I know that Markov matrices have steady state since they always have eigenvalue $\lambda = 1$. We just solve the system of equations $A\vec x = 1 \cdot \vec x$ or
$$\begin{cases} k_{a\to a} a + k_{b\to a} b = a \\ k_{a\to b} a + k_{b\to b} b = b \end{cases} $$
For instance, I see that $\begin{bmatrix}.9 & .5\\ .1 & .5 \end{bmatrix}$ expands to $ \begin{cases} .9 a + .5 b = a \\ .1 a + .5 b = b \end{cases} $ and both equations, in chorus, say that $b = a/5$.
This howerver can be derived more easily if we notice that dynamic balance is achieved when current from from $a \to b$ with coefficient .1 equals the current $b \to a$ with coeff .5. In this case we exploit only two coefficients: $.1a = .5b$, which immediately gives $b = .2a$ and proper feedback coefficients .9, .5 are ignored.
None of the approaches can be applied to
$$\begin{bmatrix}1 & .5 & 0 \\ 0 & 0 & 0 \\ 0 & .5 & 1 \end{bmatrix}$$
however. Brute force gives
$\begin{cases}a & + .5b & &=& a \\ & 0 & &=& b \\ &.5 b& + c &=& c \end{cases}$
It says that $b = 0$ and nothing more. I do not see how another approach, with dynamic current equilibrium could be used here. I have got this matrix analyzing the black-body isolation tents. The 2x2 case is identity matrix
$$\begin{bmatrix}1 & 0 \\ 0 & 1 \end{bmatrix}$$
, which does not shed light either. It says that if I had $[a_0, b_0]$ in the beginning, I will have that much in the end. In 3D case, initial condition $[a_0 b_0 c_0] = [0 1 0]$ must be ultimately transformed into $[a_0 + b_0/2,0,c_0 + b_0/2] = [b_0/2,0,b_0/2]$ but I do not see how does it happen.
In general the steady state is not unique. In your case this is to be expected because your system will stay in state $1$ whenever it begins there and will stay in state $3$ whenever it begins there. As a result, all probability in state $2$ is divided (in this case evenly) between states $1$ and $3$ in the steady state. Hence you get a two parameter family of steady states, which is what was suggested from your linear system (any distribution which assigns no probability to state $2$ is a steady state of your system). Normalization reduces that to a one parameter family of steady states.