464 Mr. J. J ; Sylvester on Inverse Orthogonalism 



would cause the constant product to become zero, we have the 

 two solutions, 



(1) a=p, d=p, b = p 2 , c=p 2 , 



(2) a = p q , d=p 2 , b = p, c==/3. 



There is thus but one single type of matrix of this order, viz. 

 Ill 

 1 p p* 

 1 p* p. 



(2) In like manner, for any prime number n there will be but 

 a single type of matrix, the interior nucleus of which is a square 

 matrix of the order (ft — 1) made up of lines or columns of terms 

 in which each line or column contains the (n — 1) powers taken 

 in definite order of the (n — 1) prime roots of unity. That such 

 a matrix is inversely orthogonal is not difficult of proof; but it 

 is less easy to establish, what I have scarcely a doubt is true (but 

 which I have not yet attempted to demonstrate), that such ma- 

 trix, when its lines and columns are permuted in every possible 

 manner, contains the complete solution of the corresponding 

 system of (n— l) 2 equations. The number of distinct systems 

 or roots satisfying these equations will be the number of distinct 

 forms which can be obtained by permuting the lines and columns 

 — in a word, the number of distinct derivatives (a word it will be 

 found hereafter useful to employ) of any given phase of the nu- 

 cleus. This number will be easily seen to be 



(»-l). {n-2)\{n-3) 2 ... I 2 ; 



for each derivative, w T hen all the permutations are taken of the 

 lines and of the columns, will appear n times repeated. For in- 

 stance, if p be a prime fifth root of unity so that 



p p* p 3 p 4 



p 2 p 4 p p 3 



p 3 p p 4 p 2 



p 4 p 3 p* p 



is the nucleus, if we take 



the columns in the order 1, 2, 3, 4, rows in the order 1, 2, 3, 4, 



or 



>) >) 



3, 1, 4, 2 



)> 



)> 



3, 4, 1, 3. 



or 



)> }) 



2, 1, 4, 3 



» 



a 



3, 1, 4, 2, 



or 



)} }) 



4, 3, 2, 1 



)> 



)3 



4; 3, 2, 1, 



the 



resulting derivative 



is in each cas 



e the 



same. 



Thus, then, 



