Determine whether $Q:\mathbb R^3 \to \mathbb R^1$ is positive definite, negative definite, or neither $Q(x_{1}, x_{2}, x_{3}) = x_{1}^{2} + 5x_{2}^{2} + 3x_{3}^{2} - 4x_{1}x_{2} + 2x_{1}x_{3} - 2x_{2}x_{3}$.
So, for $Q$ to be positive definite, we must have $Q(h)>0$ $\forall h \not= 0$. For $Q$ to be negative definite, $-Q$ must be positive definite. 
It looks positive definite to me, but I don't know how to rigorously prove that it is.
P.S. We aren't using matrices or eigenvalues to do this stuff in the book.

$Q$ is a quadratic function which can be expressed as (check yourself!)$$Q(x)=x^TAx$$ where $$A=\begin{bmatrix}
1 & -2 & 1\\
-2 & 5 & -1\\
1 &  -1 & 3
\end{bmatrix}$$ Apply Sylvester's criterion to this matrix and see that the principle minors are $1,1,1$ which are all positive and hence the matrix as well as $Q is positive definite.  
To see more easily that this matrix is positive definite apply Gershgorin circle theorem to see that all the eigenvalues will be in the intersection of the circles $|z-1|\le 3,\ |z-5|\le 3,\ |z-3|\le 2$ which clearly indicates that the eigenvalues will be positive since the matrix is symmetric.
