Different limiting distributions but they both satisfy same equations

Question

I needed to find the limiting distribution of the matrix

$$\pmatrix{ 0 & 0 & 1 \\ 1 & 0 & 0 \\ \frac{1}{2} & \frac{1}{2} & 0}$$

Instead of $\pi$ I'll use $A, B$ and $C$ because that might be easier to read.

To solve this, I got

$$A = B + \frac{1}{2}C$$ $$B = \frac{1}{2}C$$ $$C = A$$

Now to solve this, I ended up getting $A = C$ and I got $A = 2B$. I also need to satisfy $A + B + C = 1$. So, I said let $A = \frac{1}{4}, B = \frac{1}{2}, C = \frac{1}{4}$, which satisfies all my equations. But in the answers they say that $A = \frac{1}{5}, B = \frac{2}{5}, C = \frac{1}{5}$, which also satisfy the equations.

How did I get mine wrong? Obviously it matters what the limiting distribution is, so how am I supposed to tell which is the correct one?

Answer 1

Both answers are wrong...

The system of three equations can be readily rewritten as $A=2B=C$ hence the only solution such that $A+B+C=1$ is $(A,B,C)=(\frac25,\frac15,\frac25)$.

If $(A,B,C)=(\frac14,\frac12,\frac14)$ (your solution), then $A\ne2B\ne C$.

If $(A,B,C)=(\frac15,\frac25,\frac15)$ (their solution), then $A+B+C\ne1$.

Answer 2

Their answers are the essentially same as yours, just not normalized--they don't satisfy $A+B+C = 1$. As such, I would consider your answer "more correct" than theirs.

(Though, as Did points out, they're both wrong...)