A determinant from analytic geometry? I have a question regarding the following determinant:
$\begin{vmatrix}
+ax - by - cz & bx+ay & cx+az \\ 
bx+ay& -ax+by-cz & bz+cy \\
cx+az & bz+cy & -ax-by+cz
\end{vmatrix}
=
(a^2 + b^2 + c^2)(x^2 + y^2 +z^2)(ax+by+cz).$
I can prove the above equality by performing row operations and column operations. However the above equation has a lot of geometrical terms and it seems to me that this equation could have other interpretations that I am missing.
So I have two questions:
1) Can we write the given determinant as product of two determinants (or three even)?
2) Is there a conceptual proof of the above equality through linear algebra or analytic geometry?
 A: I do not know the answer to the first question. But the problem can be solved by using linear algebra.

Proof:
Define $u = (a,b,c)$, $v = (x,y,z)$ and $P = uv^T + vu^T$[Thanks erfink].The given problem asks us to prove the following: $$\det(P - (u^Tv)I) = ||u||^2 ||v||^2 u^Tv.$$ 
We note that $P$ is a $3 \times 3$ matrix and we make the following observations:


*

*$w = u \times v$ is in the null space of $P$ since $v^Tw = u^Tw=0$. So zero is an eigenvalue.

*We can assume that $a^2 + b^2 + c^2 = x^2 + y^2 + z^2 = 1$ because of homogeneity.

*Since we may assume $||u|| = ||v|| = 1$, it follows that $(uv^T + vu^T)(u+v) = (u^Tv + 1)(u+v)$, $(uv^T+vu^T)(u-v) = (u^Tv -1)(u-v).$ Thus $u^Tv + 1$, $u^Tv-1$ are eigenvalues of $P$ as well.


From these observation it follows that the characteristic polynomial is $$t(t+1-u^Tv)(t-1-u^Tv) = \det(tI - P).$$
Substituting $t = u^Tv$, we get $$u^Tv = \det(P - (u^Tv)I).\blacksquare$$ 
A: Now I can answer the first question. We can write the matrix as a product of two rectangular matrices.
$$\left[ \begin{matrix}
+ax - by - cz & bx+ay & cx+az \\ 
bx+ay& -ax+by-cz & bz+cy \\
cx+az & bz+cy & -ax-by+cz
 \end{matrix}\right] = \left[ \begin{matrix}
a & -b & c & 0\\ 
b & a & 0 & -c \\
c & 0 & -a & b \\
 \end{matrix}\right] \left[ \begin{matrix}
x & y & z\\ 
y & -x & 0\\
-z & 0 & x\\
0 & z & -y\\
 \end{matrix}\right] = A^TB$$
The determinant can be evaluated using Cauchy-Binet formula:
$$\begin{vmatrix}
a & -b & c & 0\\ 
b & a & 0 & -c \\
c & 0 & -a & b \\
 \end{vmatrix}\begin{vmatrix}
x & y & z\\ 
y & -x & 0\\
-z & 0 & x\\
0 & z & -y\\
 \end{vmatrix}= |A^TB|= A_{123}B_{123}+A_{124}B_{124}+A_{134}B_{134}+A_{234}B_{234}.$$
where $A_{ijk}$ stands for the determinant of the matrix formed by rows $i,j,k$.
Now just verify that 
$$A_{123}B_{123}=ax(a^2+b^2+c^2)(x^2+y^2+z^2).$$
$$A_{124}B_{124}=by(a^2+b^2+c^2)(x^2+y^2+z^2).$$
$$A_{134}B_{134}=cz(a^2+b^2+c^2)(x^2+y^2+z^2).$$
$$A_{234}B_{234}=0.$$
