91影视

From example 3, \(M = \left[ {\begin{array}{*{20}{c}}{0.95}&{0.03}\\{0.05}&{0.97}\end{array}} \right]\) and \({{\bf{x}}_0} = \left[ {\begin{array}{*{20}{c}}{600,000}\\{400,000}\end{array}} \right]\).

\({{\bf{x}}_{k + 1}} = M{{\bf{x}}_k}\)

From MATLAB, some of the population vectors are

\({x_5} = \left[ {\begin{array}{*{20}{c}}{523,293}\\{476,707}\end{array}} \right]\), \[{{\bf{x}}_{10}} = \left[ {\begin{array}{*{20}{c}}{472,737}\\{527,263}\end{array}} \right]\], \[\,{{\bf{x}}_{15}} = \left[ {\begin{array}{*{20}{c}}{439,417}\\{560,583}\end{array}} \right]\], \[{{\bf{x}}_{20}} = \left[ {\begin{array}{*{20}{c}}{417,456}\\{582,544}\end{array}} \right]\].

The data in thepopulation vectors shows that the city population is declining and the suburban population is increasing, but over the years, the population change is getting smaller.

If the initial population of the city is 350,000, and in the suburbs, it is 650,000, then \({{\bf{x}}_0} = \left[ {\begin{array}{*{20}{c}}{350,000}\\{650,000}\end{array}} \right]\).

Use the following MATLAB code to solve the equation \({{\bf{x}}_{k + 1}} = M{{\bf{x}}_k}\) for \(k = 19\).

\( > > M = \left[ {\begin{array}{*{20}{c}}{0.98034}&{0.00179;\,\begin{array}{*{20}{c}}{\,0.01966}&{0.99821}\end{array}}\end{array}} \right];\)

\( > > {x_0} = \left[ {38041430\;;\;\;275872610} \right];\)

\(\)\(\begin{array}{l} > > \,\,{\rm{for}}\;\;k = 1:19\\\, > > \,\,x\left( {k + 1} \right) = M * x\left( k \right)\end{array}\)

\[{{\bf{x}}_{20}} = \left[ {\begin{array}{*{20}{c}}{370,283}\\{629,717}\end{array}} \right]\]

The data in the population vectors shows that the city population is increasing at a slower rate, whereas the suburban population is decreasing.