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 7E from Chapter 1.1 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 7E
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
Given Augmented Matrix:
\[A = \left[ {\begin{array}{*{20}{c}}1&7&3&{ - 4}\\0&1&{ - 1}&3\\0&0&0&1\\0&0&1&{ - 2}\end{array}} \right]\]We have to find the general solution of the system of equations of above augmented matrix. By suitable row operations, we will convert the matrix into row reduced echelon form and then find the solution.
Step 2: Interchange Row-4 with Row-3
\[A = \left[ {\begin{array}{*{20}{c}}1&7&3&{ - 4}\\0&1&{ - 1}&3\\0&0&1&{ - 2}\\0&0&0&1\end{array}} \right]\,\,::\,\,{R_3} \Leftrightarrow {R_4}\]
Step 3: Convert the augmented system into set of equations
\[\begin{array}{r}{x_1} + 7{x_2} + 3{x_3} = - 4\\{x_2} - {x_3} = 3\\{x_3} = - 2\\0 = 1\end{array}\]
Step 4: Back substitution
From the last row: \[0 = 1\]As the above statement is never true, the given system of equations has no solution.
ANSWERS
NO SOLUTION
Given Augmented Matrix:
\[A = \left[ {\begin{array}{*{20}{c}}1&7&3&{ - 4}\\0&1&{ - 1}&3\\0&0&0&1\\0&0&1&{ - 2}\end{array}} \right]\]We have to find the general solution of the system of equations of above augmented matrix. By suitable row operations, we will convert the matrix into row reduced echelon form and then find the solution.
Step 2: Interchange Row-4 with Row-3
\[A = \left[ {\begin{array}{*{20}{c}}1&7&3&{ - 4}\\0&1&{ - 1}&3\\0&0&1&{ - 2}\\0&0&0&1\end{array}} \right]\,\,::\,\,{R_3} \Leftrightarrow {R_4}\]
Step 3: Convert the augmented system into set of equations
\[\begin{array}{r}{x_1} + 7{x_2} + 3{x_3} = - 4\\{x_2} - {x_3} = 3\\{x_3} = - 2\\0 = 1\end{array}\]
Step 4: Back substitution
From the last row: \[0 = 1\]As the above statement is never true, the given system of equations has no solution.
ANSWERS
NO SOLUTION