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Homework Help: Solving cubic equation with matrices

  1. Apr 4, 2006 #1
    I need to find b, c and d in a cubic equation of the form y=x3+bx2+cx+d (ie. a=1 if a is the coefficient of x3). I've been given three points (-2,-13), (-1,0), (1,2). If I had a fourth piece of data i could solve the equation I suppose, but with only three I've really struggled. Here's what I've got so far:

    http://img71.imageshack.us/img71/9821/matrix0iw.gif [Broken]

    (1) is just the set out
    (2) I've entered the data
    (3) I've rearranged the equation

    But when I thin input the equation on my calculator (the TI-89 Cas calc) I get a matrix in which every row has x and y in it. But that solution doesn't agree with the back of the book, which states that:
    b=-2
    c=0
    d=3

    which indicates a final equation of y=x3-2x2+3

    I've thought of using a matrix system which ommits the top row of all three matrices, but of course that equation system isn't defined.

    I'd really apprieciate any help, thanks.
     
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Apr 4, 2006 #2

    shmoe

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    3 pieces of data is enough, you'll have 3 equations and 3 unknowns. Remove the first [x^3 x^2 x 1] row from the first matrix and the first [y] row from the last (but don't change the [1 b c d]^T matrix). It's representing the equation y=x^3+bx^2+cx+d, which doesn't help you solve for b, c, and d.
     
  4. Apr 5, 2006 #3
    So your suggesting the following matrix?

    http://img97.imageshack.us/img97/8471/matrix28pd.gif [Broken]

    Becuase the matrix to the negative power is undefined, and even if it were, a matrix with 4 columns can't multiply with a matrix of 3 rows.
     
    Last edited by a moderator: May 2, 2017
  5. Apr 5, 2006 #4

    Hurkyl

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    No, he's suggesting

    Code (Text):

    /-8  4 -2  1 \ /1\   /-13\
    |-1  1 -1  1 | |b| = |  0|
    \ 1  1  1  1 / |c|   \  2/
                   \d/
     
    Obviously you can't simply invert your matrix to solve this one. So use something else. :tongue:


    Incidentally, is there any particular reason you put a known into your unknown vector? You could use a matrix inversion, if you really wanted to, and set the problem up right.
     
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