| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
An old problem of Basil Gordon's has been solved. One wanted 9 integers in a 3X3 matrix, none 0 or 1 or -1, such that the determinant of the matrix was 1 and the determinant of the matrix of the squares of these integers was also 1. Some solutions are 119 208 277 43257 7 9 9 14 16 and 18544 3 4 . 12 21 28 12376 2 3 Also of interest is the parameteric solution: -8n^2-8n 2n+1 4n -4n^2-4n n+1 2n+1 -4n^2-4n-1 n 2n-1 (Stan) What if we also want the determinant of the cubes to be 1?
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 358.1 | TOOLS::STAN | Fri Oct 18 1985 17:23 | 4 | ||
Something's wrong. I checked that matrix with the polynomials in n and found that its determinant was -1 not +1. Also, the determinant of the squares was not 1, but was 128n^4+256n^3+144n^2+16n+1. I must have copied something wrong from my source. | |||||