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Matrices and polynomials

  1. Mar 3, 2008 #1
    Hello! First post here.

    My question is, is it possible to use a matrix to solve a system where you have the same variable, but a different degree. i.e.

    [tex]2x^2 + 2x + y = 2[/tex]
    [tex]-3x^2 - 6x + 2y = 4[/tex]
    [tex] 4x^2 + 6x - 3y = 6[/tex]

    Now, I know this is possible other ways, but seeing as each of those would have multiple solutions, is it possible to solve using a matrix?
     
  2. jcsd
  3. Mar 3, 2008 #2

    Hurkyl

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    There are lots of ways you can use a matrix... I imagine, though you meant solving the matrix equation

    [tex]
    \left(
    \begin{array}{ccc}
    2 & 2 & 1 \\
    -3 & -6 & 2 \\
    4 & 6 & -3
    \end{array}
    \right)
    \left(
    \begin{array}{c}
    x^2 \\ x \\ y
    \end{array}
    \right)
    =
    \left(
    \begin{array}{c}
    2 \\ 4 \\ 6
    \end{array}
    \right)
    [/tex]

    Well, this is a matrix equation, so you can apply matrix algebra to do whatever you wanted to it. Just make sure you understand what your results mean.
     
  4. Mar 3, 2008 #3

    tiny-tim

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    too much information …

    Hi woodne! Welcome to PF! :smile:

    You have three equations, but only two variables. :frown:

    So there won't normally be a solution (unless one of the equations is redundant).

    If you ignored one equation, you'd only have a 2x3 matrix …
     
  5. Mar 3, 2008 #4
    That's what i was thinking, that it shouldn't be possible to solve that by a matrix, but my physics teacher swears there's a way to use a matrix for everything.
     
  6. Mar 3, 2008 #5

    Hurkyl

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    Er, if there is no solution then there is no solution. All ignoring a constraint achieves is vastly increasing the likelyhood you will mistakenly interpret a 'false solution' as being an actual solution.



    P.S. have either of you seriously consider how one might apply linear algebra to this problem? Or did you just think "oh, it can't work" and stop thinking?
     
    Last edited: Mar 3, 2008
  7. Mar 3, 2008 #6

    tiny-tim

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    Ask your teacher whether he/she means a square matrix - if so, ask how that can work if there aren't enough independent variables! :smile:

    Oh, and be tactful … :rolleyes:
     
  8. Mar 5, 2008 #7
    Isn't the team "square matrix" used when the dimensions are equal? (let y equal a dimension, such as y x y)
     
  9. Mar 5, 2008 #8

    tiny-tim

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    Yes, that's right - so a square matrix could only be useful for solving independent variables.

    If there are three variables, of which only two are independent (such as x^2 x and y), there won't normally be a solution for more than two equations involving them - which would only give enough numbers for a 3 x 2 matrix! :smile:
     
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