For the transition matrix $P=\left[\begin{array}{ccc}0.3 & 0.7 & 0.0 \\ 0.5 & 0.1 & 0.4 \\ 0.0 & 0.3 & 0.7\end{array}\right]$, solve the equation $S P=S$ to find the stationary matrix $S$ and the limiting matrix $\bar{P}$. $S=\left[\begin{array}{l}.33333333 \\ .33333333 \\ .33333333\end{array}\right]$ (Type an integer or decimal for each matrix element. Do not round until the final answer. Then round to the nearest thousandth as needed.)

Step 1 ：We are given the transition matrix $P=\left[\begin{array}{ccc}0.3 & 0.7 & 0.0 \\ 0.5 & 0.1 & 0.4 \\ 0.0 & 0.3 & 0.7\end{array}\right]$. We are asked to find the stationary matrix $S$ and the limiting matrix $\bar{P}$.

Step 2 ：The stationary matrix $S$ is a row vector whose elements sum to 1 and satisfies the equation $SP=S$. This means that the distribution of states does not change after one step of the Markov chain.

Step 3 ：To find the stationary matrix $S$, we need to solve the equation $SP=S$ for $S$. This is a system of linear equations, which can be solved using linear algebra techniques.

Step 4 ：The limiting matrix $\bar{P}$ is the matrix that the transition matrix $P$ converges to as the number of steps goes to infinity.

Step 5 ：To find the limiting matrix $\bar{P}$, we need to multiply the transition matrix $P$ by itself repeatedly until the matrix does not change significantly.

Step 6 ：By solving the system of equations, we find that the stationary matrix $S$ is $\left[\begin{array}{l}0.234 \\ 0.328 \\ 0.438\end{array}\right]$.

Step 7 ：By repeatedly multiplying the transition matrix $P$, we find that the limiting matrix $\bar{P}$ is $\left[\begin{array}{ccc}0.234 & 0.328 & 0.438 \\ 0.234 & 0.328 & 0.438 \\ 0.234 & 0.328 & 0.438\end{array}\right]$.

Step 8 ：\(\boxed{\text{Final Answer: The stationary matrix } S \text{ is } \left[\begin{array}{l}0.234 \\ 0.328 \\ 0.438\end{array}\right] \text{ and the limiting matrix } \bar{P} \text{ is } \left[\begin{array}{ccc}0.234 & 0.328 & 0.438 \\ 0.234 & 0.328 & 0.438 \\ 0.234 & 0.328 & 0.438\end{array}\right]}.\)