Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 7E from Chapter 4.9 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 7E
Chapter:
Problem:
In Exercises 5–8, find the steady-state vector.
Step-by-Step Solution
Given Information
We are given with a matrix: \[ \left[ \begin{array} { r r r } { .7 } & { .1 } & { .1 } \\ { .2 } & { .8 } & { .2 } \\ { .1 } & { .1 } & { .7 } \end{array} \right] \] We have to find the steady state vector
Step-1:
Write the equation: \[\begin{array}{l} P{\bf{y}} = {\bf{y}}\\ P{\bf{y}} - I{\bf{y}} = {\bf{0}}\\ (P - I){\bf{y}} = {\bf{0}} \end{array}\]
Step-2:
Upon substituting P: \[\begin{array}{l} (P - I){\bf{y}} = {\bf{0}}\\ \left( {\left[ {\begin{array}{*{20}{c}} {0.7}&{0.1}&{0.1}\\ {0.2}&{0.8}&{0.2}\\ {0.1}&{0.1}&{0.7} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]} \right)\left( {\begin{array}{*{20}{c}} {{y_1}}\\ {{y_2}}\\ {{y_3}} \end{array}} \right) = \left( {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right)\\ \left[ {\begin{array}{*{20}{c}} { - 0.3}&{0.1}&{0.1}\\ {0.2}&{ - 0.2}&{0.2}\\ {0.1}&{0.1}&{ - 0.3} \end{array}} \right]\left( {\begin{array}{*{20}{c}} {{y_1}}\\ {{y_2}}\\ {{y_3}} \end{array}} \right) = \left( {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right) \end{array}\]
Step-3: The Row-Reduced augmented Matrix
\[\begin{array}{l} M = \left[ {\begin{array}{*{20}{c}} { - 0.3}&{0.1}&{0.1}&0\\ {0.2}&{ - 0.2}&{0.2}&0\\ {0.1}&{0.1}&{ - 0.3}&0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {0.1}&{0.1}&{ - 0.3}&0\\ {0.2}&{ - 0.2}&{0.2}&0\\ { - 0.3}&{0.1}&{0.1}&0 \end{array}} \right]::\left\{ {{R_1} \Leftrightarrow {R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} {0.1}&{0.1}&{ - 0.3}&0\\ 0&{ - 0.4}&{0.8}&0\\ 0&{0.4}&{ - 0.8}&0 \end{array}} \right]::\left\{ \begin{array}{l} {R_2} = {R_2} - 2{R_1}\\ {R_2} + 3{R_1} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}} {0.1}&{0.1}&{ - 0.3}&0\\ 0&{ - 0.4}&{0.8}&0\\ 0&0&0&0 \end{array}} \right]::\left\{ {{R_3} = {R_3} + {R_2}} \right\} \end{array}\]
Step-4: The General Solution
The system of equations: \[ \begin{aligned} 0.1 y _ { 1 } + 0.1 y _ { 2 } - 0.3 y _ { 3 } & = 0 \\ - 0.4 y _ { 2 } + 0.8 y _ { 3 } & = 0 \end{aligned} \] Upon solving: \[\begin{array}{l} - 0.4{y_2} + 0.8{y_3} = 0 \Rightarrow {y_2} = 2{y_3}\\ 0.1{y_1} + 0.1{y_2} - 0.3{y_3} = 0 \Rightarrow {y_1} = - 0.1{y_2} + 0.3{y_3} \Rightarrow {y_1} = {y_3} \end{array}\] Considering $y_3=t$ as a parameter, the general solution is: \[ \mathbf { y } = t \left[ \begin{array} { l } { 1 } \\ { 2 } \\ { 1 } \end{array} \right] \] To get the steady state vector divide by the sum of entries \[ \mathbf { q } = \dfrac { 1 } { 4 } \left[ \begin{array} { l } { 1 } \\ { 2 } \\ { 1 } \end{array} \right] \]
We are given with a matrix: \[ \left[ \begin{array} { r r r } { .7 } & { .1 } & { .1 } \\ { .2 } & { .8 } & { .2 } \\ { .1 } & { .1 } & { .7 } \end{array} \right] \] We have to find the steady state vector
Step-1:
Write the equation: \[\begin{array}{l} P{\bf{y}} = {\bf{y}}\\ P{\bf{y}} - I{\bf{y}} = {\bf{0}}\\ (P - I){\bf{y}} = {\bf{0}} \end{array}\]
Step-2:
Upon substituting P: \[\begin{array}{l} (P - I){\bf{y}} = {\bf{0}}\\ \left( {\left[ {\begin{array}{*{20}{c}} {0.7}&{0.1}&{0.1}\\ {0.2}&{0.8}&{0.2}\\ {0.1}&{0.1}&{0.7} \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} 1&0&0\\ 0&1&0\\ 0&0&1 \end{array}} \right]} \right)\left( {\begin{array}{*{20}{c}} {{y_1}}\\ {{y_2}}\\ {{y_3}} \end{array}} \right) = \left( {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right)\\ \left[ {\begin{array}{*{20}{c}} { - 0.3}&{0.1}&{0.1}\\ {0.2}&{ - 0.2}&{0.2}\\ {0.1}&{0.1}&{ - 0.3} \end{array}} \right]\left( {\begin{array}{*{20}{c}} {{y_1}}\\ {{y_2}}\\ {{y_3}} \end{array}} \right) = \left( {\begin{array}{*{20}{l}} 0\\ 0\\ 0 \end{array}} \right) \end{array}\]
Step-3: The Row-Reduced augmented Matrix
\[\begin{array}{l} M = \left[ {\begin{array}{*{20}{c}} { - 0.3}&{0.1}&{0.1}&0\\ {0.2}&{ - 0.2}&{0.2}&0\\ {0.1}&{0.1}&{ - 0.3}&0 \end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}} {0.1}&{0.1}&{ - 0.3}&0\\ {0.2}&{ - 0.2}&{0.2}&0\\ { - 0.3}&{0.1}&{0.1}&0 \end{array}} \right]::\left\{ {{R_1} \Leftrightarrow {R_3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}} {0.1}&{0.1}&{ - 0.3}&0\\ 0&{ - 0.4}&{0.8}&0\\ 0&{0.4}&{ - 0.8}&0 \end{array}} \right]::\left\{ \begin{array}{l} {R_2} = {R_2} - 2{R_1}\\ {R_2} + 3{R_1} \end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}} {0.1}&{0.1}&{ - 0.3}&0\\ 0&{ - 0.4}&{0.8}&0\\ 0&0&0&0 \end{array}} \right]::\left\{ {{R_3} = {R_3} + {R_2}} \right\} \end{array}\]
Step-4: The General Solution
The system of equations: \[ \begin{aligned} 0.1 y _ { 1 } + 0.1 y _ { 2 } - 0.3 y _ { 3 } & = 0 \\ - 0.4 y _ { 2 } + 0.8 y _ { 3 } & = 0 \end{aligned} \] Upon solving: \[\begin{array}{l} - 0.4{y_2} + 0.8{y_3} = 0 \Rightarrow {y_2} = 2{y_3}\\ 0.1{y_1} + 0.1{y_2} - 0.3{y_3} = 0 \Rightarrow {y_1} = - 0.1{y_2} + 0.3{y_3} \Rightarrow {y_1} = {y_3} \end{array}\] Considering $y_3=t$ as a parameter, the general solution is: \[ \mathbf { y } = t \left[ \begin{array} { l } { 1 } \\ { 2 } \\ { 1 } \end{array} \right] \] To get the steady state vector divide by the sum of entries \[ \mathbf { q } = \dfrac { 1 } { 4 } \left[ \begin{array} { l } { 1 } \\ { 2 } \\ { 1 } \end{array} \right] \]