- #1

- 22

- 0

## Homework Statement

Examining the answers of the previous two questions, write the quadratic polynomial f(x

_{1},x

_{2},x

_{3})=x

_{1}x

_{2}−6x

_{2}

^{2}+3x

_{2}x

_{3}−3x

^{2}

_{3}in the form

f(x

_{1},x

_{2},x

_{3})=[x

_{1}x

_{2}x

_{3}]A[x

_{1}x

_{2}x

_{3}]<-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:

x

_{1}x

_{2}−6x

_{2}

^{2}+3x

_{2}x

_{3}−3x

^{2}

_{3}.

Grouped like terms:

(x

_{1}x

_{2}) + (−6x

_{2}

^{2}+ 3x

_{2}x

_{3}) + (−3x

^{2}

_{3})

Took out an x

_{1}, x

_{2}, x

_{3}form each grouping, respectively. That gives me the second step of the problem:

[(x

_{2}) (−6x

_{2}3x

_{3}) (−3x

_{3})] [x

_{1}x

_{2}x

_{3}] <-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 [x

_{1}x

_{2}x

_{3}] will result in the output of [(x

_{2}) (−6x

_{2}3x

_{3}) (−3x

_{3})].

In other words, [x

_{1}][# # #](<-column) will equal [x

_{2}].

So I end up getting matrix A =

[ 0 0 0

1 -6 0

0 3 -3]

I know A

_{1}

_{1}= 0 and A

_{2}

_{2}= -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