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Dido525

  • 3 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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  1. Dido525
    • 3 years ago
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    @zepdrix @UnkleRhaukus @saifoo.khan

  2. BluFoot
    • 3 years ago
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    First, turn it into a matrix!

  3. Dido525
    • 3 years ago
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    Umm okay...

  4. Dido525
    • 3 years ago
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    Okay.

  5. BluFoot
    • 3 years ago
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    Here do you understand how I got this? [1 6 4 2 ] [4 25 31 7 ] [2 13 7a^2 -1+a]

  6. Dido525
    • 3 years ago
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    Yeah I do.,

  7. BluFoot
    • 3 years ago
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    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?

  8. Dido525
    • 3 years ago
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    Yeah but what about that a?

  9. Dido525
    • 3 years ago
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    @BluFoot

  10. BluFoot
    • 3 years ago
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    We're going to work with that later. It shouldn't get in the way because it's in the bottom row.

  11. Dido525
    • 3 years ago
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    I have to still manipulate it like normal correct?

  12. BluFoot
    • 3 years ago
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    Yes. So first try to get 0s in the first column

  13. Dido525
    • 3 years ago
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    |dw:1358201674350:dw|

  14. Dido525
    • 3 years ago
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    No wait. That's not right.

  15. BluFoot
    • 3 years ago
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    Nope :P OK I'll give you the first step. Subtract 4 times the first row from the second row.

  16. Dido525
    • 3 years ago
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    Sorry about that. This Linear algebra is fun but totally new to me.

  17. BluFoot
    • 3 years ago
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    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...

  18. Dido525
    • 3 years ago
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    If we went 4* Row 1 we get -3 in the 2nd row for the first term.

  19. BluFoot
    • 3 years ago
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    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]

  20. Dido525
    • 3 years ago
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    Wait... Oops.

  21. Dido525
    • 3 years ago
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    KK. Let me try myslef again.

  22. Dido525
    • 3 years ago
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    I was looking at another Matrix >.< .

  23. Dido525
    • 3 years ago
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    Kk. I got what you got. Now I would go and divide Row 3 by 2 right?

  24. Dido525
    • 3 years ago
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    @BluFoot

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

  26. Dido525
    • 3 years ago
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    Okay, that works too.

  27. Dido525
    • 3 years ago
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    For row 3 I got [0 1 5+a^2 -3+a]

  28. Dido525
    • 3 years ago
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    Wait, third colums in wrong.

  29. Dido525
    • 3 years ago
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    [0 1 -1+a^2 -3+a]

  30. BluFoot
    • 3 years ago
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    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

  31. Dido525
    • 3 years ago
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    [0 1 -1+a^2 -5+a] . Woudn't it be that actually?

  32. BluFoot
    • 3 years ago
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    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]

  33. Dido525
    • 3 years ago
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    K I got [0 0 -16+a^2 -4+a]

  34. BluFoot
    • 3 years ago
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    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!!

  35. BluFoot
    • 3 years ago
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    If you've got any questions about my explanation, ask them and I'll answer later.

  36. Dido525
    • 3 years ago
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    WHat's a Pivot column?

  37. Dido525
    • 3 years ago
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    And thanks for the encouragement :) .

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