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 12E from Chapter 1.2 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 12E
Chapter:
Problem:
Find the general solutions of the systems whose augmented matrices are given in Exercises...
Step-by-Step Solution
Given Information
We are given with an augmented matrix, we have to find the general solution of the corresponding system of equations.:
Step 1: The Augmented Matrix
\[A = \left[ {\begin{array}{*{20}{c}}1&{ - 7}&0&6&5\\0&0&1&{ - 2}&{ - 3}\\{ - 1}&7&{ - 4}&2&7\end{array}} \right]\]
Step 2: Row Reduced Echelon Form
\[\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}1&{ - 7}&0&6&5\\0&0&1&{ - 2}&{ - 3}\\{ - 1}&7&{ - 4}&2&7\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 7}&0&6&5\\0&0&1&{ - 2}&{ - 3}\\0&0&{ - 4}&8&{12}\end{array}} \right]\,\,\,::\,\,\left\{ {{R_3} = {R_3} + {R_1}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 7}&0&6&5\\0&0&1&{ - 2}&{ - 3}\\0&0&0&0&0\end{array}} \right]\,\,\,::\,\,\left\{ {{R_3} = {R_3} + 4{R_2}} \right\}\end{array}\]Therefore,
\[A = \left[ {\begin{array}{*{20}{c}}1&{ - 7}&0&6&5\\0&0&1&{ - 2}&{ - 3}\\0&0&0&0&0\end{array}} \right]\,\]
Step 3: System of Equations
From the above step, we can see that above row reduced matrix has 2 pivot columns and 4 variables, so we keep 2 variables as free ($x_2$ and $x_3$).\[\begin{array}{l}{x_1} - 7{x_2} + 6{x_4} = 5 \Rightarrow {x_1} = 5 + 7{x_2} - 6{x_4}\\{x_3} - 2{x_4} = - 3 \Rightarrow {x_2} = - 3 + 2{x_4}\end{array}\]
ANSWER
Therefore,
\[\left\{ \begin{array}{l}{x_1} = 5 + 7{x_2} - 6{x_4}\\{x_2}\,\,{\rm{is}}\,\,{\rm{free}}\\{x_2} = - 3 + 2{x_4}\\{x_4}\,\,{\rm{is}}\,\,{\rm{free}}\end{array} \right.\]
We are given with an augmented matrix, we have to find the general solution of the corresponding system of equations.:
Step 1: The Augmented Matrix
\[A = \left[ {\begin{array}{*{20}{c}}1&{ - 7}&0&6&5\\0&0&1&{ - 2}&{ - 3}\\{ - 1}&7&{ - 4}&2&7\end{array}} \right]\]
Step 2: Row Reduced Echelon Form
\[\begin{array}{l}A = \left[ {\begin{array}{*{20}{c}}1&{ - 7}&0&6&5\\0&0&1&{ - 2}&{ - 3}\\{ - 1}&7&{ - 4}&2&7\end{array}} \right]\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 7}&0&6&5\\0&0&1&{ - 2}&{ - 3}\\0&0&{ - 4}&8&{12}\end{array}} \right]\,\,\,::\,\,\left\{ {{R_3} = {R_3} + {R_1}} \right\}\\ = \left[ {\begin{array}{*{20}{c}}1&{ - 7}&0&6&5\\0&0&1&{ - 2}&{ - 3}\\0&0&0&0&0\end{array}} \right]\,\,\,::\,\,\left\{ {{R_3} = {R_3} + 4{R_2}} \right\}\end{array}\]Therefore,
\[A = \left[ {\begin{array}{*{20}{c}}1&{ - 7}&0&6&5\\0&0&1&{ - 2}&{ - 3}\\0&0&0&0&0\end{array}} \right]\,\]
Step 3: System of Equations
From the above step, we can see that above row reduced matrix has 2 pivot columns and 4 variables, so we keep 2 variables as free ($x_2$ and $x_3$).\[\begin{array}{l}{x_1} - 7{x_2} + 6{x_4} = 5 \Rightarrow {x_1} = 5 + 7{x_2} - 6{x_4}\\{x_3} - 2{x_4} = - 3 \Rightarrow {x_2} = - 3 + 2{x_4}\end{array}\]
ANSWER
Therefore,
\[\left\{ \begin{array}{l}{x_1} = 5 + 7{x_2} - 6{x_4}\\{x_2}\,\,{\rm{is}}\,\,{\rm{free}}\\{x_2} = - 3 + 2{x_4}\\{x_4}\,\,{\rm{is}}\,\,{\rm{free}}\end{array} \right.\]