For which values of a will the following system have no solutions? Exactly one solution? Infinitely many solutions?
x+6y+4z=2
4x+25y+31z=7
2x+13y+(7+a^2)z=-1+a
I tried elimination but that didn't work... I don't know how I would use matricies for this either.

- anonymous

- chestercat

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- anonymous

- anonymous

First, turn it into a matrix!

- anonymous

Umm okay...

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- anonymous

Okay.

- anonymous

Here do you understand how I got this?
[1 6 4 2 ]
[4 25 31 7 ]
[2 13 7a^2 -1+a]

- anonymous

Yeah I do.,

- anonymous

Actually it should be 7+ a^2 so
[1 6 4 2 ]
[4 25 31 7 ]
[2 13 7+a^2 -1+a]
OK now do you know how to row reduce this?

- anonymous

Yeah but what about that a?

- anonymous

- anonymous

We're going to work with that later. It shouldn't get in the way because it's in the bottom row.

- anonymous

I have to still manipulate it like normal correct?

- anonymous

Yes. So first try to get 0s in the first column

- anonymous

|dw:1358201674350:dw|

- anonymous

No wait. That's not right.

- anonymous

Nope :P OK I'll give you the first step. Subtract 4 times the first row from the second row.

- anonymous

Sorry about that. This Linear algebra is fun but totally new to me.

- anonymous

Nono don't be sorry, it's not an easy course at all! I just took it last semester so it's still fresh in my mind. Everybody said it would be easier than calculus, but I think they were crazy, linear is way harder...

- anonymous

If we went 4* Row 1 we get -3 in the 2nd row for the first term.

- anonymous

OK so subtracting 4*R1 from R2 you get.
4*R1 = 4 * [ 1 6 4 2 ] = [4 24 16 8]
R2 - 4*R1 = [ 4 25 31 7 ] - [4 24 16 8] = [0 1 15 -1]
Make sense? So now your matrix looks like
[1 6 4 2 ]
[0 1 15 -1 ]
[2 13 7+a^2 -1+a]

- anonymous

Wait... Oops.

- anonymous

KK. Let me try myslef again.

- anonymous

I was looking at another Matrix >.< .

- anonymous

Kk. I got what you got. Now I would go and divide Row 3 by 2 right?

- anonymous

- anonymous

Nope you would do Row 3 - 2 times Row 1. You want to get a 0 in the first spot of row 3

- anonymous

Okay, that works too.

- anonymous

For row 3 I got [0 1 5+a^2 -3+a]

- anonymous

Wait, third colums in wrong.

- anonymous

[0 1 -1+a^2 -3+a]

- anonymous

Perfect very good! :D
[1 6 4 2 ]
[0 1 15 -1 ]
[0 1 -1+a^2 -3+a]
almost there. Now you want to get a 0 in row 3 column 2

- anonymous

[0 1 -1+a^2 -5+a] . Woudn't it be that actually?

- anonymous

Oops yes! Missed that, just skimmed over it. Good job noticing that. So now it's
[1 6 4 2 ]
[0 1 15 -1 ]
[0 1 -1+a^2 -5+a]

- anonymous

K I got [0 0 -16+a^2 -4+a]

- anonymous

Perfect! OK I have to go so I'll just go over the answer and stop giving you hints. If you want to guess for yourself do that first then read my answer!
[1 6 4 2 ]
[0 1 15 -1 ]
[0 0 a^2-16 a-4]
is equal to (by difference of squares)
[1 6 4 2 ]
[0 1 15 -1 ]
[0 0 (a+4)(a-4) a-4]
So you've got
(a+4)(a-4) * z = a-4
If the left equals 0, but the right equals a number, then there is no solution because:
0*z=1 has no solution. So when does the left equal 0 but the right equal a number? When a = -4. Then 0*z = -8, no solution.
If they both equal a number, let's say a=5 then
[1 6 4 2 ]
[0 1 15 -1 ]
[0 0 9 1 ]
You now have 3 pivot columns, which means that there is only 1 solution.
If they both equal 0, then you will have a free variable in the third column and you will have infinite solutions:
[1 6 4 2 ]
[0 1 15 -1 ]
[0 0 0 0 ]
When do they both equal 0? When a =4, infinite solutions.
Good luck in linear algebra! It's a fun class, but make sure you don't fall behind, it gets really hard!!

- anonymous

If you've got any questions about my explanation, ask them and I'll answer later.

- anonymous

WHat's a Pivot column?

- anonymous

And thanks for the encouragement :) .

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