For the transition matrix $P=\left[\begin{array}{ll}0.5 & 0.5 \\ 0.3 & 0.7\end{array}\right]$, solve the equation $S P=S$ to find the stationary matrix $S$ and the limiting matrix $\bar{P}$.
\[
S=
\]
(Type an integer or decimal for each matrix element. Round to the nearest thousandth as needed.)
\[
\overline{\mathrm{P}}=
\]
(Type an integer or decimal for each matrix element. Round to the nearest thousandth as needed.)

Step 1 ：We are given the transition matrix $P=\left[\begin{array}{ll}0.5 & 0.5 \\ 0.3 & 0.7\end{array}\right]$. 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 such that when it is multiplied by the transition matrix $P$, it remains unchanged. This means that $S$ is the eigenvector of $P$ corresponding to the eigenvalue 1. We can find $S$ by solving the system of linear equations formed by $SP=S$ and the condition that the sum of the elements of $S$ is 1 (since $S$ is a probability distribution).

Step 3 ：By solving the system of equations, we find that the stationary matrix $S$ is $S=\left[\begin{array}{ll}0.375 & 0.625\end{array}\right]$.

Step 4 ：The limiting matrix $\bar{P}$ is the matrix that we get when we multiply $P$ by itself an infinite number of times. In other words, it is the matrix that $P$ converges to as we keep multiplying it by itself. Since $P$ is a stochastic matrix (its rows sum to 1), it has a unique stationary distribution $S$, and $\bar{P}$ is the matrix all of whose rows are $S$.

Step 5 ：By performing the multiplication, we find that the limiting matrix $\bar{P}$ is $\bar{P}=\left[\begin{array}{ll}0.375 & 0.625 \\ 0.375 & 0.625\end{array}\right]$.

Step 6 ：\(\boxed{\text{Final Answer: The stationary matrix } S \text{ is } S=\left[\begin{array}{ll}0.375 & 0.625\end{array}\right] \text{ and the limiting matrix } \bar{P} \text{ is } \bar{P}=\left[\begin{array}{ll}0.375 & 0.625 \\ 0.375 & 0.625\end{array}\right]}\)