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 20E from Chapter 1.2 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 20E
Chapter:
Problem:
Choose h and k such that the system has (a) no solution, (b) a unique solution, and (c) many solutions. Give separate answers for each part...
Step-by-Step Solution
Step 1
Given system of equations:
\[\begin{array}{l}{x_1} + 3{x_2} = 2\\3{x_1} + h{x_2} = k\end{array}\]We have to find the values of h and k, such that the system has (a) No solution (b) A unique solution (c) Many Solutions.
Condition for No solution: System has an equation of the form $0 = c$, where $c \ne 0$
Condition for unique solution: There should no zero in pivot position (leading location).
Condition for many solutions: the system should have at least one free variable;
Step 2: The augmented form
\[A = \left[ {\begin{array}{*{20}{l}}1&3&2\\3&h&k\end{array}} \right]\]
Step 3: Convert to Echelon Form
\[\begin{array}{l}A = \left[ {\begin{array}{*{20}{l}}1&3&2\\3&h&k\end{array}} \right]\\A = \left[ {\begin{array}{*{20}{l}}1&3&2\\0&{h - 9}&{k - 6}\end{array}} \right]::{R_2} = {R_2} - 3{R_1}\end{array}\]
Step 4: Condition for No Solution
From the last row, if \[\begin{array}{l}h-9 = 0\\k - 6 \ne 0\end{array}\]So, \[\begin{array}{l}h-9 = 0 \Rightarrow h = 9\\k - 6 \ne 0 \Rightarrow k \ne 6\end{array}\]
Step 5: Condition for a unique Solution
The system should not have any free variables so from last row:\[h-9 \ne 0 \Rightarrow h \ne 9\]
Step 6: Condition for many Solutions
The system must have at least one free variable; So, we make the last row to 0, making infinite possible solutions.\[\begin{array}{l}h-9 = 0 \Rightarrow h = 9\\k - 6 = 0 \Rightarrow k = 6\end{array}\]
ANSWERS
(a) No Solution: $h = 9;\,\,k \ne 6$
(b) Unique Solution: $h \ne 9$
(c) Many Solutions: $h = 9;\,\,k = 6$
Given system of equations:
\[\begin{array}{l}{x_1} + 3{x_2} = 2\\3{x_1} + h{x_2} = k\end{array}\]We have to find the values of h and k, such that the system has (a) No solution (b) A unique solution (c) Many Solutions.
Condition for No solution: System has an equation of the form $0 = c$, where $c \ne 0$
Condition for unique solution: There should no zero in pivot position (leading location).
Condition for many solutions: the system should have at least one free variable;
Step 2: The augmented form
\[A = \left[ {\begin{array}{*{20}{l}}1&3&2\\3&h&k\end{array}} \right]\]
Step 3: Convert to Echelon Form
\[\begin{array}{l}A = \left[ {\begin{array}{*{20}{l}}1&3&2\\3&h&k\end{array}} \right]\\A = \left[ {\begin{array}{*{20}{l}}1&3&2\\0&{h - 9}&{k - 6}\end{array}} \right]::{R_2} = {R_2} - 3{R_1}\end{array}\]
Step 4: Condition for No Solution
From the last row, if \[\begin{array}{l}h-9 = 0\\k - 6 \ne 0\end{array}\]So, \[\begin{array}{l}h-9 = 0 \Rightarrow h = 9\\k - 6 \ne 0 \Rightarrow k \ne 6\end{array}\]
Step 5: Condition for a unique Solution
The system should not have any free variables so from last row:\[h-9 \ne 0 \Rightarrow h \ne 9\]
Step 6: Condition for many Solutions
The system must have at least one free variable; So, we make the last row to 0, making infinite possible solutions.\[\begin{array}{l}h-9 = 0 \Rightarrow h = 9\\k - 6 = 0 \Rightarrow k = 6\end{array}\]
ANSWERS
(a) No Solution: $h = 9;\,\,k \ne 6$
(b) Unique Solution: $h \ne 9$
(c) Many Solutions: $h = 9;\,\,k = 6$