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Dido525
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.
@zepdrix @UnkleRhaukus @saifoo.khan
First, turn it into a matrix!
Here do you understand how I got this? [1 6 4 2 ] [4 25 31 7 ] [2 13 7a^2 -1+a]
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?
Yeah but what about that a?
We're going to work with that later. It shouldn't get in the way because it's in the bottom row.
I have to still manipulate it like normal correct?
Yes. So first try to get 0s in the first column
No wait. That's not right.
Nope :P OK I'll give you the first step. Subtract 4 times the first row from the second row.
Sorry about that. This Linear algebra is fun but totally new to me.
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...
If we went 4* Row 1 we get -3 in the 2nd row for the first term.
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]
KK. Let me try myslef again.
I was looking at another Matrix >.< .
Kk. I got what you got. Now I would go and divide Row 3 by 2 right?
Nope you would do Row 3 - 2 times Row 1. You want to get a 0 in the first spot of row 3
For row 3 I got [0 1 5+a^2 -3+a]
Wait, third colums in wrong.
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
[0 1 -1+a^2 -5+a] . Woudn't it be that actually?
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]
K I got [0 0 -16+a^2 -4+a]
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!!
If you've got any questions about my explanation, ask them and I'll answer later.
And thanks for the encouragement :) .