Question

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

6. \(\left( {\begin{array}{*{20}{c}}{.8}&{.5}\\{.2}&{.5}\end{array}} \right)\)

Short answer

The steady-state vector is \({\mathop{\rm q}\nolimits} = \left( {\begin{array}{*{20}{c}}{.714}\\{.286}\end{array}} \right)\).

Step-by-step solution

Compute \(P - I\)

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:

\(\begin{array}{c}P - I = \left( {\begin{array}{*{20}{c}}{.8}&{.5}\\{.2}&{.5}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}1&0\\0&1\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}{ - .2}&{.5}\\{.2}&{ - .5}\end{array}} \right)\end{array}\)

Write the augmented matrix

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

\(\left( {\begin{array}{*{20}{c}}{ - .2}&{.5}&0\\{.2}&{ - .5}&0\end{array}} \right)\)

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

At row 1, multiply row 1 by \( - \frac{1}{{0.2}}\).

\( \sim \left( {\begin{array}{*{20}{c}}1&{ - \frac{5}{2}}&0\\{.2}&{ - .5}&0\end{array}} \right)\)

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

\( \sim \left( {\begin{array}{*{20}{c}}1&{ - \frac{5}{2}}&0\\0&0&0\end{array}} \right)\)

Determine the steady-state vector

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

\(\begin{array}{c}{\mathop{\rm x}\nolimits} = \left( {\begin{array}{*{20}{c}}{{{\mathop{\rm x}\nolimits} _1}}\\{{{\mathop{\rm x}\nolimits} _2}}\end{array}} \right)\\ = {{\mathop{\rm x}\nolimits} _2}\left( {\begin{array}{*{20}{c}}{\frac{5}{2}}\\1\end{array}} \right)\end{array}\)

One solution is \(\left( {\begin{array}{*{20}{c}}5\\2\end{array}} \right)\). The sum of the entries of \(\left( {\begin{array}{*{20}{c}}5\\2\end{array}} \right)\) is 7.

Multiply \({\mathop{\rm x}\nolimits} \) by \(\frac{1}{7}\) as shown below:

\(\begin{array}{c}{\mathop{\rm q}\nolimits} = \left( {\begin{array}{*{20}{c}}{\frac{5}{7}}\\{\frac{2}{7}}\end{array}} \right)\\ \approx \left( {\begin{array}{*{20}{c}}{.714}\\{.286}\end{array}} \right)\end{array}\)

Thus, the steady-state vector is \({\mathop{\rm q}\nolimits} = \left( {\begin{array}{*{20}{c}}{.714}\\{.286}\end{array}} \right)\).