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Write the complete solution as the particular solution plus any multip

  1. Feb 24, 2013 #1

    s3a

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    1. The problem statement, all variables and given/known data
    Could someone please help me do this problem?:

    “Write the complete solution as x_p plus any multiple of s in the nullspace:
    x + 3y + 3z = 1
    2x + 6y + 9z = 5
    –x – 3y + 3z = 5”

    The answer is x = x_p + x_n = {{-2},{0},{1}} + x_2 {{-3, 1, 0}}.

    2. Relevant equations
    Ax = b

    3. The attempt at a solution
    I get the {{-2},{0},{1}} and {{-3},{1},{0}} but, I don't get how the x_2 was obtained there.

    Any input would be really appreciated!
     
  2. jcsd
  3. Feb 24, 2013 #2

    LCKurtz

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    Re: Write the complete solution as the particular solution plus any mu

    Did you try solving the system by row reduction?
     
  4. Feb 24, 2013 #3

    tiny-tim

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    hi s3a! :smile:

    (try using the X2 button just above the Reply box :wink::
    i've no idea why they've written x2 :confused:

    we'd normally use C (for "constant"! :biggrin:)

    (btw, why are you using so many curly brackets, instead of fewer, ordinary brackets?)
     
  5. Feb 24, 2013 #4

    LCKurtz

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    Re: Write the complete solution as the particular solution plus any mu

    Also, are you sure the problem is copied correctly? I don't think your solution works.

    [Edit] Yes it does work, arithmetic mistake here.
     
    Last edited: Feb 24, 2013
  6. Feb 24, 2013 #5

    s3a

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    Re: Write the complete solution as the particular solution plus any mu

    Sorry. I double-posted. This website should have some sort of mechanism where you can't post twice until 10 seconds elapse or something like that.
     
  7. Feb 24, 2013 #6

    s3a

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    Re: Write the complete solution as the particular solution plus any mu

    LCKurtz, I did use row reduction. Also, thanks for telling me about the existence of the x_2 button and, yes, the solution was copied down correctly. Letting c = 1, for example, yields a solution to the system of equations so, it works. :)

    tiny-tim, the abundance of curly brackets is Wolfram Alpha's notation for emphasizing that the vectors are vertical.

    Okay, so basically x_2 = c is a constant. That pretty much helped me see through my confusion, lol!

    Thanks!
     
  8. Feb 24, 2013 #7

    LCKurtz

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    Re: Write the complete solution as the particular solution plus any mu

    So you solved it OK with row reduction and all is well, right?
     
  9. Feb 24, 2013 #8

    s3a

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    Re: Write the complete solution as the particular solution plus any mu

    Yes.

    To elaborate, I solved it augmented with b and that allowed me to get a particular solution and then took that fully-row-reduced matrix and took away the last column (=the augmented column) and replaced it with zeroes and that gave me the other part which, when all multiples of it are summed with the particular solution, all solutions are obtained.

    (Sorry for the sloppy explanation, I am sleepy as I write this.)

    (If anyone knows what that other part is called, I'd be interested in knowing.)
     
  10. Feb 24, 2013 #9

    Dick

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    Re: Write the complete solution as the particular solution plus any mu

    The part you got with the specific vector is called a 'particular solution'. The part with the parameter you got from Ax=0 is the 'homogeneous solution'.
     
  11. Feb 24, 2013 #10

    s3a

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    Re: Write the complete solution as the particular solution plus any mu

    Oh, ya ... I remember that term! Thanks!
     
  12. Feb 24, 2013 #11

    LCKurtz

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    Re: Write the complete solution as the particular solution plus any mu

    Of course, you don't have to solve it twice like that. For example if your reduced augmented matrix came out$$
    \begin{bmatrix}
    1&3&0&-2\\
    0&0&1&1\\
    0&0&0&0
    \end{bmatrix}$$that corresponds to the two equations
    x + 3y = -2
    z = 1
    So you can let y be anything, say y = c and write the solution like this$$
    \begin{bmatrix}
    x\\
    y\\
    z\end{bmatrix}=
    \begin{bmatrix}
    -2-3c\\
    c\\
    1
    \end{bmatrix}=\begin{bmatrix}
    -2\\
    0\\
    1
    \end{bmatrix}+c\begin{bmatrix}
    -3\\
    1\\
    0
    \end{bmatrix}
    $$and you have it.
     
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