826 
Proceedings of the Royal Society 
4. We come now to show that, if the complementary minors of 
all the elements of the last column of an alternant he taken, their 
product is divisible by some power of £}(abc . . .). This result, it 
will be seen, rests formally on the fact that the said minors are 
themselves alternants. 
First, let us consider the case where the indices of the minors are 
consecutive, so that each minor is the difference-product of the 
letters which it involves. 
Denoting throughout the phrase, “ Product of complementary 
minors of elements of last column” (in any alternant) by P m , we 
see that (in the case of three letters, a, b, c ) , in the case of 
| a%\ n | 
the value of P w is obviously 
| d'Fc 2 | or £*(abc ) . 
Again, in the case of 
| a°6W | 
the factor ( a - b) obviously occurs only in the cofactors of the 
elements c n and d n , and similarly with regard to the other factors in 
the difference-product of a, b, c, d. 
Thus we have 
P m | a°b 1 c 2 d n | 2 or £(abcd ) . 
So, in the case of 
| a°b l c 2 d 3 e n | , 
since the factor ( a - b) obviously occurs only in the cofactors of the 
elements c n , d n , and e n , and similarly with regard to the other 
factors in the difference-product of a, b, c, d, e, we have 
P m = | oPFfid^e^ | 3 or 7f-(abcde ) . 
While, generally (the number of letters a, b, c, . . . 7c, l being m), 
in the case of 
| a% l c 2 d 3 . . . k m ~H n | 
since the factor ( a — b ) occurs in the cofactors of the elements 
