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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.
 one year ago
 one year 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.
 one year ago
 one year ago

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Dido525Best ResponseYou've already chosen the best response.0
@zepdrix @UnkleRhaukus @saifoo.khan
 one year ago

BluFootBest ResponseYou've already chosen the best response.1
First, turn it into a matrix!
 one year ago

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

BluFootBest ResponseYou've already chosen the best response.1
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?
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
Yeah but what about that a?
 one year ago

BluFootBest ResponseYou've already chosen the best response.1
We're going to work with that later. It shouldn't get in the way because it's in the bottom row.
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
I have to still manipulate it like normal correct?
 one year ago

BluFootBest ResponseYou've already chosen the best response.1
Yes. So first try to get 0s in the first column
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
No wait. That's not right.
 one year ago

BluFootBest ResponseYou've already chosen the best response.1
Nope :P OK I'll give you the first step. Subtract 4 times the first row from the second row.
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
Sorry about that. This Linear algebra is fun but totally new to me.
 one year ago

BluFootBest ResponseYou've already chosen the best response.1
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...
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
If we went 4* Row 1 we get 3 in the 2nd row for the first term.
 one year ago

BluFootBest ResponseYou've already chosen the best response.1
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]
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
KK. Let me try myslef again.
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
I was looking at another Matrix >.< .
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
Kk. I got what you got. Now I would go and divide Row 3 by 2 right?
 one year ago

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

Dido525Best ResponseYou've already chosen the best response.0
For row 3 I got [0 1 5+a^2 3+a]
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
Wait, third colums in wrong.
 one year ago

BluFootBest ResponseYou've already chosen the best response.1
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
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
[0 1 1+a^2 5+a] . Woudn't it be that actually?
 one year ago

BluFootBest ResponseYou've already chosen the best response.1
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]
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
K I got [0 0 16+a^2 4+a]
 one year ago

BluFootBest ResponseYou've already chosen the best response.1
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^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!!
 one year ago

BluFootBest ResponseYou've already chosen the best response.1
If you've got any questions about my explanation, ask them and I'll answer later.
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
WHat's a Pivot column?
 one year ago

Dido525Best ResponseYou've already chosen the best response.0
And thanks for the encouragement :) .
 one year ago
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