How do I get to this formula for the area of a triangle I am new here and new student to geometry.
In my geometry skript there is a task:
Show that the formula for the area $F$ of a triangle with sidelengths $a,b,c $ is given by
$$F^2 = - \frac{1}{16} \det \begin{pmatrix} 0 & c^2 & b^2 & 1 \\ c^2 & 0 & a^2 & 1 \\ b^2 & a^2 & 0 & 1 \\ 1& 1& 1& 0 \end{pmatrix}$$
This drives me crazy because I cant see how to get there.
I do know how to get to
$F= \frac12 \det \begin{pmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \end{pmatrix} $
and $ F = \frac12 \det \begin{pmatrix} 1 & 1 & 1 \\ a & b & c \end{pmatrix} $
I also thought about using Heron's formula for the square of a triangle:
$ F^2 = s(s-a)(s-b)(s-c) $ where $s= \frac{a+b+c}{2} $
I did multiplied it all out and can't seem to find a pattern..
Hope this question is appropriate.
Maybe here is someone who has an idea :)

Start with
\begin{pmatrix} 0 & c^2 & b^2 & 1 \\ c^2 & 0 & a^2 & 1 \\ b^2 & a^2 & 0 & 1 \\ 1& 1& 1& 0 \end{pmatrix}
Perform the column transformations $C_2 \to C_2 - C_1$ and $C_3 \to C_3- C_1$ which don't change the determinant.
\begin{pmatrix} 0 & c^2 & b^2 & 1 \\ c^2 & -c^2 & a^2-c^2 & 1 \\ b^2 & a^2-b^2 & -b^2 & 1 \\ 1& 0& 0& 0 \end{pmatrix}
Perform the row transformations $R_2 \to R_2 - R_1$ and $R_3 \to R_3 - R_1$ which also leave the determinant unchanged.
\begin{pmatrix} 0 & c^2 & b^2 & 1 \\ c^2 & -2c^2 & a^2-c^2-b^2 & 0 \\ b^2 & a^2-b^2-c^2 & -2b^2 & 0 \\ 1& 0& 0& 0 \end{pmatrix}
A simple Laplace expansion, first along the fourth row then along the third column of the resulting submatrix, will tell you that the determinant of this matrix is in fact the determinant of the central matrix
\begin{pmatrix} -2c^2 & a^2-c^2-b^2\\ a^2-b^2-c^2 & -2b^2 \end{pmatrix}
which is just equal to $$
4b^2c^2 - (a^2-b^2-c^2)^2 = (2bc)^2 - (a^2-b^2-c^2)^2\\=(a^2-b^2+2bc-c^2)(a^2-b^2-2bc-c^2) \\= (a^2 - (b+c)^2) (a^2-(b-c)^2) \\
= (a-b-c)(a+b+c)(a+b-c)(a+c-b)
$$
Recognizing the semi-perimeter $2S = a+b+c$, we just have $$
a-b-c = 2(a-S) \\
a+b+c = 2S \\
(a+b-c) = 2(S-c)\\
(a+c-b) = 2(S-b)
$$
which leads to $$
(a-b-c)(a+b+c)(a+b-c)(a+c-b) = 16S(S-c)(S-b)(a-S) = -16F^2
$$
where $F$ is the area of the triangle with sides $a,b,c$ and $F^2 = S(S-a)(S-b)(S-c)$ by Heron's formula. That is, the determinant at the start equals $-16F^2$, as desired.
