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Q27E

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Found in: Page 336

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# 27: a. Based on your answers in Exercises 24 and 25, find closed formulas for the components of the dynamical system$\overline{\mathbf{x}}\left(t+1\right){\mathbf{=}}\left[\begin{array}{cc}0.5& 0.25\\ 0.5& 0.75\end{array}\right]\overline{\mathbf{x}}{\mathbf{\left(}}{\mathbit{t}}{\mathbf{\right)}}$with initial value $\stackrel{\mathbf{\to }}{{\mathbf{x}}_{\mathbf{0}}}{\mathbf{=}}\stackrel{\mathbf{\to }}{{\mathbf{e}}_{\mathbf{1}}}$. Then do the same for the initial value $\stackrel{\mathbf{\to }}{{\mathbf{x}}_{\mathbf{0}}}{\mathbf{=}}\stackrel{\mathbf{\to }}{{\mathbf{e}}_{\mathbf{2}}}$. Sketch the two trajectories.b. Consider the matrix${\mathbit{A}}{\mathbf{=}}\left[\begin{array}{cc}0.5& 0.25\\ 0.5& 0.75\end{array}\right]$ .Using technology, compute some powers of the matrix A, say, A2, A5, A10, . . . . What do you observe? Diagonalize matrix A to prove your conjecture. (Do not use Theorem 2.3.11, which we have not provenyet.)c. If ${\mathbf{A}}{\mathbf{=}}\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$ is an arbitrary positive transition matrix, what can you say about the powers At as t goes to infinity? Your result proves Theorem 2.3.11c for the special case of a positive transition matrix of size 2 × 2.

1. Formulas for the dynamical system :$c\left(t\right)=\frac{1}{3}.{1}^{t}+\frac{2}{3}0.{25}^{t}=\frac{1}{3}+\frac{2}{3}0.{25}^{t}\phantom{\rule{0ex}{0ex}}r\left(t\right)=\frac{2}{3}.{1}^{t}-\frac{2}{3}0.{25}^{t}=\frac{1}{3}-\frac{2}{3}0.{25}^{t}$

1. ${A}^{n}=\left[\begin{array}{cc}\frac{1}{3}& \frac{1}{3}\\ \frac{2}{3}& \frac{2}{3}\end{array}\right]$
2. $\mathrm{l}\underset{\mathrm{t}\to \infty }{\mathrm{lim}}{\mathrm{A}}^{\mathrm{t}}=\frac{1}{\mathrm{b}+\mathrm{c}}\left[\begin{array}{cc}\mathrm{b}& \mathrm{b}\\ \mathrm{c}& \mathrm{c}\end{array}\right]$
See the step by step solution

## Step 1: Transition Matrix

Transition matrix may refer to: The matrix associated with a change of basis for a vector space. Stochastic matrix, a square matrix used to describe the transitions of a Markov chain. State-transition matrix, a matrix whose product with the state vector at an initial time.

## Step 2: Finding closed formulas for the dynamical system and computing some powers of the matrix :

(a) Clearly we see that,

${v}_{1}=\left[\begin{array}{c}0.25\\ 0.5\end{array}\right]\phantom{\rule{0ex}{0ex}}{v}_{2}=\left[\begin{array}{c}1\\ -1\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{\lambda }_{1}=1,{\lambda }_{2}=0.25\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{x}_{0}=\left[\begin{array}{c}1\\ 0\end{array}\right]=\frac{4}{3}\left[\begin{array}{c}0.25\\ 0.5\end{array}\right]+\frac{2}{3}\left[\begin{array}{c}1\\ -1\end{array}\right]\phantom{\rule{0ex}{0ex}}c\left(t\right)=\frac{1}{3}.{1}^{t}+\frac{2}{3}0.{25}^{t}=\frac{1}{3}+\frac{2}{3}0.{25}^{t}\phantom{\rule{0ex}{0ex}}r\left(t\right)=\frac{2}{3}.{1}^{t}-\frac{2}{3}0.{25}^{t}=\frac{1}{3}-\frac{2}{3}0.{25}^{t}$

(b) We compute,

${A}^{2}=\left[\begin{array}{cc}0.375& 0.3125\\ 0.625& 0.6875\end{array}\right]\phantom{\rule{0ex}{0ex}}{A}^{5}=\left[\begin{array}{cc}0.333984& 0.333008\\ 0.666016& 0.666992\end{array}\right]\phantom{\rule{0ex}{0ex}}{A}^{n}=\left[\begin{array}{cc}\frac{1}{3}& \frac{1}{3}\\ \frac{2}{3}& \frac{2}{3}\end{array}\right]$

(c) We can see that for t>0 applies that,

${A}^{t}={\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]}^{t}\phantom{\rule{0ex}{0ex}}\mathrm{l}\underset{t\to \infty }{\mathrm{lim}}{A}^{t}=\frac{1}{b+c}\left[\begin{array}{cc}b& b\\ c& c\end{array}\right]$

Hence, the final answer is : (a) $c\left(t\right)=\frac{1}{3}.{1}^{t}+\frac{2}{3}0.{25}^{t}=\frac{1}{3}+\frac{2}{3}0.{25}^{t}\phantom{\rule{0ex}{0ex}}r\left(t\right)=\frac{2}{3}.{1}^{t}-\frac{2}{3}0.{25}^{t}=\frac{1}{3}-\frac{2}{3}0.{25}^{t}$

(b) ${A}^{n}=\left[\begin{array}{cc}\frac{1}{3}& \frac{1}{3}\\ \frac{2}{3}& \frac{2}{3}\end{array}\right]$

(c $\mathrm{l}\underset{t\to \infty }{\mathrm{lim}}{A}^{t}=\frac{1}{b+c}\left[\begin{array}{cc}b& b\\ c& c\end{array}\right]$