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 1E from Chapter 1.5 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 1E
Chapter:
Problem:
In Exercises 1-4, determine if the system has a nontrivial solution. Try to use as few row operations as possible...
Step-by-Step Solution
Step 1
Given System of Equations:
\[\begin{array}{l}2{x_1} - 5{x_2} + 8{x_3} = 0\\ - 2{x_1} - 7{x_2} + {x_3} = 0\\4{x_1} + 2{x_2} + 7{x_3} = 0\end{array}\]We have to determine if the system has a nontrivial solution.
Step 2: The Augmented Form
\[M = \left[ {\begin{array}{*{20}{c}}2&{ - 5}&8&0\\{ - 2}&{ - 7}&1&0\\4&2&7&0\end{array}} \right]\]
Step 3: Reduced Echelon Form
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}2&{ - 5}&8&0\\{ - 2}&{ - 7}&1&0\\4&2&7&0\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}2&{ - 5}&8&0\\0&{ - 12}&9&0\\0&{12}&{ - 9}&0\end{array}} \right]\,\,\,\left\{ {{R_3} = {R_3} - 2{R_1};\,\,{R_2} = {R_2} + {R_1}} \right\}\\\\ = \left[ {\begin{array}{*{20}{c}}2&{ - 5}&8&0\\0&{ - 12}&9&0\\0&0&0&0\end{array}} \right]\,\,\,\left\{ {{R_3} = {R_3} + {R_2}} \right\}\end{array}\]
Step 4: general Solution
From the reduced echelon form, we can see that the last row has all 0 entries, making one free variable. If there is any free variable in system of equation, the system, has a non trivial solution.
ANSWER
NON-TRIVAL SOLUTION
Given System of Equations:
\[\begin{array}{l}2{x_1} - 5{x_2} + 8{x_3} = 0\\ - 2{x_1} - 7{x_2} + {x_3} = 0\\4{x_1} + 2{x_2} + 7{x_3} = 0\end{array}\]We have to determine if the system has a nontrivial solution.
Step 2: The Augmented Form
\[M = \left[ {\begin{array}{*{20}{c}}2&{ - 5}&8&0\\{ - 2}&{ - 7}&1&0\\4&2&7&0\end{array}} \right]\]
Step 3: Reduced Echelon Form
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}2&{ - 5}&8&0\\{ - 2}&{ - 7}&1&0\\4&2&7&0\end{array}} \right]\\\\ = \left[ {\begin{array}{*{20}{c}}2&{ - 5}&8&0\\0&{ - 12}&9&0\\0&{12}&{ - 9}&0\end{array}} \right]\,\,\,\left\{ {{R_3} = {R_3} - 2{R_1};\,\,{R_2} = {R_2} + {R_1}} \right\}\\\\ = \left[ {\begin{array}{*{20}{c}}2&{ - 5}&8&0\\0&{ - 12}&9&0\\0&0&0&0\end{array}} \right]\,\,\,\left\{ {{R_3} = {R_3} + {R_2}} \right\}\end{array}\]
Step 4: general Solution
From the reduced echelon form, we can see that the last row has all 0 entries, making one free variable. If there is any free variable in system of equation, the system, has a non trivial solution.
ANSWER
NON-TRIVAL SOLUTION