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 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
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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Dido525
 one year ago
Best ResponseYou've already chosen the best response.0@zepdrix @UnkleRhaukus @saifoo.khan

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

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

BluFoot
 one year ago
Best ResponseYou've already chosen the best response.1Actually 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?

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0Yeah but what about that a?

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

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0I have to still manipulate it like normal correct?

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

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0No wait. That's not right.

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

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0Sorry about that. This Linear algebra is fun but totally new to me.

BluFoot
 one year ago
Best ResponseYou've already chosen the best response.1Nono 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...

Dido525
 one year 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.

BluFoot
 one year ago
Best ResponseYou've already chosen the best response.1OK 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]

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0KK. Let me try myslef again.

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0I was looking at another Matrix >.< .

Dido525
 one year 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?

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

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0For row 3 I got [0 1 5+a^2 3+a]

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0Wait, third colums in wrong.

BluFoot
 one year ago
Best ResponseYou've already chosen the best response.1Perfect 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

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0[0 1 1+a^2 5+a] . Woudn't it be that actually?

BluFoot
 one year ago
Best ResponseYou've already chosen the best response.1Oops 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]

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0K I got [0 0 16+a^2 4+a]

BluFoot
 one year ago
Best ResponseYou've already chosen the best response.1Perfect! 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!!

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

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0WHat's a Pivot column?

Dido525
 one year ago
Best ResponseYou've already chosen the best response.0And thanks for the encouragement :) .
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