Short Answer

27: a. Based on your answers in Exercises 24 and 25, find closed formulas for the components of the dynamical system
$\bar{x} (t + 1) = [\begin{matrix} 0.5 & 0.25 \\ 0.5 & 0.75 \end{matrix}] \bar{x} (t)$
with initial value $\vec{x_{0}} = \vec{e_{1}}$ . Then do the same for the initial value $\vec{x_{0}} = \vec{e_{2}}$ . Sketch the two trajectories.
b. Consider the matrix
$A = [\begin{matrix} 0.5 & 0.25 \\ 0.5 & 0.75 \end{matrix}]$
.
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 proven
yet.)
c. If $A = [\begin{matrix} a & b \\ c & d \end{matrix}]$
is an arbitrary positive transition matrix, what can you say about the powers A^tas t goes to infinity? Your result proves Theorem 2.3.11c for the special case of a positive transition matrix of size 2 × 2.

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

$A^{n} = [\begin{matrix} \frac{1}{3} & \frac{1}{3} \\ \frac{2}{3} & \frac{2}{3} \end{matrix}]$
$l \lim_{t \to \infty} A^{t} = \frac{1}{b + c} [\begin{matrix} b & b \\ c & c \end{matrix}]$

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

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,

$v_{1} = [\begin{matrix} 0.25 \\ 0.5 \end{matrix}] v_{2} = [\begin{matrix} 1 \\ - 1 \end{matrix}] λ_{1} = 1, λ_{2} = 0.25 x_{0} = [\begin{matrix} 1 \\ 0 \end{matrix}] = \frac{4}{3} [\begin{matrix} 0.25 \\ 0.5 \end{matrix}] + \frac{2}{3} [\begin{matrix} 1 \\ - 1 \end{matrix}] c (t) = \frac{1}{3} . 1^{t} + \frac{2}{3} 0 . 25^{t} = \frac{1}{3} + \frac{2}{3} 0 . 25^{t} r (t) = \frac{2}{3} . 1^{t} - \frac{2}{3} 0 . 25^{t} = \frac{1}{3} - \frac{2}{3} 0 . 25^{t}$

(b) We compute,

$A^{2} = [\begin{matrix} 0.375 & 0.3125 \\ 0.625 & 0.6875 \end{matrix}] A^{5} = [\begin{matrix} 0.333984 & 0.333008 \\ 0.666016 & 0.666992 \end{matrix}] A^{n} = [\begin{matrix} \frac{1}{3} & \frac{1}{3} \\ \frac{2}{3} & \frac{2}{3} \end{matrix}]$

$A^{t} = {[\begin{matrix} a & b \\ c & d \end{matrix}]}^{t} l \lim_{t \to \infty} A^{t} = \frac{1}{b + c} [\begin{matrix} b & b \\ c & c \end{matrix}]$

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

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

(c $l \lim_{t \to \infty} A^{t} = \frac{1}{b + c} [\begin{matrix} b & b \\ c & c \end{matrix}]$

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Linear Algebra With Applications

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Step 1: Transition Matrix

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

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

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