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Dido525 Group Title

Linear algebra question?

  • one year ago
  • one year ago

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  1. Dido525 Group Title
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    • one year ago
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  2. Dido525 Group Title
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    Anyone want to help me get started? I am pretty sure I can do the rest on my own.

    • one year ago
  3. Cecily Group Title
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    Represent the system in matrix form

    • one year ago
  4. Dido525 Group Title
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    Yep, that's what I did.

    • one year ago
  5. Dido525 Group Title
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    @Cecily

    • one year ago
  6. Dido525 Group Title
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    How do I go on from there?

    • one year ago
  7. Dido525 Group Title
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    @zepdrix @Outkast3r09

    • one year ago
  8. Dido525 Group Title
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    @dumbcow

    • one year ago
  9. sirm3d Group Title
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    take the determinant of the coefficient matrix, and set it equal to zero.

    • one year ago
  10. Dido525 Group Title
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    Woah woah!!! We haven't learned what a determinant is yet. :P .

    • one year ago
  11. sirm3d Group Title
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    okay. how about gauss-jordan reduction?

    • one year ago
  12. Dido525 Group Title
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    Yeah. We learned that.

    • one year ago
  13. sirm3d Group Title
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    i got something easier than gauss-jordan. multiply equation 2 by (-1), and equation 3 by 2, then add the two equations.

    • one year ago
  14. Dido525 Group Title
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    Elimination?

    • one year ago
  15. sirm3d Group Title
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    yup.

    • one year ago
  16. Dido525 Group Title
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    3y-3z=-a+2 ?

    • one year ago
  17. sirm3d Group Title
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    not 3y+3z = -a + 2 ?

    • one year ago
  18. Dido525 Group Title
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    Yeah. My mistake.

    • one year ago
  19. Dido525 Group Title
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    Allright, what next?

    • one year ago
  20. sirm3d Group Title
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    use that equation with equation 1 to eliminate both y and z.

    • one year ago
  21. Dido525 Group Title
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    1 sec.

    • one year ago
  22. Dido525 Group Title
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    THat dosen't eliminate it. It makes it bigger.

    • one year ago
  23. sirm3d Group Title
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    bx + 3y + 3z = a 3y + 3z = 2 - a subtract.

    • one year ago
  24. Dido525 Group Title
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    I used substitution instead. Should till be valid right?

    • one year ago
  25. sirm3d Group Title
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    or if you wish to substitute...

    • one year ago
  26. sirm3d Group Title
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    that's valid too.

    • one year ago
  27. sirm3d Group Title
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    bx + (2-a) =a bx = 2a - 2

    • one year ago
  28. Dido525 Group Title
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    |dw:1358650895030:dw|

    • one year ago
  29. sirm3d Group Title
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    you're not to eliminate a, but y and z.

    • one year ago
  30. Dido525 Group Title
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    Right.

    • one year ago
  31. Dido525 Group Title
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    Lets us elimination then lol.

    • one year ago
  32. Dido525 Group Title
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    bx=2a-2?

    • one year ago
  33. sirm3d Group Title
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    yup.

    • one year ago
  34. Dido525 Group Title
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    Now would I solve for a and substitute this into equation 2?

    • one year ago
  35. sirm3d Group Title
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    now if you put b = 1, and a any real number other than 1, say 3, 0x = 4 0 = 4 which is clearly false.

    • one year ago
  36. sirm3d Group Title
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    err. i mean put b = 0.

    • one year ago
  37. Dido525 Group Title
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    Why can I put 1?

    • one year ago
  38. Dido525 Group Title
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    for a*

    • one year ago
  39. Dido525 Group Title
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    Ohh never mind. I see why.

    • one year ago
  40. Dido525 Group Title
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    For infinitely many solutions I would make 0=0 right?

    • one year ago
  41. sirm3d Group Title
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    right about that.

    • one year ago
  42. Dido525 Group Title
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    Okay what about a unique solution? @sirm3d

    • one year ago
  43. sirm3d Group Title
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    unique solution when b is not equal to zero.

    • one year ago
  44. Dido525 Group Title
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    and a is 1 right?

    • one year ago
  45. sirm3d Group Title
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    any value of a for as long as b not zero will yield a unique solution

    • one year ago
  46. Dido525 Group Title
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    Thanks so much!

    • one year ago
  47. sirm3d Group Title
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    yw.

    • one year ago
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