$
\def \J {\Delta}
\def \half {\frac{1}{2}}
\def \kwart {\frac{1}{4}}
\def \hieruit {\quad \Longrightarrow \quad}
\def \slechts {\quad \Longleftrightarrow \quad}
\def \norm {\frac{1}{\sigma \sqrt{2\pi}} \; }
\def \EN {\quad \mbox{and} \quad}
\def \OF {\quad \mbox{or} \quad}
\def \wit {\quad \mbox{;} \quad}
\def \spekhaken {\iint}
\newcommand{\dq}[2]{\displaystyle \frac{\partial #1}{\partial #2}}
\newcommand{\oq}[2]{\partial #1 / \partial #2}
\newcommand{\qq}[3]{\frac{\partial^2 #1}{{\partial #2}{\partial #3}}}
\def \erf {\operatorname{Erf}}
$
Extreme Cases
When looking for a sensible meaning of the determinant
$\sigma_{xx}\sigma_{yy} - \sigma_{xy}^2$, it can be proved in the first place
that:
$$
\sigma_{xx}\sigma_{yy} - \sigma_{xy}^2 \ge 0
$$
Which is a direct consequence of Schwartz inequality:
$$
(\vec{x}\cdot\vec{y})^2 \le (\vec{x}\cdot\vec{x})(\vec{y}\cdot\vec{y})
\slechts \left(\sum_k w_k x_k y_k\right)^2 \le
\left(\sum_k w_k x_k^2\right) \left(\sum_k w_k y_k^2\right)
$$
Remember the Ellipse of Inertia:
$$
\sigma_{yy} (x - \mu_x)^2 - 2 \sigma_{xy} (x - \mu_x)(y - \mu_y)
+ \sigma_{xx} (y - \mu_y)^2 = \sigma_{xx}\sigma_{yy} - \sigma_{xy}^2
$$
For $\sigma_{xx}\sigma_{yy} - \sigma_{xy}^2 = 0$ it reduces to:
$$
\sigma_{yy} (x - \mu_x)^2
\pm 2 \sqrt{\sigma_{xx}\sigma_{yy}} (x - \mu_x)(y - \mu_y)
+ \sigma_{xx} (y - \mu_y)^2 = 0
$$ $$
\slechts \sqrt{\sigma_{yy}}(x-\mu_x)\pm\sqrt{\sigma_{xx}}(y-\mu_y) = 0
$$
Thus the ellipse is degenerated to a straight line segment.
Next consider the following fact:
$$
\left(\frac{\sigma_{xx}-\sigma_{yy}}{2}\right)^2 + \sigma_{xy}^2 \ge 0
$$
On the other hand:
$$
\left(\frac{\sigma_{xx}-\sigma_{yy}}{2}\right)^2 + \sigma_{xy}^2
= \left(\frac{\sigma_{xx}+\sigma_{yy}}{2}\right)^2 -
\left(\sigma_{xx}\sigma_{yy} - \sigma_{xy}^2\right) \ge 0
$$ $$
\hieruit 0 \le \sigma_{xx}\sigma_{yy} - \sigma_{xy}^2 \le
\left(\frac{\sigma_{xx}+\sigma_{yy}}{2}\right)^2
$$
And the maximum is obtained for:
$$
\left(\frac{\sigma_{xx}-\sigma_{yy}}{2}\right)^2 + \sigma_{xy}^2 = 0
\slechts \sigma_{xx} = \sigma_{yy} \EN \sigma_{xy} = 0
$$ $$
\slechts \left(\frac{\sigma_{xx}+\sigma_{yy}}{2}\right)^2 = \sigma_{xx}^2 =
\sigma_{xx}\sigma_{yy} - \sigma_{xy}^2
$$
Remember the Ellipse of Inertia:
$$
\sigma_{yy} (x - \mu_x)^2 - 2 \sigma_{xy} (x - \mu_x)(y - \mu_y)
+ \sigma_{xx} (y - \mu_y)^2 = \sigma_{xx}\sigma_{yy} - \sigma_{xy}^2
$$
Which is now reduced to a circle:
$$
\sigma_{xx} (x - \mu_x)^2
+ \sigma_{xx} (y - \mu_y)^2 = \sigma_{xx}^2 \slechts
(x - \mu_x)^2 + (y - \mu_y)^2 = \sigma_{xx}
$$
Summarizing:
Determinant $ = $ minimum then Ellipse of Inertia $ = $ Straight Line Segment.
Determinant $ = $ maximum then Ellipse of Inertia $ = $ Circle.
Thus the determinant $\sigma_{xx}\sigma_{yy} - \sigma_{xy}^2$, eventually
divided by the trace
$(\sigma_{xx}+\sigma_{yy})/2$, is a $0 \le $ measure
$ \le 1$ for the "roundness" of the ellipse.