Finite Math and Applied Calculus (6th Edition)

Published by Brooks Cole
ISBN 10: 1133607705
ISBN 13: 978-1-13360-770-0

Chapter 7 - Section 7.7 - Markov Systems - Exercises - Page 531: 11

Answer

(a) $P^2=\begin{bmatrix} .25 & .75 \\ 0 & 1 \\ \end{bmatrix}$ (b) After one step: $vP=\begin{bmatrix} .5 & .5 \\ \end{bmatrix}$ After two steps: $vP^2=\begin{bmatrix} .25 & .75 \\ \end{bmatrix}$ After three steps: $vP^3=\begin{bmatrix} .125 & .875 \\ \end{bmatrix}$

Work Step by Step

(a) $P^2=\begin{bmatrix} .5 & .5 \\ 0 & 1 \\ \end{bmatrix}\begin{bmatrix} .5 & .5 \\ 0 & 1 \\ \end{bmatrix}=\begin{bmatrix} .25+0 & .25+.5 \\ 0+0 & 0+1 \\ \end{bmatrix}=\begin{bmatrix} .25 & .75 \\ 0 & 1 \\ \end{bmatrix}$ (b) $P^3=P^2P=\begin{bmatrix} .25 & .75 \\ 0 & 1 \\ \end{bmatrix}\begin{bmatrix} .5 & .5 \\ 0 & 1 \\ \end{bmatrix}=\begin{bmatrix} .125+0 & .125+.75 \\ 0+0 & 0+1 \\ \end{bmatrix}=\begin{bmatrix} .125 & .875 \\ 0 & 1 \\ \end{bmatrix}$ After one step: $vP=\begin{bmatrix} 1 & 0 \\ \end{bmatrix}\begin{bmatrix} .5 & .5 \\ 0 & 1 \\ \end{bmatrix}=\begin{bmatrix} .5 & .5 \\ \end{bmatrix}$ After two steps: $vP^2=\begin{bmatrix} 1 & 0 \\ \end{bmatrix}\begin{bmatrix} .25 & .75 \\ 0 & 1 \\ \end{bmatrix}=\begin{bmatrix} .25 & .75 \\ \end{bmatrix}$ After three steps: $vP^3=\begin{bmatrix} 1 & 0 \\ \end{bmatrix}\begin{bmatrix} .125 & .875 \\ 0 & 1 \\ \end{bmatrix}=\begin{bmatrix} .125 & .875 \\ \end{bmatrix}$
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