Solving cubic equation with matrices

[Beam me down]

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

(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.

 

[shmoe]

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.

 

[Beam me down]

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.

So your suggesting the following matrix?

http://img97.imageshack.us/img97/8471/matrix28pd.gif

because 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.

 

[Hurkyl]

So your suggesting the following matrix?

No, he's suggesting

/-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.


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.