Finding a Unique Solution to a System of Equations

Click For Summary

Homework Help Overview

The discussion revolves around finding a unique solution to a system of equations, specifically examining the conditions under which the system remains consistent or inconsistent based on the parameters a and b.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants explore the implications of different values for a and b on the uniqueness of solutions, questioning the conditions that lead to unique, infinite, or no solutions. They discuss specific cases and the potential ambiguity in the wording of the problem statement.

Discussion Status

Participants are actively engaging with the problem, clarifying conditions for solutions and expressing uncertainty about the wording in the problem statement. Some guidance has been provided regarding the implications of certain values for a and b, but no consensus has been reached on the interpretation of all conditions.

Contextual Notes

There is mention of specific conditions that lead to unique or inconsistent solutions, as well as a potential oversight regarding the completeness of the listed cases. The discussion reflects an ongoing examination of the problem's parameters and their implications.

The Head
Messages
137
Reaction score
2
Homework Statement
Find when the system of equations is unique:

x-y-2z-2w= 3
y+z+w= 4a+3
z+3w= -4a-4
(-a+2)w= b+4a^2-4a-7
Relevant Equations
Full Rank = Unique
It makes sense that a=2 would cause problems because then we wouldn't have a matrix of full rank and we'd be unable to determine a value for w. But the key also says that when b+4a^2-4a-7≠0. Why is that an issue? For example, if a=1, that just says implies that w=0. Through back-subsitution, we get z=-8, y=15, x=2. And the solution: (2, 15,-8, 0) is unique still because it's the only possible solution, right?

What am I missing here?
 
  • Like
Likes   Reactions: Delta2
Physics news on Phys.org
I think you are correct.

If ##a = 2## then there are no solutions, unless also ##b + 4a^2 -4a -7 = 0## when there are infinitely many. Is that what the book means?
 
  • Like
Likes   Reactions: The Head
Hmm, yeah, maybe it was just the wording that is a little ambiguous. It says for a=2 AND b≠-1, the solution is inconsistent (makes sense, because then you get something like 0=1), for a≠2 AND b+4a^2-4a-7≠0, it is unique, and for a=2 AND b=-1 there are infinitely many solutions (also makes sense, since then you get 0=0). As long as I'm not missing something fundamental and the basic conditions I'm looking for are correct, I'm not too worried about it.

Thanks for helping out with this. I appreciate it.
 
  • Like
Likes   Reactions: Delta2
The Head said:
Hmm, yeah, maybe it was just the wording that is a little ambiguous. It says for a=2 AND b≠-1, the solution is inconsistent (makes sense, because then you get something like 0=1), for a≠2 AND b+4a^2-4a-7≠0, it is unique, and for a=2 AND b=-1 there are infinitely many solutions (also makes sense, since then you get 0=0). As long as I'm not missing something fundamental and the basic conditions I'm looking for are correct, I'm not too worried about it.

Thanks for helping out with this. I appreciate it.
But that list leaves out a≠2 AND b+4a^2-4a-7=0, and as you say that case does yield a unique solution. So you are right, the "AND b+4a^2-4a-7≠0" is entirely redundant.
 
  • Like
Likes   Reactions: The Head

Similar threads

  • · Replies 2 ·
Replies
2
Views
1K
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 27 ·
Replies
27
Views
2K
  • · Replies 7 ·
Replies
7
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 9 ·
Replies
9
Views
2K
Replies
3
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K