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anonymous

  • one year ago

Find the general solutions of 1 0 -5 0 -8 3 0 1 4 -1 0 6 0 0 0 0 1 0 0 0 0 0 0 0 Last question I'm going to ask, I promise.

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  1. anonymous
    • one year ago
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    Is there even a solution? The third row doesn't make a lot of sense 0 + 0 + 0 + 0 + 1 = 0?

  2. ganeshie8
    • one year ago
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    is that the augmented matrix in row echelon form ?

  3. anonymous
    • one year ago
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    I'm not sure. The problem doesn't say.

  4. anonymous
    • one year ago
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    Wait! okay, it is the augmented matrices.

  5. ganeshie8
    • one year ago
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    good, how many pivots are there ?

  6. anonymous
    • one year ago
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    I don't know. How do I tell if they are pivots?

  7. ganeshie8
    • one year ago
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    in echelon form, pivot is the first "non zero" number in a row

  8. anonymous
    • one year ago
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    So there would be three pivots then.

  9. ganeshie8
    • one year ago
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    Right, `1` 0 -5 0 -8 3 0 `1` 4 -1 0 6 0 0 0 0 `1` 0 0 0 0 0 0 0 you have 3 pivot columns and 2 free columns lets find a particular solution by letting free variables = 0

  10. anonymous
    • one year ago
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    Okay.

  11. ganeshie8
    • one year ago
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    just for definiteness, call the variables : x1, x2, x3, x4, x5 so to find a particular solution you let the free variables x3 = x4 = 0 and solve for the pivot variables x1, x2, x5

  12. ganeshie8
    • one year ago
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    x1 x2 x3 x4 x5 b `1` 0 -5 0 -8 3 0 `1` 4 -1 0 6 0 0 0 0 `1` 0 0 0 0 0 0 0 ^ ^ free columns

  13. ganeshie8
    • one year ago
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    notice that the corresponding system is ``` x5 = 0 x2 + 4x3 - x4 = 6 x1 - 5x3 - 8x5 = 3 ``` let x3 = x4 = 0 and solve remaining variables

  14. anonymous
    • one year ago
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    Okay, so how do I solve for the pivot points?

  15. ganeshie8
    • one year ago
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    plugin x3=x4=0 in above system

  16. ganeshie8
    • one year ago
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    then you will have 3 equations and 3 unknowns in triangular form which is a piece of cake to solve by back substitution

  17. anonymous
    • one year ago
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    so.. x2 = 5 x5 = 0 x1=0

  18. ganeshie8
    • one year ago
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    try again

  19. ganeshie8
    • one year ago
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    the corresponding system is ``` x5 = 0 x2 + 4x3 - x4 = 6 x1 - 5x3 - 8x5 = 3 ``` letting x3 = x4 = 0, the system becomes ``` x5 = 0 x2 = 6 x1 - 8x5 = 3 ``` yes ?

  20. anonymous
    • one year ago
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    Yes.

  21. ganeshie8
    • one year ago
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    solve it

  22. ganeshie8
    • one year ago
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    its trivial, but you have to do it

  23. anonymous
    • one year ago
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    I'm trying to do it right now, give me a moment.

  24. anonymous
    • one year ago
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    x5 = 0 x2 + 4(0) - 0 = 6 -> x2 = 6 x1 - 5(0) - 8(0) = 3 -> x1 = 3

  25. ganeshie8
    • one year ago
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    so \(\begin{pmatrix} 3\\6\\0\\0\\0 \end{pmatrix}\) is a particular solution we still need to find the null solution, familiar with the process ?

  26. anonymous
    • one year ago
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    Nope

  27. ganeshie8
    • one year ago
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    null solution is the solution to following system : ``` x5 = 0 x2 + 4x3 - x4 = 0 x1 - 5x3 - 8x5 = 0 ```

  28. ganeshie8
    • one year ago
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    Notice that I have put 0s on the right hand side now

  29. ganeshie8
    • one year ago
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    again we mess with the free variables to cookup the null solution : since we have two free variables, we expect to get two independent null solutions

  30. anonymous
    • one year ago
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    Do I use the values for the system that we came up with earlier?

  31. ganeshie8
    • one year ago
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    let x3 = 0, x4 = 1 and find one null solution let x3 = 1, x4 = 0 and find the other null solution

  32. ganeshie8
    • one year ago
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    null solution is the solution to following system : ``` x5 = 0 x2 + 4x3 - x4 = 0 x1 - 5x3 - 8x5 = 0 ``` plugin x3 = 0, x4 = 1 and solve the other variables

  33. anonymous
    • one year ago
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    x2 - 1 = 0 x2 = 1

  34. ganeshie8
    • one year ago
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    yes, keep going find the remaining two variables

  35. anonymous
    • one year ago
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    x1 = 0

  36. anonymous
    • one year ago
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    Unless you wanted me to use "let x3 = 1, x4 = 0 and find the other null solution" to solve x1 - 5x3 - 8x5 = 0

  37. anonymous
    • one year ago
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    Because then x1 would equal 5

  38. ganeshie8
    • one year ago
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    one thing at a time

  39. ganeshie8
    • one year ago
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    null solution is the solution to following system : ``` x5 = 0 x2 + 4x3 - x4 = 0 x1 - 5x3 - 8x5 = 0 ``` plugin x3 = 0, x4 = 1 and solve the other variables this gives a null solution \(\begin{pmatrix} 0\\1\\0\\1\\0\end{pmatrix}\), yes ?

  40. anonymous
    • one year ago
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    Yeah.

  41. ganeshie8
    • one year ago
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    save it, find the other null solution also similarly

  42. ganeshie8
    • one year ago
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    null solution is the solution to following system : ``` x5 = 0 x2 + 4x3 - x4 = 0 x1 - 5x3 - 8x5 = 0 ``` plugin x3 = 1, x4 = 0 and solve the other variables

  43. anonymous
    • one year ago
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    Okay, so then x1 = 1 and x2 = -1

  44. ganeshie8
    • one year ago
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    try again

  45. anonymous
    • one year ago
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    Thanks for your help ganeshie8, but I have to go to sleep, it's really late here and I have to wake up for class. Have a wonderful evening, and I'm sorry for not being about to finish the problem.

  46. ganeshie8
    • one year ago
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    we're mostly done just wait one minute

  47. ganeshie8
    • one year ago
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    null solution is the solution to following system : ``` x5 = 0 x2 + 4x3 - x4 = 0 x1 - 5x3 - 8x5 = 0 ``` plugin x3 = 1, x4 = 0 and solve the other variables that gives the null solution \(\begin{pmatrix} 5\\-4\\1\\0\\0\end{pmatrix}\)

  48. ganeshie8
    • one year ago
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    so the nullspace is given by \[\large c_1\begin{pmatrix} 0\\1\\0\\1\\0\end{pmatrix} + c_2\begin{pmatrix} 5\\-4\\1\\0\\0\end{pmatrix}\]

  49. ganeshie8
    • one year ago
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    then the final general solutions is given by \(x_p + x_{null}\) : \[x_{\text{general }} = \begin{pmatrix} 3\\6\\0\\0\\0 \end{pmatrix} +\large c_1\begin{pmatrix} 0\\1\\0\\1\\0\end{pmatrix} + c_2\begin{pmatrix} 5\\-4\\1\\0\\0\end{pmatrix} \]

  50. anonymous
    • 11 months ago
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    Closing now as it is no longer helpful.

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