Question: In Exercises5-8, find the steady-state vector.

8. \(\left[ {\begin{array}{*{20}{c}}{.7}&{.2}&{.2}\\0&{.2}&{.4}\\{.3}&{.6}&{.4}\end{array}} \right]\)

The equation \(P{\mathop{\rm x}\nolimits} = {\mathop{\rm x}\nolimits} \) can be solved by rewriting itas \(\left( {P - I} \right){\mathop{\rm x}\nolimits} = 0\).

Compute \(P - I\) as shown below:

\(\begin{array}{c}P - I = \left[ {\begin{array}{*{20}{c}}{.7}&{.2}&{.2}\\0&{.2}&{.4}\\{.3}&{.6}&{.4}\end{array}} \right] - \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&0\\0&0&1\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - .3}&{.2}&{.2}\\0&{ - .8}&{.4}\\{.3}&{.6}&{ - .6}\end{array}} \right]\end{array}\)

Write the augmented matrix for the homogeneous system \(\left( {P - I} \right){\mathop{\rm x}\nolimits} = 0\)as shown below:

\(\left[ {\begin{array}{*{20}{c}}{ - .3}&{.2}&{.2}&0\\0&{ - .8}&{.4}&0\\{.3}&{.6}&{ - .6}&0\end{array}} \right]\)

Perform an elementary row operation to produce the row-reduced echelon form of the matrix.

At row 1, multiply row 1 by \( - \frac{1}{{0.3}}\).

\( \sim \left[ {\begin{array}{*{20}{c}}1&{ - 0.666}&{ - 0.666}&0\\0&{ - .8}&{.4}&0\\{.3}&{.6}&{ - .6}&0\end{array}} \right]\)

At row 3, multiply row 1 by 0.3 and subtract it from row 3.

\( \sim \left[ {\begin{array}{*{20}{c}}1&{ - 0.666}&{ - 0.666}&0\\0&{ - .8}&{.4}&0\\0&{0.8}&{ - 0.4}&0\end{array}} \right]\)

At row 2, multiply row 2 by \( - \frac{1}{{0.8}}\).

\( \sim \left[ {\begin{array}{*{20}{c}}1&{ - 0.666}&{ - 0.666}&0\\0&1&{ - \frac{1}{2}}&0\\0&{0.8}&{ - 0.4}&0\end{array}} \right]\)

At row 1, multiply row 2 by 0.666 and add it to row 1. At row 3, multiply row 2 by 0.8 and subtract it from row 3.

\( \sim \left[ {\begin{array}{*{20}{c}}1&0&{ - 1}&0\\0&1&{ - \frac{1}{2}}&0\\0&0&0&0\end{array}} \right]\)

The general solution of the equation \(\left( {P - I} \right){\mathop{\rm x}\nolimits} = 0\)is shown below:

\(\begin{array}{c}{\mathop{\rm x}\nolimits} = \left[ {\begin{array}{*{20}{c}}{{{\mathop{\rm x}\nolimits} _1}}\\{{{\mathop{\rm x}\nolimits} _2}}\\{{{\mathop{\rm x}\nolimits} _3}}\end{array}} \right]\\ = {{\mathop{\rm x}\nolimits} _3}\left[ {\begin{array}{*{20}{c}}1\\{\frac{1}{2}}\\1\end{array}} \right]\end{array}\)

One solution is \(\left[ {\begin{array}{*{20}{c}}2\\1\\2\end{array}} \right]\). The sum of the entries of \(\left[ {\begin{array}{*{20}{c}}2\\1\\2\end{array}} \right]\) is 5.

Multiply \({\mathop{\rm x}\nolimits} \) by \(\frac{1}{5}\) to obtain the steady-state vector as shown below:

\(\begin{array}{c}{\mathop{\rm q}\nolimits} = \left[ {\begin{array}{*{20}{c}}{\frac{2}{5}}\\{\frac{1}{5}}\\{\frac{2}{5}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{.4}\\{.2}\\{.4}\end{array}} \right]\end{array}\)

Thus, the steady-state vector is \({\mathop{\rm q}\nolimits} = \left[ {\begin{array}{*{20}{c}}{.4}\\{.2}\\{.4}\end{array}} \right]\).