For the transition matrix $P=\left[\begin{array}{cc}0.7 & 0.3 \\ 0.3 & 0.7\end{array}\right]$, solve the equation $\mathrm{SP}=\mathrm{S}$ to find the stationary matrix $\mathrm{S}$ and the limiting matrix $\overline{\mathrm{P}}$. \[ S= \] (Type an integer or decimal for each matrix element. Round to the nearest thousandth as needed.)
Solution
Step 1 ：Given the transition matrix $P=\left[\begin{array}{cc}0.7 & 0.3 \\ 0.3 & 0.7\end{array}\right]$, we are to find the stationary matrix $S$ and the limiting matrix $\overline{P}$.
Step 2 ：The stationary matrix $S$ is a row vector that satisfies the equation $SP=S$, where $P$ is the transition matrix. This means that the stationary matrix is the eigenvector of the transition matrix corresponding to the eigenvalue 1.
Step 3 ：To find the stationary matrix, we can solve the system of linear equations formed by $SP=S$ and the condition that the sum of the elements in $S$ is 1 (since $S$ is a probability distribution).
Step 4 ：The limiting matrix $\overline{P}$ is the matrix that we get when we raise the transition matrix $P$ to a large power. In other words, it's the matrix that describes the long-term behavior of the system. The limiting matrix is a matrix where every row is the stationary distribution.
Step 5 ：By solving the system of equations, we find that the stationary matrix $S$ is $\left[\begin{array}{cc}0.5 & 0.5\end{array}\right]$.
Step 6 ：By raising the transition matrix $P$ to a large power, we find that the limiting matrix $\overline{P}$ is $\left[\begin{array}{cc}0.25 & 0.25 \\ 0.25 & 0.25\end{array}\right]$.
Step 7 ：\(\boxed{\text{Final Answer: The stationary matrix } S \text{ is } \left[\begin{array}{cc}0.5 & 0.5\end{array}\right] \text{ and the limiting matrix } \overline{P} \text{ is } \left[\begin{array}{cc}0.25 & 0.25 \\ 0.25 & 0.25\end{array}\right]}.\)
