Science:Math Exam Resources/Courses/MATH152/April 2016/Question B 04 (d)/Solution 1

We first use the Matrix diagonalization method to write the matrix $P$ in the following form:

$P=QAQ^{-1},$ where $A={\begin{pmatrix}\lambda _{1}&0\\0&\lambda _{2}\end{pmatrix}}$ and $V={\begin{pmatrix}v_{1}{\Big |}v_{2}\end{pmatrix}}.$

From part (a) and (c), we have $\lambda _{1}=0.7$ and $v_{1}={\begin{pmatrix}1\\-1\end{pmatrix}}.$ Also, $\lambda _{2}=1$ and $v_{2}={\begin{pmatrix}2\\1\end{pmatrix}}.$ So, we have $Q={\begin{pmatrix}1&2\\-1&1\end{pmatrix}},$ and therefore, $Q^{-1}={\frac {1}{3}}{\begin{pmatrix}1&-2\\1&1\end{pmatrix}}.$

We finally get that

$P={\begin{pmatrix}0.9&0.2\\0.1&0.8\end{pmatrix}}=QAQ^{-1}={\begin{pmatrix}1&2\\-1&1\end{pmatrix}}{\begin{pmatrix}0.7&0\\0&1\end{pmatrix}}{\begin{pmatrix}{\frac {1}{3}}&-{\frac {2}{3}}\\{\frac {1}{3}}&{\frac {1}{3}}\end{pmatrix}},$

which means for any time $t$ we have

$P^{t}={\begin{pmatrix}0.9&0.2\\0.1&0.8\end{pmatrix}}^{t}=QA^{t}Q^{-1}={\begin{pmatrix}1&2\\-1&1\end{pmatrix}}{\begin{pmatrix}0.7&0\\0&1\end{pmatrix}}^{t}{\begin{pmatrix}{\frac {1}{3}}&-{\frac {2}{3}}\\{\frac {1}{3}}&{\frac {1}{3}}\end{pmatrix}}={\begin{pmatrix}1&2\\-1&1\end{pmatrix}}{\begin{pmatrix}0.7^{t}&0\\0&1^{t}\end{pmatrix}}{\begin{pmatrix}{\frac {1}{3}}&-{\frac {2}{3}}\\{\frac {1}{3}}&{\frac {1}{3}}\end{pmatrix}}.$

Now we are ready to find $P^{t}$ for large $t$ by computing $\lim \limits _{t\to \infty }P^{t}.$ So,

$\lim \limits _{t\to \infty }P^{t}=\lim \limits _{t\to \infty }QA^{t}Q^{-1}=\lim \limits _{t\to \infty }{\begin{pmatrix}1&2\\-1&1\end{pmatrix}}{\begin{pmatrix}0.7^{t}&0\\0&1^{t}\end{pmatrix}}{\begin{pmatrix}{\frac {1}{3}}&-{\frac {2}{3}}\\{\frac {1}{3}}&{\frac {1}{3}}\end{pmatrix}}={\begin{pmatrix}1&2\\-1&1\end{pmatrix}}{\begin{pmatrix}0&0\\0&1\end{pmatrix}}{\begin{pmatrix}{\frac {1}{3}}&-{\frac {2}{3}}\\{\frac {1}{3}}&{\frac {1}{3}}\end{pmatrix}}={\begin{pmatrix}1&2\\-1&1\end{pmatrix}}{\begin{pmatrix}0&0\\{\frac {1}{3}}&{\frac {1}{3}}\end{pmatrix}}={\frac {1}{3}}{\begin{pmatrix}2&2\\1&1\end{pmatrix}}.$

Now by applying this matrix on our initial population ${\begin{pmatrix}50\\50\end{pmatrix}},$ we get

${\frac {1}{3}}{\begin{pmatrix}2&2\\1&1\end{pmatrix}}{\begin{pmatrix}50\\50\end{pmatrix}}={\begin{pmatrix}{\frac {200}{3}}\\{\frac {100}{3}}\end{pmatrix}}.$

Therefore, $\color {blue}{\frac {100}{3}}$ million people will be living in the suburbs far in the future.