2 & = 0, \tag{2}\\ 4\pi_1 - 6\pi_2 + \pi_3 & = 0, \tag{3}\\ 2\pi_2 - \pi_3 & = 0, \tag{4} \end{align} where summing $(3)$ and $(4)$ gives $$\pi_1 = \pi_2,$$ which can then be substituted into $(2)$ to return $$\pi_0 = \pi_1.$$ Thus, we have $$\begin{aligned} \pi_0 & = \pi_1\\ & = \pi_2 \end{aligned}$$ and from $(4)$, we have $$\begin{aligned} \pi_3 & = 2\pi_2\\ & = 2\pi_0. \end{aligned}$$ Finally, recalling that $$\pi_0 + \pi_1 + \pi_2 + \pi_3 = 1$$ means that $$\pi_Q = \left(\frac 1 5, \frac 1 5, \frac 1 5, \frac 2 5\right).$$
However, this $\pi_Q$ clearly does not satisfy $$\pi_QQ = 0.$$ Did I go wrong when I assumed that there are no absorbing states when in fact, there could be? If so, then how should I solve for the stationary distribution? Do I leave my answer in terms of $q_0, q_1, q_2$ and $q_3$? Any intuitive explanations will be greatly appreciated :)
As mentioned kindly in a comment, I simply went wrong in my verification step. Now thinking back, I believe I was erroneously checking for $$\pi_QQ = 1$$ and not $$\pi_QQ = 0.$$