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Matrices Question

  • Thread starter danizh
  • Start date
16
0
Please help me solve this problem associated with matrices; I can't seem to figure it out.

The following reduced augmented matrix represents a system of equations of three plans:

1 0 0 | 3
0 1 0 | 4
0 0 (k^2)-k | k

For what value(s) of k will this system have no solution?
:eek:
 
Last edited:

Answers and Replies

TD
Homework Helper
1,020
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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 independant, then you have a 2x3 system (which has infinite solutions).

For k = 1, it has no solution :smile:
 
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16
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Awesome, thanks for the help! :smile:
Wouldn't "+1" have no solution as well?
 
TD
Homework Helper
1,020
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I mean +1 lol, it has a solution for -1 :wink:

I'll correct.

---

Done.

Since, for k = -1, the last row becomes: 0 0 2 | 1 => 2z = 1 => z = 1/2
 
16
0
But (-1)^2 - 1 = 0 as well. So there are no solutions for "+/- 1," I suppose.
Please correct me if I am wrong. :uhh:
 
TD
Homework Helper
1,020
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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.
 
16
0
Sorry, I missed that. :tongue2:
Thanks for the help!
 
TD
Homework Helper
1,020
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No problem :smile:
 
16
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I don't think that's true, since anything greater than 2 seems to have no solution as well.
 
TD
Homework Helper
1,020
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danizh said:
I don't think that's true, since anything greater than 2 seems to have no solution as well.
How is that?

Take k = 3, the last row becomes: 0 0 6 | 3 => 6z = 3 <=> z = 1/2
 

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