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Clarence

  • one year ago

How would I go abouts solving a 3x4 matrix system?

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  1. clarence
    • one year ago
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    \[\left[\begin{matrix}5 & 3 & -9 & 2 \\ -4 & -4 & -8 & 5 \\ 21 & 11 & -53 & 15\end{matrix}\right]\]

  2. clarence
    • one year ago
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    Specifically one such as above, except with a line down the middle before the last column?

  3. ganeshie8
    • one year ago
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    \[\left[\begin{matrix}5 & 3 & -9 &|& 2 \\ -4 & -4 & -8 &|& 5 \\ 21 & 11 & -53 &|& 15\end{matrix}\right]\] like this ?

  4. clarence
    • one year ago
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    Yes, how did you manage to do that on here by the way?

  5. ganeshie8
    • one year ago
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    you may right click and see latex commands : |dw:1440239723612:dw|

  6. clarence
    • one year ago
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    Oh, okay then.

  7. ganeshie8
    • one year ago
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    btw, we call it a 3x3 system the size of matrix \(A\) is \(3\times 3\) in the system of equations : \[Ax=b\]

  8. ganeshie8
    • one year ago
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    you're just attaching that "b" also as the last column to get that 3x4 matrix that 3x4 matrix with "b" attached is called "augmented matrix"

  9. ganeshie8
    • one year ago
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    familiar with elimination/row opearations ?

  10. clarence
    • one year ago
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    Familiar but not well enough to solve the problem at hand unfortunately.

  11. ganeshie8
    • one year ago
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    first see that the solutions are not affected by row operations because you're just adding/subtracting a multiple of one row from the other... this doesn't change the solutions

  12. ganeshie8
    • one year ago
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    if you have x + y = 2 x+2y = 3 then, adding one equation to other won't change the solutions because you're adding the same thing to both sides of a particular equation

  13. ganeshie8
    • one year ago
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    first convince yourself that below two systems will have same solutions : system1 ``` x + y = 2 x+2y = 3 ``` system2 ``` x + y = 2 2x+3y = 5 ```

  14. clarence
    • one year ago
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    Okay..

  15. ganeshie8
    • one year ago
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    your goal in elimination is to get a triangular matrix

  16. clarence
    • one year ago
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    So, not 4x + 6y = 10?

  17. ganeshie8
    • one year ago
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    that is a valid row operation too, you're allowed to multiply entire row by a constant

  18. ganeshie8
    • one year ago
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    |dw:1440240774738:dw|

  19. ganeshie8
    • one year ago
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    you want to change the starting matrix to above form

  20. ganeshie8
    • one year ago
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    start by working the element at 2,1 \[\left[\begin{matrix}5 & 3 & -9 &|& 2 \\ -4 & -4 & -8 &|& 5 \\ 21 & 11 & -53 &|& 15\end{matrix}\right]\] adding 4/5 times first row to second row does the job, right ?

  21. clarence
    • one year ago
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    Yes, at least I think so anyway

  22. ganeshie8
    • one year ago
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    do it

  23. clarence
    • one year ago
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    So now the second and third row match?

  24. ganeshie8
    • one year ago
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    no, what do you get after doing that first step

  25. clarence
    • one year ago
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    \[\left[\begin{matrix}5 & 3 & -9 & | 2 \\ 21 & 11 & -53 & |15 \\ 21 & 11 & -53 & |15\end{matrix}\right]\]

  26. ganeshie8
    • one year ago
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    how ?

  27. clarence
    • one year ago
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    5 times the 1st row added into the 2nd row?

  28. ganeshie8
    • one year ago
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    yeah you could do that!

  29. clarence
    • one year ago
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    Was I not supposed to do that?

  30. ganeshie8
    • one year ago
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    that is not a standard method, but what you did is a better one..

  31. ganeshie8
    • one year ago
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    so lets keep going

  32. ganeshie8
    • one year ago
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    for second step, maybe subtract second row from third row

  33. ganeshie8
    • one year ago
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    \[\left[\begin{matrix}5 & 3 & -9 & | 2 \\ 21 & 11 & -53 & |15 \\ 21 & 11 & -53 & |15\end{matrix}\right]\] R3-R2 gives : \[\left[\begin{matrix}5 & 3 & -9 & | 2 \\ 21 & 11 & -53 & |15 \\ 0& 0& 0& |0\end{matrix}\right]\]

  34. clarence
    • one year ago
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    Okay then, what's the next step?

  35. ganeshie8
    • one year ago
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    next look at the element at 2,1 position you want to make it 0 so that you get a triangular matrix

  36. clarence
    • one year ago
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    So 21 needs to be 0?

  37. ganeshie8
    • one year ago
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    yes how to do that using row operations ?

  38. clarence
    • one year ago
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    Multiply the 2nd row by 5 and then add 21x1st row to the second row?

  39. ganeshie8
    • one year ago
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    Multiply the 2nd row by 5 and then "subtract" 21x1st row to the second row?

  40. ganeshie8
    • one year ago
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    you mean that right ?

  41. clarence
    • one year ago
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    Yes, sorry I was thinking in terms of -21

  42. ganeshie8
    • one year ago
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    that works perfectly! do it

  43. clarence
    • one year ago
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    \[\left[\begin{matrix}5 & 3 & -9 & |2 \\ 0 & -8 & -76 & |33 \\ 0 & 0 & 0 & |0\end{matrix}\right]\]

  44. ganeshie8
    • one year ago
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    Looks good! so above reduced system is equivalent to : ``` 5x + 3y - 9z = 2 - 8y - 76z = 33 ```

  45. ganeshie8
    • one year ago
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    let \(z=0\) and find a particular solution

  46. clarence
    • one year ago
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    x-y=7?

  47. ganeshie8
    • one year ago
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    how ? plugin \(z=0\) in the bottom equation first

  48. clarence
    • one year ago
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    Oh, I just plugged 0 in for z for both equations, my bad.

  49. ganeshie8
    • one year ago
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    the advantage of triangular matrix is just that solving is trivial from bottom

  50. clarence
    • one year ago
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    So y = -33/8 for the bottom solution?

  51. ganeshie8
    • one year ago
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    yes, find x also

  52. ganeshie8
    • one year ago
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    plugin z=0, y=-33/8 in first equation

  53. clarence
    • one year ago
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    x = 23/8

  54. ganeshie8
    • one year ago
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    right, so our particular solution is (23/8, -33/8, 0)

  55. ganeshie8
    • one year ago
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    save that, we need to find nullspace

  56. ganeshie8
    • one year ago
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    for nullspace, you want to solve : ``` 5x + 3y - 9z = 0 - 8y - 76z = 0 ```

  57. ganeshie8
    • one year ago
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    plugin \(z=1\) and solve y, x values

  58. clarence
    • one year ago
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    x = 15/2 & y = -19/2?

  59. ganeshie8
    • one year ago
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    Yes, so the nullspace is `t(15/2, -19/2, 1)`

  60. ganeshie8
    • one year ago
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    the complete solution is given by `particular` + `nullspace`

  61. clarence
    • one year ago
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    So the one with z being 0 + what we just got?

  62. ganeshie8
    • one year ago
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    Complete solution : ``` (23/8, -33/8, 0) + t(15/2, -19/2, 1) ```

  63. clarence
    • one year ago
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    Oh I see, that makes sense!

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