PARALLELEPIPED VOLUME

Nigel Nettheim

This item first appeared in Mathematics Magazine, Vol 43 No. 4, September 1970, pp. 229-230.
Published by the Mathematical Association of America.

Problem 750. [January, 1970] Proposed by Charles W. Trigg, San Diego, California.

Evaluate the determinant

| a2 + b2 - c2 - d2    2bc - 2ad    2bd + 2ac |
| 2bc + 2ad    a2 - b2 + c2 - d2    2cd - 2ab |
| 2bd - 2ac    2cd + 2ab    a2 - b2 - c2 + d2 |

Solution by Nigel F. Nettheim, Bureau of the Census, Washington, D. C.

The determinant equals the volume of the parallelepiped formed by the three vectors whose Cartesian coordinates are given by the rows of the matrix. The squared length of the first vector is

(a2 + b2 + c2 + d2)2

and the squared length of each of the remaining vectors is the same by symmetry. The inner product between the first pair of vectors is 0 and the inner product between each of the remaining pairs is the same by symmetry. Therefore the parallelepiped is a cube with volume

± (a2 + b2 + c2 + d2)3 ;

since the coefficient of a6 is +1, the positive sign must be chosen, so the determinant equals

(a2 + b2 + c2 + d2)3 .


By the same method more general results could be obtained. For example, let

sn = x12 + x22 + . . . + xn2 ,

let In be the identity matrix of order n and let An be the matrix of order n with (i, j) element

(-1) i + j + 1 xi xj ;

then, for n = 1, 2, . . . we have

| sn In + 2 An | = - snn .
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