Answer:
Both get the same results that is,
[tex]\left[\begin{array}{ccc}140\\160\\200\end{array}\right][/tex]
Step-by-step explanation:
Given :
[tex]\bf M=\left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right][/tex]
and initial population,
[tex]\bf P=\left[\begin{array}{ccc}130\\300\\70\end{array}\right][/tex]
a) - After two times, we will find in each position.
[tex]P_2=[P].[M]^2=[P].[M].[M][/tex]
[tex]M^2=\left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right]\times \left[\begin{array}{ccc}\frac{1}{5}&\frac{1}{5}&\frac{2}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{1}{5}\\\frac{2}{5}&\frac{2}{5}&\frac{2}{5}\end{array}\right][/tex]
[tex]=\frac{1}{25} \left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right][/tex]
[tex]\therefore\;\;\;\;\;\;\;\;\;\;\;P_2=\left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right] \times\left[\begin{array}{ccc}130\\300\\70\end{array}\right] = \left[\begin{array}{ccc}140\\160\\200\end{array}\right][/tex]
b) - With in migration process, 500 people are numbered. There will be after a long time,
[tex]After\;inifinite\;period=[M]^n.[P][/tex]
[tex]Then,\;we\;get\;the\;same\;result\;if\;we\;measure [M]^n=\frac{1}{25} \left[\begin{array}{ccc}7&7&7\\8&8&8\\10&10&10\end{array}\right][/tex]
[tex]=\left[\begin{array}{ccc}140\\160\\200\end{array}\right][/tex]
