Linear Algebra - Quadratic polynomial to Matrix

[Snoogx]

Homework Statement

Examining the answers of the previous two questions, write the quadratic polynomial f(x₁,x₂,x₃)=x₁x₂−6x₂²+3x₂x₃−3x²₃ in the form
f(x₁,x₂,x₃)=[x₁x₂x₃]A[x₁x₂x₃]<-this last group is a column matrix
where A is a symmetric matrix.

Homework Equations

Matrix multiplication

The Attempt at a Solution

So the previous problems had me start with the matrix, then multiply by the row then column vectors to get a polynomial. Then here it wants me to work backwards to get matrix A.

I started with:
x₁x₂−6x₂²+3x₂x₃−3x²₃.
Grouped like terms:
(x₁x₂) + (−6x₂² + 3x₂x₃) + (−3x²₃)
Took out an x₁, x₂, x₃ form each grouping, respectively. That gives me the second step of the problem:
[(x₂) (−6x₂ 3x₃) (−3x₃)] [x₁ x₂ x₃] <-again, column matrix
So, working backwards, I now have to find the values of the elements of matrix A so when multiplied by row vector [x₁ x₂ x₃] will result in the output of [(x₂) (−6x₂ 3x₃) (−3x₃)].
In other words, [x₁][# # #](<-column) will equal [x₂].
So I end up getting matrix A =
[ 0 0 0
1 -6 0
0 3 -3]

I know A₁₁ = 0 and A₂₂ = -6 are correct. But when I input the rest of the matrix in they homework system tells me I'm wrong.
Could anyone look through this and see if I made a mistake somewhere? Or maybe the system has the wrong key and my answer is correct? Any insight and help is greatly appreciated.

Thanks

 

[lanedance]

that matrix is not symmetric for a start

 

[lanedance]

now say you have found a matrix B that is not symmetric, but gives you teh correct polynomial. you can always write it in terms of a symmetric (S) and skew-symmetric (P) part
B = S+P

consider the polynomial
p(x) = x^TBx = x^TSx + x^TPx =

as p(x) is a scalar function, it will be equal to its transpose
p(x) = p(x)^T = (x^TSx)^T + (x^TPx)^T = (x^TS^T x)] + (x^TP^Tx)
= (x^TS x) - (x^TPx)

which implies x^TPx = 0 that only the symmetric part of the matrix is important in defining the polynomial

so if you are happy you matrix is correct (have you checked it gives the right result?)

you can find the symmetric and anti symmetric parts as follows

B = S+P
S = (B+B^T)/2
P = (B-B^T)/2

 

[Snoogx]

Thank you for the explanation lanedance. I didn't know about symmetric matrices.

 

[lanedance]

no worries - just to add, for most polynomials there is actually an infinite way they can be written interms of a matrix x^TBx, however if you restrict yourself to symmetric matrix , which appear to be the natural choice based on teh above, then is is one unique solution