I have
$A = $ $ \left[\begin{array}{rrr} 2 & \alpha & -1 \\ \alpha & 2 & 1 \\ -1 & 1 & 4 \end{array}\right] $
and I want to find all $\alpha$ such that $A$ is positive definite.
I tried
$ x^tAx = $ $ \left[\begin{array}{r} x & y & z \end{array}\right] $ $ \left[\begin{array}{rrr} 2 & \alpha & -1 \\ \alpha & 2 & 1 \\ -1 & 1 & 4 \end{array}\right] $ $ \left[\begin{array}{r} x \\ y \\ z \end{array}\right] $
$=$ $ \left[\begin{array}{r} 2x + \alpha y - z & \alpha x + 2y + z & -x + y + 4z \end{array}\right] $ $ \left[\begin{array}{r} x \\ y \\ z \end{array}\right] $
$= 2x^2 + \alpha xy - xz + \alpha xy + 2y^2 + yz - xz + yz + 4z^2$
$= 2 \alpha xy + 2x^2 + 2y^2 - 2xz + 2yz + 4z^2$
and I wanted to solve the inequality $2 \alpha xy + 2x^2 + 2y^2 - 2xz + 2yz + 4z^2 > 0$ for $\alpha$, but I wasn't sure what to do next.
Am I doing this correctly?