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 2E from Chapter 1.5 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 2E
Chapter:
Problem:
Determine if the system has a nontrivial solution. Try to use as few row operations as possible.x1 - 3x2 + 7x3 = 0-2x1 - x2 + 4x3 = 0x1 + 2x2 + 9x3 = 0
Step-by-Step Solution
Given Information
We are given following equations\[\begin{array}{l}{x_1} - 3{x_2} + 7{x_3} = 0\\ - 2{x_1} + {x_2} - 4{x_3} = 0\\{x_1} + 2{x_2} + 9{x_3} = 0\end{array}\]We have to find whether the system has a non trivial solution or not
Step-1: The Augmented matrix
Form an augmented matrix [A 0]. \[M = \left[ {\begin{array}{*{20}{c}}1&{ - 3}&7&0\\{ - 2}&1&{ - 4}&0\\1&2&9&0\end{array}} \right]\]
Step-2: Row- Reduced form of Augmented Matrix
Row-Reduce the Matrix: \[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&{ - 3}&7&0\\{ - 2}&1&{ - 4}&0\\1&2&9&0\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 3}&7&0\\0&{ - 5}&{10}&0\\0&5&2&0\end{array}} \right]::\,\,\left\{ \begin{array}{l}{R_2} = {R_2} + 2{R_1},\\{R_3} = {R_3} - {R_1}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 3}&7&0\\0&{ - 5}&{10}&0\\0&0&{12}&0\end{array}} \right]::\,\,\left\{ {{R_3} = {R_3} + {R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 3}&7&0\\0&1&{ - 2}&0\\0&0&1&0\end{array}} \right]::\,\,\left\{ \begin{array}{l}{R_2} = - \dfrac{{{R_2}}}{5}\\{R_3} = \dfrac{{{R_3}}}{{12}}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 3}&0&0\\0&1&0&0\\0&0&1&0\end{array}} \right]::\,\,\left\{ \begin{array}{l}{R_1} = {R_1} - 7{R_3}\\{R_2} = {R_2} + 2{R_3}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&0&0&0\\0&1&0&0\\0&0&1&0\end{array}} \right]::\,\,\left\{ {{R_1} = {R_1} + 3{R_2}} \right\}\end{array}\]As we can see that there is no free variable,
The system has only trivial solution
We are given following equations\[\begin{array}{l}{x_1} - 3{x_2} + 7{x_3} = 0\\ - 2{x_1} + {x_2} - 4{x_3} = 0\\{x_1} + 2{x_2} + 9{x_3} = 0\end{array}\]We have to find whether the system has a non trivial solution or not
Step-1: The Augmented matrix
Form an augmented matrix [A 0]. \[M = \left[ {\begin{array}{*{20}{c}}1&{ - 3}&7&0\\{ - 2}&1&{ - 4}&0\\1&2&9&0\end{array}} \right]\]
Step-2: Row- Reduced form of Augmented Matrix
Row-Reduce the Matrix: \[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&{ - 3}&7&0\\{ - 2}&1&{ - 4}&0\\1&2&9&0\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 3}&7&0\\0&{ - 5}&{10}&0\\0&5&2&0\end{array}} \right]::\,\,\left\{ \begin{array}{l}{R_2} = {R_2} + 2{R_1},\\{R_3} = {R_3} - {R_1}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 3}&7&0\\0&{ - 5}&{10}&0\\0&0&{12}&0\end{array}} \right]::\,\,\left\{ {{R_3} = {R_3} + {R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 3}&7&0\\0&1&{ - 2}&0\\0&0&1&0\end{array}} \right]::\,\,\left\{ \begin{array}{l}{R_2} = - \dfrac{{{R_2}}}{5}\\{R_3} = \dfrac{{{R_3}}}{{12}}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 3}&0&0\\0&1&0&0\\0&0&1&0\end{array}} \right]::\,\,\left\{ \begin{array}{l}{R_1} = {R_1} - 7{R_3}\\{R_2} = {R_2} + 2{R_3}\end{array} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&0&0&0\\0&1&0&0\\0&0&1&0\end{array}} \right]::\,\,\left\{ {{R_1} = {R_1} + 3{R_2}} \right\}\end{array}\]As we can see that there is no free variable,
The system has only trivial solution