- #1
danizh
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Please help me solve this problem associated with matrices; I can't seem to figure it out.
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Matrices Question: Solving a System with No Solution for k | Help Needed
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Homework Help Overview
The discussion revolves around a problem involving a system of equations represented by a reduced augmented matrix. Participants are exploring the conditions under which this system has no solution, particularly focusing on the parameter k.
Discussion Character
- Exploratory, Assumption checking, Problem interpretation
Approaches and Questions Raised
- Participants discuss the implications of the z-coefficient becoming zero while the constant remains non-zero, leading to no solutions. They explore specific values of k, such as 0, 1, and -1, and question the independence of equations in the system.
Discussion Status
The discussion is active, with participants providing insights and corrections regarding the values of k that lead to no solutions. There is an ongoing examination of various cases and interpretations, but no explicit consensus has been reached.
Contextual Notes
Participants are navigating the complexities of the system's equations and the conditions under which they may or may not be independent. There is a focus on the implications of k being equal to specific values and the resulting structure of the augmented matrix.
- #2
TD
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If the z-coëfficiënt becomes 0 while the constant isn't 0, your system won't have a solution.
So [tex]k^2 - k = 0 \Leftrightarrow k = 0 \vee k = 1[/tex].
But for k = 0, the constant is 0 as well so the last equation is no longer independent, then you have a 2x3 system (which has infinite solutions).
For k = 1, it has no solution
So [tex]k^2 - k = 0 \Leftrightarrow k = 0 \vee k = 1[/tex].
But for k = 0, the constant is 0 as well so the last equation is no longer independent, then you have a 2x3 system (which has infinite solutions).
For k = 1, it has no solution
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- #3
danizh
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Awesome, thanks for the help!
Wouldn't "+1" have no solution as well?
Wouldn't "+1" have no solution as well?
- #4
TD
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I mean +1 lol, it has a solution for -1 
I'll correct.
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Done.
Since, for k = -1, the last row becomes: 0 0 2 | 1 => 2z = 1 => z = 1/2
I'll correct.
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Done.
Since, for k = -1, the last row becomes: 0 0 2 | 1 => 2z = 1 => z = 1/2
- #5
danizh
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But (-1)^2 - 1 = 0 as well. So there are no solutions for "+/- 1," I suppose.
Please correct me if I am wrong.
Please correct me if I am wrong.
- #6
TD
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Watch out, mind your minus! For k = -1, you get:
[tex]k^2 - k \Rightarrow \left( { - 1} \right)^2 - \left( { - 1} \right) = 1 + 1 = 2[/tex]
Which is what I wrote at the end of my previous post.
[tex]k^2 - k \Rightarrow \left( { - 1} \right)^2 - \left( { - 1} \right) = 1 + 1 = 2[/tex]
Which is what I wrote at the end of my previous post.
- #7
danizh
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Sorry, I missed that. 
Thanks for the help!
Thanks for the help!
- #8
TD
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No problem
- #9
danizh
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I don't think that's true, since anything greater than 2 seems to have no solution as well.
- #10
TD
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How is that?danizh said:
Take k = 3, the last row becomes: 0 0 6 | 3 => 6z = 3 <=> z = 1/2