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.
 A: Compute the gradient and solve for $\nabla Q = 0$ to look for the minimum
$$\left\{
\begin{matrix}
2x_1 - 4x_2 + 2x_3 &= 0 \\
-4x_1 + 10x_2 -2x_3 &= 0 \\
2x_1 - 2x_2 + 6x_3 &= 0
\end{matrix} \right.$$
has a unique solution $(0,0,0)$. Clearly this point is not a maximum. To see that it is neither a saddle you have to compute the Hessian (we already computed it, it's the matrix associated to the system above).
The hessian is (obviously) symmetric and all eigenvalues are positive, so you have a minimum.
A: $Q$ 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.
