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 15E from Chapter 1.5 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 15E
Chapter:
Problem:
Follow the method of Example to describe the solutions of the following system in parametric vector form. Also, give a geometric description of the solution set and compare it to that in Exercise...
Step-by-Step Solution
Given information
We are given with following system of Equations. \[\begin{array}{l}{x_1} + 3{x_2} + {x_3} = 1\\ - 4{x_1} - 9{x_2} + 2{x_3} = - 1\\ - 3{x_2} - 6{x_3} = - 3\end{array}\]We have to find the solution of above equations in parametric form
Let, \[\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}1&3&1\\{ - 4}&{ - 9}&2\\0&{ - 3}&{ - 6}\end{array}} \right]\\b = \left[ {\begin{array}{*{20}{c}}1\\{ - 1}\\{ - 3}\end{array}} \right]\end{array}\]
Step 1: The Augmented Matrix
\[M = \left[ {A\,\,\,\,b} \right] = \left[ {\begin{array}{*{20}{c}}1&3&1&1\\{ - 4}&{ - 9}&2&{ - 1}\\0&{ - 3}&{ - 6}&{ - 3}\end{array}} \right]\]
Step 2: Row Reduced Augmented Matrix
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&3&1&1\\{ - 4}&{ - 9}&2&{ - 1}\\0&{ - 3}&{ - 6}&{ - 3}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&3&1&1\\0&3&6&3\\0&{ - 3}&{ - 6}&{ - 3}\end{array}} \right]\,\,\,::\,\,\left\{ {{R_2} = {R_2} + 4{R_1}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&3&1&1\\0&3&6&3\\0&0&0&0\end{array}} \right]\,\,\,::\,\,\left\{ {{R_3} = {R_3} + {R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&3&1&1\\0&1&2&1\\0&0&0&0\end{array}} \right]\,\,\,::\,\,\left\{ {{R_2} = \dfrac{{{R_2}}}{3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&0&{ - 5}&{ - 2}\\0&1&2&1\\0&0&0&0\end{array}} \right]\,\,\,::\,\,\left\{ {{R_1} = {R_1} - 3{R_2}} \right\}\end{array}\]We can see that there are only 2 pivot rows for 3 variables, so we assume one variable as free:
Step 3: The system of Equations
The two corresponding equations are \[\begin{array}{l}{x_1} - 5{x_3} = - 2\\{x_2} + 2{x_3} = 1\end{array}\]By considering $x_3$ as free.\[\begin{array}{l}{x_1} = - 2 + 5{x_3}\\{x_2} = 1 - 2{x_3}\\{x_3}\,\,{\rm{is}}\,\,{\rm{free}}\end{array}\]
Step 4: The General Solution
\[\begin{array}{l}x = \left[ {\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 2 + 5{x_3}}\\{1 - 2{x_3}}\\{{x_3}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{5{x_3}}\\{ - 2{x_3}}\\{{x_3}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\0\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}5\\{ - 2}\\1\end{array}} \right]\end{array}\]Therefore, the General Solution is
\[x = \left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\0\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}5\\{ - 2}\\1\end{array}} \right]\]
Step 5: Geometric Description
The solution set is an equation of line in 3d obtained by intersection of planes given by:\[\begin{array}{l}{x_1} + 3{x_2} + {x_3} = 1\\ - 4{x_1} - 9{x_2} + 2{x_3} = - 1\end{array}\]https://imgur.com/HYbDMON
https://imgur.com/wzZ74Yj
https://imgur.com/GJoymHk
We are given with following system of Equations. \[\begin{array}{l}{x_1} + 3{x_2} + {x_3} = 1\\ - 4{x_1} - 9{x_2} + 2{x_3} = - 1\\ - 3{x_2} - 6{x_3} = - 3\end{array}\]We have to find the solution of above equations in parametric form
Let, \[\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}1&3&1\\{ - 4}&{ - 9}&2\\0&{ - 3}&{ - 6}\end{array}} \right]\\b = \left[ {\begin{array}{*{20}{c}}1\\{ - 1}\\{ - 3}\end{array}} \right]\end{array}\]
Step 1: The Augmented Matrix
\[M = \left[ {A\,\,\,\,b} \right] = \left[ {\begin{array}{*{20}{c}}1&3&1&1\\{ - 4}&{ - 9}&2&{ - 1}\\0&{ - 3}&{ - 6}&{ - 3}\end{array}} \right]\]
Step 2: Row Reduced Augmented Matrix
\[\begin{array}{l}M = \left[ {\begin{array}{*{20}{c}}1&3&1&1\\{ - 4}&{ - 9}&2&{ - 1}\\0&{ - 3}&{ - 6}&{ - 3}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&3&1&1\\0&3&6&3\\0&{ - 3}&{ - 6}&{ - 3}\end{array}} \right]\,\,\,::\,\,\left\{ {{R_2} = {R_2} + 4{R_1}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&3&1&1\\0&3&6&3\\0&0&0&0\end{array}} \right]\,\,\,::\,\,\left\{ {{R_3} = {R_3} + {R_2}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&3&1&1\\0&1&2&1\\0&0&0&0\end{array}} \right]\,\,\,::\,\,\left\{ {{R_2} = \dfrac{{{R_2}}}{3}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&0&{ - 5}&{ - 2}\\0&1&2&1\\0&0&0&0\end{array}} \right]\,\,\,::\,\,\left\{ {{R_1} = {R_1} - 3{R_2}} \right\}\end{array}\]We can see that there are only 2 pivot rows for 3 variables, so we assume one variable as free:
Step 3: The system of Equations
The two corresponding equations are \[\begin{array}{l}{x_1} - 5{x_3} = - 2\\{x_2} + 2{x_3} = 1\end{array}\]By considering $x_3$ as free.\[\begin{array}{l}{x_1} = - 2 + 5{x_3}\\{x_2} = 1 - 2{x_3}\\{x_3}\,\,{\rm{is}}\,\,{\rm{free}}\end{array}\]
Step 4: The General Solution
\[\begin{array}{l}x = \left[ {\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 2 + 5{x_3}}\\{1 - 2{x_3}}\\{{x_3}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\0\end{array}} \right] + \left[ {\begin{array}{*{20}{c}}{5{x_3}}\\{ - 2{x_3}}\\{{x_3}}\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\0\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}5\\{ - 2}\\1\end{array}} \right]\end{array}\]Therefore, the General Solution is
\[x = \left[ {\begin{array}{*{20}{c}}{ - 2}\\1\\0\end{array}} \right] + {x_3}\left[ {\begin{array}{*{20}{c}}5\\{ - 2}\\1\end{array}} \right]\]
Step 5: Geometric Description
The solution set is an equation of line in 3d obtained by intersection of planes given by:\[\begin{array}{l}{x_1} + 3{x_2} + {x_3} = 1\\ - 4{x_1} - 9{x_2} + 2{x_3} = - 1\end{array}\]https://imgur.com/HYbDMON
https://imgur.com/wzZ74Yj
https://imgur.com/GJoymHk