Let $ q : \mathbb{R}^3 \to \mathbb{R}^3 $ be the quadratic form defined in the standard basis as $ q(x_1, x_2, x_3) = x_1^2+3x_2^2 +2x_1x_2+2x_2x_3 + 4\lambda x_1x_3 $ where $ \lambda \in \mathbb{R} $. I want to find for what values of $ \lambda $ is $ q $ positive definite. Is my answer correct?
We can notice that $ [q]_{(e)} = \begin{pmatrix} 1 & 1 & 2\lambda \\ 1 & 3 & 1 \\ 2\lambda &1 & 0 \end{pmatrix} $.
The form is positive-definite if and only if the determinant of every leading minor is greater than $ 0 $.
$ \left|\left[1\right]\right| = 1 \gt 0 $.
$ \begin{vmatrix} 1 & 1 \\ 1 & 3 \end{vmatrix} = 2 \gt 0$
We'll check for what values of $ \lambda $ $ \ \det[q] \gt 0$, we'll expand the determinant from the bottom row.
$ \left|[q]\right|= 2\lambda \begin{vmatrix} 1 & 2\lambda \\ 3 & 1 \end{vmatrix} - \begin{vmatrix} 1 & 2\lambda \\ 1 & 1 \end{vmatrix} = 2\lambda(1-6\lambda)-(1-2\lambda) = 2\lambda-12\lambda^2-1+2\lambda=-12\lambda^2+4\lambda-1. $
Using some analysis, we can find that $ \frac{d}{d\lambda} (-12\lambda^2+4\lambda-1) =0 \iff-24\lambda+4=0 \iff \lambda=\frac{1}{6} $.
$ -12(\frac{1}{6})^2 +4\frac{1}{6} - 1 < 0$ and because $ -12 \lt 0 $, we know that $ \frac{1}{6} $ is the maximum of $ -12\lambda^2 +4\lambda - 1 $, and therefore $ det[q] < 0 $ for all $ \lambda $, and therefore the form is not positive-definite for any $ \lambda $.