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

7. \(\left( {\begin{array}{*{20}{c}}{.7}&{.1}&{.1}\\{.2}&{.8}&{.2}\\{.1}&{.1}&{.7}\end{array}} \right)\)

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

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

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

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

\(\left( {\begin{array}{*{20}{c}}{ - .3}&{.1}&{.1}&0\\{.2}&{ - .2}&{.2}&0\\{.1}&{.1}&{ - .3}&0\end{array}} \right)\)

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

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

\( \sim \left( {\begin{array}{*{20}{c}}1&{ - .333}&{ - .3333}&0\\{.2}&{ - .2}&{.2}&0\\{.1}&{.1}&{ - .3}&0\end{array}} \right)\)

At row 2, multiply row 1 by 0.2 and subtract it from row 2. At row 3, multiply row 1 by 0.1 and subtract it from row 3.

\( \sim \left( {\begin{array}{*{20}{c}}1&{ - .333}&{ - .3333}&0\\0&{ - 0.1333}&{0.2666}&0\\0&{0.1333}&{0.2666}&0\end{array}} \right)\)

At row 2, multiply row 2 by \( - \frac{1}{{0.1333}}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&{ - .333}&{ - .3333}&0\\0&1&{ - 2}&0\\0&{0.1333}&{0.2666}&0\end{array}} \right)\)

At row 1, multiply row 2 by 0.333 and add it to row 1. At row 3, multiply row 2 by 0.1333 and subtract it from row 3.

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

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

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

One solution is \(\left( {\begin{array}{*{20}{c}}1\\2\\1\end{array}} \right)\). The sum of the entries of \(\left( {\begin{array}{*{20}{c}}1\\2\\1\end{array}} \right)\) is 4.

Multiply \({\mathop{\rm x}\nolimits} \) by \(\frac{1}{4}\) as shown below:

\(\begin{array}{c}{\mathop{\rm q}\nolimits} = \left( {\begin{array}{*{20}{c}}{\frac{1}{4}}\\{\frac{1}{2}}\\{\frac{1}{4}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{.25}\\{.5}\\{.25}\end{array}} \right)\end{array}\)

Thus, the steady-state vector is \({\mathop{\rm q}\nolimits} = \left( {\begin{array}{*{20}{c}}{.25}\\{.5}\\{.25}\end{array}} \right)\).