How to solve for the n-step state probability vector for the Markov chain Could you help me to solve for the $n$-step state probability vector for the Markov chain given below. Assume that the system starts in $S_1$ (the other two are $S_2$ and $S_3$).
$$
        \begin{bmatrix}
        0.5 & 0.5 & 0 \\
        0 & 0 & 1 \\
        1 & 0 & 0 \\
        \end{bmatrix}
$$
This is what I tried

 A: Allow me a divagation. 
Say we have a discrete Markov chain $X=\{X_n\}$ with space of possible states $S$ and transition matrix $P=(p_{xy})_{x,y\in S}$. A vector $\pi$ with non-negative entries is said to be a stationary distribution of $X$ if $\sum_{x\in S}\pi(x)=1$ and $\pi P = \pi$, that is if it satisfies $\pi(y) = \sum_{x\in S}p_{xy}\pi(x)$ for every $y\in S$
Let $\pi$ be a stationary distribution, then for every $n\in\mathbb N$
$$\pi P^n = (\pi P)P^{n-1} = \pi P^{n-1}=\ldots=\pi P = \pi$$
Thus, if $X$ has initial distribution $\pi$, for every $n\in\mathbb N$ we have that
$$\mathcal P(X_n=y)=\sum_{x\in X}\pi(x)p_{xy}^{(n)} =\sum_{x\in X}\pi(x)p_{xy}=\pi(y)$$
On the other hand, if the distribution of $X_n$ is independent of the time $n$, then the initial distribution $\pi_0$ is such that 
$$\mathcal \pi_0(y)= P(X_0=y) =\mathcal P(X_1=y)=\sum_{x\in S}\pi_0(x)p_{xy}$$ 
thus, $\pi_0$ is an stationary distribution.
In summary, the distribution of $X_n$ is independent of $n$ iff the initial distribution is a stationary distribution. Also, it sounds somewhat plausible that $\lim_{n\rightarrow\infty}p_{xy}^{(n)}=\pi(y)$.
One more definition: the period of a state $x$ is defined as 
$$d_x=\gcd\{n\geq1:p_{xx}^{(n)}\}$$
It can be shown that if $x$ and $y$ are communicated (they lead to each other) then $d_x =d_y$, so when $X$ is irreducible it has a unique period. If such period happens to be 1, then we say $X$ is aperiodic. 
Actually, there is a strong result (whose proof is rather long) that if $X$ is irreducible and positive recurrent, then it has a stationary distribution $\pi$ and it is unique. Moreover, if $X$ aperiodic, then $\lim_{n\rightarrow\infty}p_{xy}^{(n)}=\pi(y)$
Now, let's get back to your problem. It has transition matrix 
$$P=\begin{pmatrix}
\frac12 &\frac12&0\\
0&0&1\\
1&0&0
\end{pmatrix}
$$
We're interested in finding $\pi$, its stationary distribution. That is, we solve a simple system of equations
$$\begin{align}
\pi(1) & = & \frac12\pi(1) & &+\pi(3) \\
\pi(2) & = &\frac12\pi(1)\\
\pi(3) & = &&\pi(2)\\
1 & = & \pi(1)&+\pi(2)&+\pi(3)
\end{align}
$$
which has solution $\pi=(\frac12,\frac14,\frac14)$
Certainly your chain is irreducible and positive recurrent, and $d_1=1$, so it is aperiodic as well. Then
$\lim_{n\rightarrow\infty}p_{xy}^{(n)}=\pi(y)$ and
$$\lim_{n\rightarrow\infty}P^n=\begin{pmatrix}
\frac12&\frac14&\frac14\\
\frac12&\frac14&\frac14\\
\frac12&\frac14&\frac14
\end{pmatrix}
$$
