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 10E from Chapter 1.1 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 10E
Chapter:
Problem:
In Exercises 7-10, the augmented matrix of a linear system has been reduced by row operations to the form shown. In each case, continue the appropriate row operations and describe the solution set of the original system.
Step-by-Step Solution
Step 1
We are given with a row reduced matrix form.:
\[M = \left[ {\begin{array}{*{20}{r}}1&{ - 2}&0&3&{ - 2}\\0&1&0&{ - 4}&7\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{array}} \right]\]We have to find the solution set using the appropriate steps.
Step 2: The Augmented matrix Form
\[\left[ {\begin{array}{*{20}{c}}1&{ - 2}&0&3\\0&1&0&{ - 4}\\0&0&1&0\\0&0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 2}\\7\\6\\{ - 3}\end{array}} \right]\]
Step 3: System of Equations
\[\begin{array}{l}{x_1} - 2{x_2} + 3{x_4} = - 2\\{x_2} - 4{x_4} = 7\\{x_3} = 6\\{x_4} = - 3\end{array}\]
Step 4: The Back-Substitution
From last two rows, we have\[\begin{array}{l}{x_3} = 6\\{x_4} = - 3\end{array}\]Now substitute in row-2, to get value of $x_2$\[\begin{array}{l}{x_2} - 4\left( { - 3} \right) = 7\\{x_2} + 12 = 7\\{x_2} = 7 - 12\\{x_2} = - 5\end{array}\]Now substitute in row-1, to get value of $x_1$\[\begin{array}{l}{x_1} - 2\left( { - 5} \right) + 3\left( { - 3} \right) = - 2\\{x_1} + 10 - 9 = - 2\\{x_1} + 1 = - 2\\{x_1} = - 3\end{array}\]
ANSWERS
\[\left[ {\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 5}\\6\\{ - 3}\end{array}} \right]\]
We are given with a row reduced matrix form.:
\[M = \left[ {\begin{array}{*{20}{r}}1&{ - 2}&0&3&{ - 2}\\0&1&0&{ - 4}&7\\0&0&1&0&6\\0&0&0&1&{ - 3}\end{array}} \right]\]We have to find the solution set using the appropriate steps.
Step 2: The Augmented matrix Form
\[\left[ {\begin{array}{*{20}{c}}1&{ - 2}&0&3\\0&1&0&{ - 4}\\0&0&1&0\\0&0&0&1\end{array}} \right]\left[ {\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 2}\\7\\6\\{ - 3}\end{array}} \right]\]
Step 3: System of Equations
\[\begin{array}{l}{x_1} - 2{x_2} + 3{x_4} = - 2\\{x_2} - 4{x_4} = 7\\{x_3} = 6\\{x_4} = - 3\end{array}\]
Step 4: The Back-Substitution
From last two rows, we have\[\begin{array}{l}{x_3} = 6\\{x_4} = - 3\end{array}\]Now substitute in row-2, to get value of $x_2$\[\begin{array}{l}{x_2} - 4\left( { - 3} \right) = 7\\{x_2} + 12 = 7\\{x_2} = 7 - 12\\{x_2} = - 5\end{array}\]Now substitute in row-1, to get value of $x_1$\[\begin{array}{l}{x_1} - 2\left( { - 5} \right) + 3\left( { - 3} \right) = - 2\\{x_1} + 10 - 9 = - 2\\{x_1} + 1 = - 2\\{x_1} = - 3\end{array}\]
ANSWERS
\[\left[ {\begin{array}{*{20}{l}}{{x_1}}\\{{x_2}}\\{{x_3}}\\{{x_4}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 3}\\{ - 5}\\6\\{ - 3}\end{array}} \right]\]