(1 point) Find the vector of stable probabilities for the Markov chain whose transition matrix is
\[
\left[\begin{array}{ccc}
0 & 1 & 0 \\
0.2 & 0.1 & 0.7 \\
0 & 1 & 0
\end{array}\right]
\]
\[
W=[
\]
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Step 1 ：The stable probabilities for a Markov chain are the probabilities that the system will be in each state in the long run. They are found by solving the equation \(\pi P = \pi\), where P is the transition matrix and \(\pi\) is the row vector of stable probabilities. This equation represents the condition that the probabilities do not change from one step to the next.

Step 2 ：The equation \(\pi P = \pi\) can be rewritten as \((\pi P - \pi) = 0\), or \(\pi(P - I) = 0\), where I is the identity matrix. This is a system of linear equations, which can be solved to find \(\pi\).

Step 3 ：The transition matrix P is given by \[P = \left[\begin{array}{ccc} 0 & 1 & 0 \\ 0.2 & 0.1 & 0.7 \\ 0 & 1 & 0 \end{array}\right]\]

Step 4 ：The identity matrix I is \[I = \left[\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}\right]\]

Step 5 ：Subtracting I from P gives \[P - I = \left[\begin{array}{ccc} -1 & 1 & 0 \\ 0.2 & -0.9 & 0.7 \\ 0 & 1 & -1 \end{array}\right]\]

Step 6 ：The system of equations \(\pi(P - I) = 0\) is then \[-\pi_1 + \pi_2 = 0\\ 0.2\pi_1 - 0.9\pi_2 + 0.7\pi_3 = 0\\ \pi_2 - \pi_3 = 0\]

Step 7 ：Adding the condition that the probabilities must sum to 1, \(\pi_1 + \pi_2 + \pi_3 = 1\), gives a system of four equations in three unknowns. This system can be solved to find the stable probabilities.

Step 8 ：Solving this system gives \(\pi_1 = 0\), \(\pi_2 = 0.5\), and \(\pi_3 = 0.5\). Therefore, the vector of stable probabilities is \[\boxed{[0, 0.5, 0.5]}\]