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 17E from Chapter 1.2 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 17E
Chapter:
Problem:
Determine the value(s) of h such that the matrix is the augmented matrix of a consistent linear system...
Step-by-Step Solution
Step 1
Given Augmented Matrix:
\[A = \left[ {\begin{array}{*{20}{l}}2&3&h\\4&6&7\end{array}} \right]\]We have to find the value of such that the system of equations given by above matrix is consistent. The only case when the system is inconsistent is, when the system has no solution.
Step 2: Row Reduced Echelon Form
Subtract 2 times of Row-1 from Row-2\[\begin{array}{l}A = \left[ {\begin{array}{*{20}{l}}2&3&h\\4&6&7\end{array}} \right]\\\\A = \left[ {\begin{array}{*{20}{l}}2&3&h\\0&0&{7 - 2h}\end{array}} \right]\end{array}\]The system has a solution if in the last row all elements equal to zero: \[\begin{array}{l}7 - 2h = 0\\h = \dfrac{7}{2}\end{array}\]
Step 3: ANSWER
\[h = \dfrac{7}{2}\]
Given Augmented Matrix:
\[A = \left[ {\begin{array}{*{20}{l}}2&3&h\\4&6&7\end{array}} \right]\]We have to find the value of such that the system of equations given by above matrix is consistent. The only case when the system is inconsistent is, when the system has no solution.
Step 2: Row Reduced Echelon Form
Subtract 2 times of Row-1 from Row-2\[\begin{array}{l}A = \left[ {\begin{array}{*{20}{l}}2&3&h\\4&6&7\end{array}} \right]\\\\A = \left[ {\begin{array}{*{20}{l}}2&3&h\\0&0&{7 - 2h}\end{array}} \right]\end{array}\]The system has a solution if in the last row all elements equal to zero: \[\begin{array}{l}7 - 2h = 0\\h = \dfrac{7}{2}\end{array}\]
Step 3: ANSWER
\[h = \dfrac{7}{2}\]