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anonymous

  • one year ago

help me plz again

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  1. anonymous
    • one year ago
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  2. anonymous
    • one year ago
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    @Nnesha

  3. anonymous
    • one year ago
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    @paki

  4. anonymous
    • one year ago
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    @dan815

  5. anonymous
    • one year ago
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    @jim_thompson5910

  6. jim_thompson5910
    • one year ago
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    how far did you get? were you able to set up the matrix?

  7. anonymous
    • one year ago
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    i was not i was so confused on this like i dont know where to start

  8. jim_thompson5910
    • one year ago
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    well you first pull out the coefficients of each equation. Do you see what they are?

  9. anonymous
    • one year ago
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    i do not sorry

  10. jim_thompson5910
    • one year ago
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    they are simply the numbers in front of the variables

  11. jim_thompson5910
    • one year ago
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    example: `4x` has the coefficient of `4`

  12. jim_thompson5910
    • one year ago
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    the `-y` is really `-1y`

  13. anonymous
    • one year ago
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    so four and five right

  14. jim_thompson5910
    • one year ago
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    along the first equation, what are the 3 coefficients?

  15. anonymous
    • one year ago
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    4 5 and 1x right

  16. jim_thompson5910
    • one year ago
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    read across the row

  17. jim_thompson5910
    • one year ago
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    not column

  18. jim_thompson5910
    • one year ago
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    do you see what I mean?

  19. anonymous
    • one year ago
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    ohh ok i got you hold on

  20. anonymous
    • one year ago
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    so 4 and three for the first one right

  21. jim_thompson5910
    • one year ago
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    don't forget about the -1

  22. jim_thompson5910
    • one year ago
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    the first equation is really `4x - 1y + 3z = 12`

  23. anonymous
    • one year ago
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    ohh yeah sorry lol

  24. anonymous
    • one year ago
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    so the seciond one is 1x+4y+6z=12

  25. jim_thompson5910
    • one year ago
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    -32 on the right side, not 12

  26. anonymous
    • one year ago
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    opps my bad i looked at it wrong haha i acciedently looked up at the end sorry

  27. anonymous
    • one year ago
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    5x+3y+9z=20 for the last

  28. jim_thompson5910
    • one year ago
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    So if we just focus on the coefficients on the left side, we have this 3x3 matrix \[ \left[\begin{array}{ccc} 4 & -1 & 3\\ 1 & 4 & 6\\ 5 & 3 & 9 \end{array} \right] \]

  29. jim_thompson5910
    • one year ago
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    agreed? or no?

  30. anonymous
    • one year ago
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    yes i agree

  31. jim_thompson5910
    • one year ago
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    the next matrix is simply the matrix of variables written as a column like this \[ \left[\begin{array}{ccc} x\\ y\\ z \end{array} \right] \]

  32. jim_thompson5910
    • one year ago
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    and then finally we have a third matrix of the values on the right side of the equation the next matrix is simply the matrix of variables written as a column like this \[ \left[\begin{array}{ccc} 12\\ -32\\ 20 \end{array} \right] \]

  33. jim_thompson5910
    • one year ago
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    putting this all together, we form this matrix equation \[ \left[\begin{array}{ccc} 4 & -1 & 3\\ 1 & 4 & 6\\ 5 & 3 & 9 \end{array} \right] * \left[\begin{array}{ccc} x\\ y\\ z \end{array} \right] = \left[\begin{array}{ccc} 12\\ -32\\ 20 \end{array} \right] \] making sense so far?

  34. anonymous
    • one year ago
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    yeah

  35. jim_thompson5910
    • one year ago
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    next we form an augmented matrix by ignoring the variables temporarily. We just combine the first and third matrix like this \[ \left[\begin{array}{ccc|c} 4 & -1 & 3 & 12\\ 1 & 4 & 6 & -32\\ 5 & 3 & 9 & 20 \end{array} \right] \] the vertical bar separates the two matrices

  36. anonymous
    • one year ago
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    makes since now

  37. jim_thompson5910
    • one year ago
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    this is where gauss jordan elimination begins

  38. anonymous
    • one year ago
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    ok

  39. jim_thompson5910
    • one year ago
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    do you know how to do gauss jordan elimination? or no?

  40. anonymous
    • one year ago
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    i really dont i have treid and i keep faling at it this is whay im here now hahaha

  41. jim_thompson5910
    • one year ago
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    that's ok

  42. jim_thompson5910
    • one year ago
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    The first thing I would do is swap rows 1 and 2. We denote this with notation R1 <--> R2. I'm doing this to make the element in row1,column1 be a `1` \[ \left[\begin{array}{ccc|c} 4 & -1 & 3 & 12\\ 1 & 4 & 6 & -32\\ 5 & 3 & 9 & 20 \end{array} \right] \] \[ \left[\begin{array}{ccc|c} 1 & 4 & 6 & -32\\ 4 & -1 & 3 & 12\\ 5 & 3 & 9 & 20 \end{array} \right] \begin{array}{c} \phantom\\ R_1 \leftrightarrow R_2\\ \phantom\\ \end{array} \] I couldn't get the thing to line up properly, but the `R1 <--> R2` should either be in row 1 or row 2

  43. jim_thompson5910
    • one year ago
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    let me know when you're ready for the next step

  44. anonymous
    • one year ago
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    OK GIVE ME A SECOND TO GET THIS DOWN sorry for the caps i for got i had them on

  45. anonymous
    • one year ago
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    OKim ready

  46. jim_thompson5910
    • one year ago
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    Now we want to make that `4` in row2,column1 to be a `0`. We do this by computing `R2-4*R1` and replacing all of R2 with that result \[ \left[\begin{array}{ccc|c} 1 & 4 & 6 & -32\\ 0 & -17 & -21 & 140\\ 5 & 3 & 9 & 20 \end{array} \right] \begin{array}{c} \\ R_2 - 4R_1 \rightarrow R_2\\ \\ \end{array} \]

  47. anonymous
    • one year ago
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    OK GOTCHA

  48. jim_thompson5910
    • one year ago
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    Next we want to make that `5` in row3,column1 to be a `0`. We do this by computing `R3-5*R1` and replacing all of R3 with that result \[ \left[\begin{array}{ccc|c} 1 & 4 & 6 & -32\\ 0 & -17 & -21 & 140\\ 0 & -17 & -21 & 180\\ \end{array} \right] \begin{array}{c} \\ \\ R_3 - 5R_1 \rightarrow R_3\\ \end{array} \]

  49. anonymous
    • one year ago
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    ok i got that

  50. jim_thompson5910
    • one year ago
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    We could divide the second row by -17 to make that first '-17' in that row equal to '1', but I'm going to skip that step. Instead I'm going to subtract R2 and R3 and replace R3 R3-R2 --> R3 \[ \left[\begin{array}{ccc|c} 1 & 4 & 6 & -32\\ 0 & -17 & -21 & 140\\ 0 & 0 & 0 & 40\\ \end{array} \right] \begin{array}{c} \\ \\ R_3 - R_2 \rightarrow R_3\\ \end{array} \]

  51. jim_thompson5910
    • one year ago
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    The last row of this matrix \[ \left[\begin{array}{ccc|c} 1 & 4 & 6 & -32\\ 0 & -17 & -21 & 140\\ 0 & 0 & 0 & 40\\ \end{array} \right] \] tells us that `0x + 0y + 0z = 40` which essentially turns into `0+0+0 = 40` and that turns into `0 = 40`

  52. anonymous
    • one year ago
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    wow that was alot

  53. jim_thompson5910
    • one year ago
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    but `0 = 40` is NEVER true regardless of what x,y,z are. Since we have a contradiction, we have an inconsistent system that has NO solutions at all.

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