<s>[INST] Using A Determinant to Solve a Linear System I recently asked how to determine the standard form equation for an ellipse,
$\frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1$, where the ellipse has a given eccentricity
$e$ and passes through three points $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$.  I got a fantastic
answer here, but there's some magic in the answer I don't understand.
To recap the solution, I can start with the general ellipse (actually conic) equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$, which has six unknowns ($A$, $B$, ..., $F$), and then take advantage of the constraints on the problem to set $B = 0$ and $A$ as a constant factor $(1-e)^2$ of $C$.  This leave me with four unknowns ($A/C$, $D$, $E$, and $F$) which can be solved with three points using the system:
$$
\begin{bmatrix}
(1-e^2)x^2 + y^2 & x & y & 1\\
(1-e^2)x_1^2 + y_1^2 & x_1 & y_1 & 1\\
(1-e^2)x_2^2 + y_2^2 & x_2 & y_2 & 1\\
(1-e^2)x_3^2 + y_3^2 & x_3 & y_3 & 1
\end{bmatrix}
\begin{bmatrix}
A/C\\
D\\
E\\
F
\end{bmatrix}
=
\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}
$$
The linked answer somehow converts this system to the determinant equation
$$
\begin{vmatrix}
  (1-e^2)x^2+y^2 & x & y & 1 \\
  (1-e^2)x_1^2+y_1^2 & x_1 & y_1 & 1 \\
  (1-e^2)x_2^2+y_2^2 & x_2 & y_2 & 1 \\
  (1-e^2)x_3^2+y_3^2 & x_3 & y_3 & 1
\end{vmatrix}=0
$$
which allows me to solve
$$
A/C = \begin{vmatrix}x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1\end{vmatrix}
\\D = -
  \begin{vmatrix}
    (1-e^2)x_1^2+y_1^2 & y_1 & 1 \\
    (1-e^2)x_2^2+y_2^2 & y_2 & 1 \\
    (1-e^2)x_3^2+y_3^2 & y_3 & 1
  \end{vmatrix}
\\etc...  
$$
I see the mechanics of computing $A/C$, $D$, ..., but I don't understand why I get to build the determinant formula from the original linear system.  I thought I was maybe onto something with Cramer's rule, but that goes nowhere quickly.  Can anyone explain or point me at an answer?
 [/INST] At this stage:
$\begin{bmatrix}
(1-e^2)x^2 + y^2 & x & y & 1\\
(1-e^2)x_1^2 + y_1^2 & x_1 & y_1 & 1\\
(1-e^2)x_2^2 + y_2^2 & x_2 & y_2 & 1\\
(1-e^2)x_3^2 + y_3^2 & x_3 & y_3 & 1
\end{bmatrix}
\begin{bmatrix}
A/C\\
D\\
E\\
F
\end{bmatrix}
=
\begin{bmatrix}
0\\
0\\
0\\
0
\end{bmatrix}$
If this has a solution, either we have the trival solution: $A = D = E = F = 0$ or the matrix is "singular."
If the matrix is singular, then its determinant must be equal to 0.  And $A/C, D, E, F$ are in the kernel of the matrix.
</s>