# Matrices Question

1. Sep 10, 2005

### danizh

Last edited: Sep 10, 2005
2. Sep 10, 2005

### TD

If the z-coëfficiënt becomes 0 while the constant isn't 0, your system won't have a solution.

So $$k^2 - k = 0 \Leftrightarrow k = 0 \vee k = 1$$.

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

Last edited: Sep 10, 2005
3. Sep 10, 2005

### danizh

Awesome, thanks for the help!
Wouldn't "+1" have no solution as well?

4. Sep 10, 2005

### TD

I mean +1 lol, it has a solution for -1

I'll correct.

---

Done.

Since, for k = -1, the last row becomes: 0 0 2 | 1 => 2z = 1 => z = 1/2

5. Sep 10, 2005

### danizh

But (-1)^2 - 1 = 0 as well. So there are no solutions for "+/- 1," I suppose.
Please correct me if I am wrong. :uhh:

6. Sep 10, 2005

### TD

Watch out, mind your minus! For k = -1, you get:

$$k^2 - k \Rightarrow \left( { - 1} \right)^2 - \left( { - 1} \right) = 1 + 1 = 2$$

Which is what I wrote at the end of my previous post.

7. Sep 10, 2005

### danizh

Sorry, I missed that. :tongue2:
Thanks for the help!

8. Sep 10, 2005

### TD

No problem

9. Sep 11, 2005

### danizh

I don't think that's true, since anything greater than 2 seems to have no solution as well.

10. Sep 12, 2005

### TD

How is that?

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