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Let $$Q(x,y,z) = – 2x^2 + 6xy + 8y^2 + z^2.$$ Find the symmetric matrix associated with this quadratic form. Use the determinant method to determine whether the quadratic form is positive definite, positive semi-definite, negative definite, negative semi-definite, or indefinite.

I am pretty stuck on this and have no idea where to start or what it should look like. All help would be appreciated.

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All right. First and foremost, every quadratic form is represented by a symmetric matrix by definition of a quadratic form. We find the symmetric form using this trick: Look at the form you are given. Leave each coefficient of $x^2,y^2,z^2$ as it is. Take the coefficient of $xy,yz,zx$, divide it by $2$. Now consider the matrix with the new coefficients rearranged as follows: $$ \begin{pmatrix} x^2 & xy & xz \\ xy & y^2 & yz \\ xz & yz & z^2 \end{pmatrix} $$ I claim this is the symmetric matrix for your form.

We'll see it work in your case. $-2x^2+6xy+8y^2+1z^2$ is your form, so you leave the $-2,8,1$ as they are, and divide the $6$ by $2$ to get $3$. The other coefficients are $0$, so you can leave them, and your matrix will be: $$ \begin{pmatrix} -2 & 3 & 0 \\ 3 & 8 & 0 \\ 0 & 0 & 1 \end{pmatrix} $$ To see why, compute the matrix $X^{T}AX$, where $A$ is the matrix above, and $X=[X,Y,Z]$ as a column vector. See that you get $-2x^2+6xy+8y^2+z^2$.

Now we analyze the given matrix. For finding the eigenvalues, we compute the characteristic polynomial = $(x-1)((x-8)(x+2) - 3*3) = (x-1)(x^2-6x-25)$. It's is clear that one root of the characteristic polynomial is $1$, while the other quadratic can be rewritten as $x^2-6x+9-34 = (x-3)^2-34$, so the values of $x$ are $3 \pm \sqrt{34}$, which give one positive and one negative eigenvalue. Hence the eigenvalues are all real, but some are positive and some negative, so your form is probably indefinite.

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  • $\begingroup$ $A$ is certainly indefinite by virtue of its having at least one negative eigenvalue. $\endgroup$ – Robert Howard Dec 11 '17 at 21:04

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