Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
We have solutions for your book!
See our solution for Question 12E from Chapter 4.9 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 12E
Chapter:
Problem:
Refer to Exercise 2. Which food will the animal prefer after many trials? Reference:
Step-by-Step Solution
Given Information
We ave to find the steady-state vector for the Markov chain in Exercise 2.
Step-1:
The stochastic Matrix is: \[ P = \left[ \begin{array} { l l } { 0.7 } & { 0.6 } \\ { 0.3 } & { 0.4 } \end{array} \right] \]
Step-2:
The $P-I$ matrix is: \[\begin{array}{l} P - I = \left[ {\begin{array}{*{20}{c}} {0.50}&{0.25}&{0.25}\\ {0.25}&{0.50}&{0.25}\\ {0.25}&{0.25}&{0.50} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} { - 0.50}&{0.25}&{0.25}\\ {0.25}&{ - 0.50}&{0.25}\\ {0.25}&{0.25}&{ - 0.50} \end{array}} \right] \end{array}\]
Step-3:
The augmented Matrix for the system is: \[ \left[ \begin{array} { c c c | c } { - 0.50 } & { 0.25 } & { 0.25 } & { 0 } \\ { 0.25 } & { - 0.50 } & { 0.25 } & { 0 } \\ { 0.25 } & { 0.25 } & { - 0.50 } & { 0 } \end{array} \right] \] Row-Reduce the augmented matrix: \[\begin{array}{l} M = \left[ {\begin{array}{*{20}{c}} { - 0.50}&{0.25}&{0.25}&0\\ {0.25}&{ - 0.50}&{0.25}&0\\ {0.25}&{0.25}&{ - 0.50}&0 \end{array}} \right]\\ \left[ {\begin{array}{*{20}{c}} { - 0.50}&{0.25}&{0.25}&0\\ 0&{ - 0.375}&{0.375}&0\\ 0&{0.375}&{ - 0.375}&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{\rm{ Row }}2 + (0.5){\rm{ Row }}1}\\ {{\rm{ Row }}3 + (0.5){\rm{ Row }}1} \end{array}} \right\}\\ \left[ {\begin{array}{*{20}{c}} { - 0.50}&{0.25}&{0.25}&0\\ 0&{ - 0.375}&{0.375}&0\\ 0&0&0&0 \end{array}} \right]::\left\{ { Row 3 + (1) Row 2} \right\} \end{array}\]
Step-4:
The system of equations is: \[ \begin{aligned} - 0.5 y _ { 1 } + 0.25 y _ { 2 } + 0.25 y _ { 3 } & = 0 \\ - 0.375 y _ { 2 } + 0.375 y _ { 3 } & = 0 \end{aligned} \] From the second equation: \[ \begin{aligned} - 0.375 y _ { 2 } + 0.375 y _ { 3 } & = 0 \\ y _ { 2 } & = y _ { 3 } \end{aligned} \] From the first equation: \[ \begin{aligned} - 0.50 y _ { 1 } + 0.25 y _ { 2 } + 0.25 y _ { 3 } & = 0 \\ - 0.50 y _ { 1 } + 0.25 \left( y _ { 3 } \right) + 0.25 y _ { 3 } & = 0 \\ - 0.50 y _ { 1 } + 0.50 y _ { 3 } & = 0 \\ y _ { 1 } & = y _ { 3 } \end{aligned} \] Therefore, the general solution is:
\[ \mathbf { y } = t \left[ \begin{array} { l } { 1 } \\ { 1 } \\ { 1 } \end{array} \right] \]
Step-5:
The steady-state vector is: \[ \begin{aligned} \mathrm { y } & = \dfrac { 1 } { 3 } \left[ \begin{array} { l } { 1 } \\ { 1 } \\ { 1 } \end{array} \right] \\ & = \left[ \begin{array} { c } { 1 / 3 } \\ { 1 / 3 } \\ { 1 / 3 } \end{array} \right] \end{aligned} \] Therefore,
Each food will be preferred equally by the animals after many trails.
We ave to find the steady-state vector for the Markov chain in Exercise 2.
Step-1:
The stochastic Matrix is: \[ P = \left[ \begin{array} { l l } { 0.7 } & { 0.6 } \\ { 0.3 } & { 0.4 } \end{array} \right] \]
Step-2:
The $P-I$ matrix is: \[\begin{array}{l} P - I = \left[ {\begin{array}{*{20}{c}} {0.50}&{0.25}&{0.25}\\ {0.25}&{0.50}&{0.25}\\ {0.25}&{0.25}&{0.50} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} { - 0.50}&{0.25}&{0.25}\\ {0.25}&{ - 0.50}&{0.25}\\ {0.25}&{0.25}&{ - 0.50} \end{array}} \right] \end{array}\]
Step-3:
The augmented Matrix for the system is: \[ \left[ \begin{array} { c c c | c } { - 0.50 } & { 0.25 } & { 0.25 } & { 0 } \\ { 0.25 } & { - 0.50 } & { 0.25 } & { 0 } \\ { 0.25 } & { 0.25 } & { - 0.50 } & { 0 } \end{array} \right] \] Row-Reduce the augmented matrix: \[\begin{array}{l} M = \left[ {\begin{array}{*{20}{c}} { - 0.50}&{0.25}&{0.25}&0\\ {0.25}&{ - 0.50}&{0.25}&0\\ {0.25}&{0.25}&{ - 0.50}&0 \end{array}} \right]\\ \left[ {\begin{array}{*{20}{c}} { - 0.50}&{0.25}&{0.25}&0\\ 0&{ - 0.375}&{0.375}&0\\ 0&{0.375}&{ - 0.375}&0 \end{array}} \right]::\left\{ {\begin{array}{*{20}{l}} {{\rm{ Row }}2 + (0.5){\rm{ Row }}1}\\ {{\rm{ Row }}3 + (0.5){\rm{ Row }}1} \end{array}} \right\}\\ \left[ {\begin{array}{*{20}{c}} { - 0.50}&{0.25}&{0.25}&0\\ 0&{ - 0.375}&{0.375}&0\\ 0&0&0&0 \end{array}} \right]::\left\{ { Row 3 + (1) Row 2} \right\} \end{array}\]
Step-4:
The system of equations is: \[ \begin{aligned} - 0.5 y _ { 1 } + 0.25 y _ { 2 } + 0.25 y _ { 3 } & = 0 \\ - 0.375 y _ { 2 } + 0.375 y _ { 3 } & = 0 \end{aligned} \] From the second equation: \[ \begin{aligned} - 0.375 y _ { 2 } + 0.375 y _ { 3 } & = 0 \\ y _ { 2 } & = y _ { 3 } \end{aligned} \] From the first equation: \[ \begin{aligned} - 0.50 y _ { 1 } + 0.25 y _ { 2 } + 0.25 y _ { 3 } & = 0 \\ - 0.50 y _ { 1 } + 0.25 \left( y _ { 3 } \right) + 0.25 y _ { 3 } & = 0 \\ - 0.50 y _ { 1 } + 0.50 y _ { 3 } & = 0 \\ y _ { 1 } & = y _ { 3 } \end{aligned} \] Therefore, the general solution is:
\[ \mathbf { y } = t \left[ \begin{array} { l } { 1 } \\ { 1 } \\ { 1 } \end{array} \right] \]
Step-5:
The steady-state vector is: \[ \begin{aligned} \mathrm { y } & = \dfrac { 1 } { 3 } \left[ \begin{array} { l } { 1 } \\ { 1 } \\ { 1 } \end{array} \right] \\ & = \left[ \begin{array} { c } { 1 / 3 } \\ { 1 / 3 } \\ { 1 / 3 } \end{array} \right] \end{aligned} \] Therefore,
Each food will be preferred equally by the animals after many trails.