Hessian of quadratic objective function I have the quadratic function
$$f(\boldsymbol{x}) = \frac{3}{2} \left (x_{1}^{2}+x_{2}^{2} \right) + (1+a) x_{1} x_{2} - \left(x_{1} + x_{2} \right) + b$$
where $a, b \in \Bbb R$ are unknown parameters. Then,
$$f(\boldsymbol{x})=\frac{1}{2} \boldsymbol{x}^{\top}\left[\begin{array}{cc}
3 & 1+a \\
1+a & 3
\end{array}\right] \boldsymbol{x}-\boldsymbol{x}^{\top}\left[\begin{array}{l}
1 \\
1
\end{array}\right]+b$$
I am trying to check for what values of $a$ and $b$ a global minimizer will exist, and I saw the solution says the following:

The unique global minimizer exists if and only if the Hessian is positive definite, which holds if and only if $(1+a)^{2}<9$ (by Sylvester's criterion). Hence, the largest set of values of $a$ and $b$ such that the global minimizer of $f$ exists is given by $-4<a<2$ and $b \in \mathbb{R}$ (unrestricted).

Why is the Hessian of $f$ just the following?
\begin{bmatrix}
3 & 1+a \\
1+a & 3
\end{bmatrix}
Which theorem does this come from? Or am I misunderstanding the solution?
 A: Given function is
$$f(\boldsymbol{x})=\frac{3}{2}\left(x_{1}^{2}+x_{2}^{2}\right)+(1+a) x_{1} x_{2}-\left(x_{1}+x_{2}\right)+b$$
This yields
$$
\nabla f(\mathbf{x}) = \left[ \begin{array}{c}
3 x_1 + (1 + a) x_2 - 1 \\
3 x_2 + (1 + a) x_1 - 1
\end{array} \right]
$$
The critical points of $f$ are at the points where
$$
\nabla f(\mathbf{x}) = 0
$$
Then we find the Hessian matrix as
$$
H = Hf(\mathbf{x}) = \left[ \begin{array}{cc}
3 & 1 + a \\
1 + a & b \\
\end{array} \right]
$$
A sufficient condition for global minimization of $f$ at a critical point $\mathbf{x}^\star$ is that $H$
is positive definite.
Sylvester's criterion: A necessary and sufficient condition for $H$ to be positive definite everywhere is that its principal minors should be positive.
This gives us:
$$
\Delta_1 = 3 > 0
$$
$$
\Delta_2 = \mbox{det}(H) = 9 - (1 + a)^2 > 0
$$
$\mbox{det}(H) > 0$ also ensures that  $f$ has a unique critical point $\mathbf{x^\star}$.
Thus, $f$ will have a global minimum at $\mathbf{x}^\star$  if
$$
(1 + a)^2 < 9 \ \mbox{or} \  |1 + a| < 3 \ \mbox{or} -3 < 1 + a < 3
$$
Hence,   if $a \in (-4, 2)$, then $f$ has a global minimum at  $\mathbf{x}^\star$.
