Answer
(a) $\left[\begin{array}{ll}0.94 & 0.04 \\ 0.06 & 0.96\end{array}\right] \mathbf{x}_{k}=\mathbf{x}_{k+1}$
(b) $\left[\begin{array}{l}8920800 \\ 1879200\end{array}\right]$
Work Step by Step
This solution appears in both the sections for the 4 th and 5 th editions of Lay's Linear Algebra and Its Applications since the only difference between the questions is that the years considered in the 4 th edition are 2010 and 2012 while in the 5 th edition they are 2015 and 2017 . I write this for the 5 th edition, thus if you're using the 4 th, just substitute the correct years in your work. It doesn't actually affect the final answer though since in both cases the difference of the years is 2
Let $p_{c}$ be the population of the city (in a given year) and $p_{s}$ be the population of the suburbs (in the same year). The first sentence of the exercise statement says us that the populations $p_{c}^{\prime}$ and $p_{s}^{\prime}$ for the next year will be
\[
\begin{array}{l}
0.94 p_{c}+0.04 p_{s}=p_{c}^{\prime} \\
0.06 p_{c}+0.96 p_{s}=p_{s}^{\prime}
\end{array}
\]
we used the fact that if $6 \%$ (resp. $4 \%$ ) moves then $94 \%$ (resp. $96 \%$ ) must stay put.
We will sum this up in the matrix equation
\[
\left[\begin{array}{ll}
0.94 & 0.04 \\
0.06 & 0.96
\end{array}\right]\left[\begin{array}{l}
p_{c} \\
p_{s}
\end{array}\right]=\left[\begin{array}{l}
p_{c}^{\prime} \\
p_{s}^{\prime}
\end{array}\right]
\]
Written in the usual difference equation notation this is
\[
\left[\begin{array}{ll}
0.94 & 0.04 \\
0.06 & 0.96
\end{array}\right] \mathbf{x}_{k}=\mathbf{x}_{k+1}
\]
Now put $\mathbf{x}_{0}=\left[\begin{array}{c}\text { population in the city in } 2015 \\ \text { population in the suburbs in } 2015\end{array}\right]$ be the $2 \times 1$ population vector for 2015 . Then
\[
\mathbf{x}_{1}=\left[\begin{array}{ll}
0.94 & 0.04 \\
0.06 & 0.96
\end{array}\right] \mathbf{x}_{0}
\]
