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.

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