In linear algebra, geometry, and trigonometry, the Cayley–Menger determinant is a formula for the content, i.e. the higher-dimensional volume, of a -dimensional simplex in terms of the squares of all of the distances between pairs of its vertices. The determinant is named after Arthur Cayley and Karl Menger.
Let be points in -dimensional Euclidean space, often with . These points are the vertices of an n-dimensional simplex: a triangle when ; a tetrahedron when , and so on. Let be the distances between and , for . The content, i.e. the n-dimensional volume of this simplex, denoted by , can be expressed as a function of determinants of certain matrices, as follows:
This is the Cayley–Menger determinant. For it is a symmetric polynomial in the 's and is thus invariant under permutation of these quantities. This fails for , but it is always invariant under permutation of the vertices.
A proof of the second equation can be found. From the second equation, the first can be derived by elementary row and column operations:
then exchange the first and last column, gaining a , and multiply each of its inner rows by .
Generalization to hyperbolic and spherical geometry
There are spherical and hyperbolic generalizations. A proof can be found here .
In a spherical space of dimension and constant curvature , any points satisfy
where , and is the spherical distance between points .
In a hyperbolic space of dimension and constant curvature , any points satisfy
where , and is the hyperbolic distance between points .
In the case of , we have that is the area of a triangle and thus we will denote this by . By the Cayley–Menger determinant, where the triangle has side lengths , and ,
The result in the third line is due to the Fibonacci identity. The final line can be rewritten to obtain Heron's formula for the area of a triangle given three sides, which was known to Archimedes prior.
In the case of , the quantity gives the volume of a tetrahedron, which we will denote by . For distances between and given by , the Cayley–Menger determinant gives
Finding the circumradius of a simplex
Given a nondegenerate n-simplex, it has a circumscribed n-sphere, with radius . Then the (n+1)-simplex made of the vertices of the n-simplex and the center of the n-sphere is degenerate. Thus, we have
In particular, when , this gives the circumradius of a triangle in terms of its edge lengths.