Q:

Proof of De Gua's Theorem using Cauchy Binet

I was trying to prove the generalized version of De Gua's theorem using the Cauchy-Binet formula, and I ran into a bit of trouble.
We are considering a right-angled simplex in $\Bbb R^n$ whose corners are the origin and the coordinates $a_i e_i$ (where $a_i > 0$ and $e_1,\dots,e_n$ denotes the canonical basis of $\Bbb R^n$).
The ($n$-dimensional) volume of a simplex whose corners are the origin and (column-vectors) $v_1,\dots,v_n$ can be computed with the formula
$$
V = \frac 1{n!}\det \pmatrix{v_1 & \cdots & v_n}
$$
Similarly, the ($(n-1)$-dimensional) volume of a simplex whose corners are $v_1,\dots,v_n$ can be computed via $V^2 = \frac 1{((n-1)!)^2}\det M^TM$ where
$$
M = \pmatrix{v_1 & v_2 & \cdots & v_n\\ 1 & 1 & \cdots & 1}
$$
(or so I believe). Note that the above $M$ is not square: it has size $(n+1) \times n$. With the above in mind: let $A$ denote the diagonal matrix
$$
A = \pmatrix{a_1 \\ & \ddots \\ && a_n}
$$
and let $x = (1,1,\dots,1)^T \in \Bbb R^n$.
De Gua's theorem should tell us that
$$
((n-1)!)^2 V^2 = \det \left[\pmatrix{A\\x^T}^T\pmatrix{A\\x^T}\right] = \sum_{i=1}^n (P/a_i)^2
$$
where $P = \det A = a_1 a_2 \cdots a_n$.  However, applying the Cauchy-Binet formula yields
$$
\det \left[\pmatrix{A\\x^T}^T\pmatrix{A\\x^T}\right] = P^2 + \sum_{i=1}^n (P/a_i)^2.
$$

The question:
So, where did I go wrong?  I suspect that there's an issue with the $M^TM$ formula I give above, but I'm not confident that this is the issue.  If that is the issue, I'm not sure if the above proof is salvageable.
Any feedback here is appreciated.

A:

An alternative proof that uses a different formula: the ($(n-1)$-dimensional) volume of a simplex whose corners are $v_1,\dots,v_n$ can be computed via $V^2 = \frac 1{((n-1)!)^2}\det M^TM$ where
$$
M = \pmatrix{v_2 - v_1 & v_3 - v_1 & \cdots & v_n - v_1}
$$
With this in mind: if we now take $A$ to be the $n \times (n-1)$ matrix given by
$$
A = \pmatrix{-a_1 & \cdots & -a_1 \\a_2\\&\ddots \\ && a_n}
$$
then the desired volume satisfies
$$
[(n-1)!]^2 V^2 = \sum_{i=1}^n \det(A_i)^2
$$
where $A_i$ is the matrix $A$ with the $i$th row deleted.  We compute $\det(A_1) = a_2 a_3 \cdots a_n = \frac{P}{a_1}$, and
$$
\det \pmatrix{A_n} = \det \pmatrix{-a_1 & \cdots & -a_1 &-a_1\\a_2\\&\ddots \\ && a_{n-1}&0} = \pm \frac{P}{a_n}
$$
and similarly, we have $\det(A_k) = \pm\frac{P}{a_k}$ for all $1 \leq k \leq n$.  So, now we end up with
$$
((n-1)!)^2 V^2 = \det \left[\pmatrix{A\\x^T}^T\pmatrix{A\\x^T}\right] = \sum_{i=1}^n (P/a_i)^2
$$
as desired.

As for where I went wrong: it is true that the volume of the $n$-simplex with corners $v_0,v_1,\dots,v_n \in \Bbb R^n$ satisfies
$$
n!V = \det\pmatrix{v_1 - v_0 & \cdots & v_n - v_0} = \det \pmatrix{v_0 & v_1 & \cdots & v_n\\ 1 & 1 & \cdots & 1}.
$$
It is also true that the volume of a $k$-simplex with corners $v_0,v_1,\dots,v_k$ (with $k \leq n$) will satisfy $(k!V)^2 = \det M^TM$, where
$$
M = \pmatrix{v_1 - v_0 & \cdots & v_k - v_0}.
$$
However, it is not true that $\det(M^TM) = \det(Q^TQ)$, where
$$
Q = \pmatrix{v_0 & v_1 & \cdots & v_k\\ 1 & 1 & \cdots & 1}.
$$
There exists a matrix $S$ with $\det(S) = 1$ such that
$$
QS = \pmatrix{v_0 & M\\1 & 0}.
$$
We can then note that $\det[(QS)^T(QS)] = \det[S(Q^TQ)S^T] = \det(Q^TQ)$.  So, we have
$$
\det(Q^TQ) = \det[(QS)^T(QS)] = \det\pmatrix{v_0^Tv_0 + 1 & v_0^TM\\ M^Tv_0 & M^TM}
$$

