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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:


    (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:

    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.
  2. jcsd
  3. Apr 4, 2006 #2


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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?


    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.
  5. Apr 5, 2006 #4


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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/
    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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