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anonymous
 4 years ago
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
 4 years ago
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.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0@zepdrix @UnkleRhaukus @saifoo.khan

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0First, turn it into a matrix!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Here do you understand how I got this? [1 6 4 2 ] [4 25 31 7 ] [2 13 7a^2 1+a]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Actually 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
 4 years ago
Best ResponseYou've already chosen the best response.0Yeah but what about that a?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0We're going to work with that later. It shouldn't get in the way because it's in the bottom row.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I have to still manipulate it like normal correct?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes. So first try to get 0s in the first column

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358201674350:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0No wait. That's not right.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Nope :P OK I'll give you the first step. Subtract 4 times the first row from the second row.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry about that. This Linear algebra is fun but totally new to me.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Nono 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
 4 years ago
Best ResponseYou've already chosen the best response.0If we went 4* Row 1 we get 3 in the 2nd row for the first term.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0OK 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
 4 years ago
Best ResponseYou've already chosen the best response.0KK. Let me try myslef again.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I was looking at another Matrix >.< .

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Kk. I got what you got. Now I would go and divide Row 3 by 2 right?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Nope you would do Row 3  2 times Row 1. You want to get a 0 in the first spot of row 3

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Okay, that works too.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0For row 3 I got [0 1 5+a^2 3+a]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Wait, third colums in wrong.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Perfect 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
 4 years ago
Best ResponseYou've already chosen the best response.0[0 1 1+a^2 5+a] . Woudn't it be that actually?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Oops 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
 4 years ago
Best ResponseYou've already chosen the best response.0K I got [0 0 16+a^2 4+a]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Perfect! 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^216 a4] is equal to (by difference of squares) [1 6 4 2 ] [0 1 15 1 ] [0 0 (a+4)(a4) a4] So you've got (a+4)(a4) * z = a4 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
 4 years ago
Best ResponseYou've already chosen the best response.0If you've got any questions about my explanation, ask them and I'll answer later.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0WHat's a Pivot column?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0And thanks for the encouragement :) .
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