Finding all $\alpha$ such that a matrix is positive definite 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?

Maybe this helps: 
A hermitian matrix is positive definite $\Leftrightarrow$ all leading principal minors are positive.
So $\left|\begin{array}{r}
    2
  \end{array}\right|$, $\left|\begin{array}{rr}
    2 & \alpha \\
    \alpha & 2
  \end{array}\right|$ and  $\left|\begin{array}{rrr}
    2 & \alpha & -1 \\
    \alpha & 2 & 1 \\
    -1 & 1 & 4
  \end{array}\right|$ have to be positive. 
This gives us 
$$\begin{align}&\text{I:}\quad 2>0\\
&\text{II:}\quad 4-\alpha^2>0\\
&\text{III:}\quad -4\alpha^2 - 2\alpha +12>0
\end{align}
$$
Try to solve this!
