# Simpler expression for a certain determinant.

A question in elementary linear algebra, while considering the Cayley-Menger Determinant:

Given an $n\times n$ matrix $M$, consider $$\tilde{M}=\begin{pmatrix} M & (1,1,\cdots, 1)^\top \\ (1,1,\cdots, 1)& 0\end{pmatrix}$$ Is it possible to express $\det(\tilde{M})$ in terms of $\det(M)$ and some simpler terms?

You may assume that $M$ is symmetric. (This problem, in its original setting in my research, has the condition that $M$ is symmetric. But it'll be more interesting to solve the general case.)

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One can certainly specialize the determinant formula for block matrices to the bordering case:

$$\begin{vmatrix}\mathbf M&\mathbf e\\\mathbf e^\top&0\end{vmatrix}=-(\mathbf e^\top\mathbf M^{-1}\mathbf e)\det\mathbf M$$

where $\mathbf e$ is the column vector whose entries are all $1$'s.

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Oh I didn't know such a formula. This is amazing. Thanks! –  progressiveforest Jul 29 '12 at 6:07
No problem. Block matrix manipulation isn't exactly common knowledge these days... –  Ｊ. Ｍ. Jul 29 '12 at 6:14
Look up the Schur complement of a matrix. –  copper.hat Jul 29 '12 at 6:24
@copper, yes, that's another way of looking at this. :) –  Ｊ. Ｍ. Jul 29 '12 at 9:37
Of course that formula only works if $M$ is invertible. –  celtschk Aug 20 '12 at 13:56