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 1.1 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 2E
Chapter:
Problem:
Solve each system in Exercises 1-4 by using elementary row operations on the equations or on the augmented matrix. Follow the systematic elimination procedure described in this section...
Step-by-Step Solution
Given Information
Given set of Equations:
\[\left\{ \begin{array}{l}2{x_1} + 4{x_2} = -4\\ 5{x_1} + 7{x_2} = 11\end{array} \right.\]We have to Solve the system of equations given above by using elementary row operations on the equations or on the augmented matrix. We will prefer the augmented matrix method.
Step 2
Convert the given system of Equations into augmented matrix form. \[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}2&4\\5&7\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 4}\\{11}\end{array}} \right]\\\\A = \left[ {\begin{array}{*{20}{c}}2&4&{ - 4}\\5&7&{11}\end{array}} \right]\end{array}\]
Step 3: ${R_2} = {R_2} - \frac{5}{2}{R_1}$
\[A = \left[ {\begin{array}{*{20}{c}}2&4&{ - 4}\\0&{ - 3}&{21}\end{array}} \right]\]
Step 4: ${R_2} = -{R_2}/3 $
\[A = \left[ {\begin{array}{*{20}{c}}2&4&{ - 4}\\0&1&{ - 7}\end{array}} \right]\]
Step 5: ${R_1} = {R_1} -4{R_2}$
\[A = \left[ {\begin{array}{*{20}{c}}2&0&{24}\\0&1&{ - 7}\end{array}} \right]\]
Step 6: Write the reduced form in system of equations
\[\begin{array}{l}2{x_1} = 24\\{x_2} = - 7\end{array}\]
Step 7
From the last row, we have\[{x_2} = -7\]Now put the value of $x_2$ in first row
\[\begin{array}{l}2{x_1} = 24\\{x_1} = \frac{{24}}{2}\\{x_1} = 12\end{array}\]
ANSWERS
\[\left[ \begin{array}{l}{x_1} = 12\\{x_2} = -7\end{array} \right]\]
Given set of Equations:
\[\left\{ \begin{array}{l}2{x_1} + 4{x_2} = -4\\ 5{x_1} + 7{x_2} = 11\end{array} \right.\]We have to Solve the system of equations given above by using elementary row operations on the equations or on the augmented matrix. We will prefer the augmented matrix method.
Step 2
Convert the given system of Equations into augmented matrix form. \[\begin{array}{l}\left[ {\begin{array}{*{20}{c}}2&4\\5&7\end{array}} \right]\left[ {\begin{array}{*{20}{c}}{{x_1}}\\{{x_2}}\end{array}} \right] = \left[ {\begin{array}{*{20}{c}}{ - 4}\\{11}\end{array}} \right]\\\\A = \left[ {\begin{array}{*{20}{c}}2&4&{ - 4}\\5&7&{11}\end{array}} \right]\end{array}\]
Step 3: ${R_2} = {R_2} - \frac{5}{2}{R_1}$
\[A = \left[ {\begin{array}{*{20}{c}}2&4&{ - 4}\\0&{ - 3}&{21}\end{array}} \right]\]
Step 4: ${R_2} = -{R_2}/3 $
\[A = \left[ {\begin{array}{*{20}{c}}2&4&{ - 4}\\0&1&{ - 7}\end{array}} \right]\]
Step 5: ${R_1} = {R_1} -4{R_2}$
\[A = \left[ {\begin{array}{*{20}{c}}2&0&{24}\\0&1&{ - 7}\end{array}} \right]\]
Step 6: Write the reduced form in system of equations
\[\begin{array}{l}2{x_1} = 24\\{x_2} = - 7\end{array}\]
Step 7
From the last row, we have\[{x_2} = -7\]Now put the value of $x_2$ in first row
\[\begin{array}{l}2{x_1} = 24\\{x_1} = \frac{{24}}{2}\\{x_1} = 12\end{array}\]
ANSWERS
\[\left[ \begin{array}{l}{x_1} = 12\\{x_2} = -7\end{array} \right]\]