448 Transactions of the South African Philosophical Society. 



question vanishes. We thus have the theorem : In any axisym- 

 metric determinant A which has (1, r) equal to zero for all values of 

 r except n-1 and n, the cof actor of A in the norm of 



(1, n-1) J[n-1, n-1] -f (l,n) J[n, n] 

 is 0, — in other words, the norm vanishes. 



(in.) 



4. If all the elements of the first row vanish except the first and 

 last, the norm then is 



which 



(1,1)11,1] -(1,»)1»,»1 



= (1, 1)»|(2, 2)(3, 3) ... (n, n)\- (1,»)».(1, 1)|(2,2)(3, 3) ... (n-1, n-l)\ 



= (1,1){(1,1)|(2,2)(3,S) ... (n,n)\ - (l,n)-|(2 l 2)(3,3) ... (n^l, n-l)\) 



= (1,1) A. 



Hence in any axisymmetric determinant A ivhich has (1, r) equal to 

 zero for all values of r except 1 and n the cof actor of A in the norm 

 of (1, 1) \/[TJl] + (l,n) J[n, n] is the element (1, 1). (IV.) 



5. Passing now to those cases in which A has three non-zero 

 elements in the first row, we naturally begin by considering the 

 case where 



a b c 



b d e 



c e f 



Here the fact already known from §2 that the norm of 

 a Jk + b JD + c >j¥ — that is, 



N(a ij'df-e" 2 + b Jaf- c 2 + c J ad - b 2 ) 



— is divisible by A is itself an aid in finding the cofactor. For A 

 being altered and the norm not being altered by changing the sign 

 of b or c or e or all of them, it follows that the norm must be 

 divisible by 



a b - c 



b d e or A' say. 



c e f , 



Since, however, the norm of Jx + sly + Jz is of the second degree 

 in x, y, z, our norm must be of the 8th degree in the elements of A : 



