of Edinburgh, Session 1885-86. 
569 
renferment le meme nombre de termes, et pr^cisement les 
memes, puisqu’elles renferment tous les termes qui peuvent 
resulter de la combinaison des n lettres a, b, c , d , e, &c., 
disposees entre elles de tontes les manieres possibles ; il ne pent 
done y avoir de difference entre deux resultantes, que dans les 
signes de chacun de leurs termes ; or, il est visible que la 
premiere resultante donne la seconde, si l’on change dans la 
premiere b en c, et reciproquement c en b ; mais ce ebangement 
augmente ou diminue d’une unite le nombre des variations de 
chaque terme ; d’ou il suit que dans la seconde resultante, tous 
les termes dont le nombre des variations est impair, auront le 
signe + , et les autres le signe partant, cette seconde 
resultante n’est que la premiere, prise n^gativement. 
“ Il est visible pareillement que . . . . ” &c. 
The proof is thus seen to consist in establishing (1) that the terms 
of the one “ resultant” must, apart from sign, be the same as those of 
the other ; and (2) that the terms of the one resultant are either all 
affected with the same sign as the like terms of the other, or are all 
affected with the opposite sign, the comparison of sign being made 
by comparing the number of variations. 
After this, the theorem that when two letters are alike the result- 
ant vanishes is established in a way different from Vandermonde’s, 
but not more satisfactory, viz., by considering what B^zout’s rule 
would lead to in that case. 
Application is then made to the problem of elimination, and to 
the solution of a set of linear simultaneous equations, the mode of 
treatment being again different from Vandermonde’s, but this time 
with better cause. He says — 
11 Je suppose maintenant que l’on ait les trois Equations 
0 = 1 a.fj, + ^b.f + hj./u/', 
0 = 2 a.[x + 2 b.f + 2 c.[x'\ 
0 = d a./x + 3 b./x' + 8 c./u/', 
je forme d’abord la resultante des trois lettres a , b, c, suivant 
l’ordre a , b , c, ce qui donne, 
l a. 2 b. s c — l a. 2 c.% + l c. 2 a.% — l b. 2 a. s c + 1 5. 2 c. 3 a — l c 2 b 2 a. 
ou 
1 a.^b,*c - 2 c. 3 6] + 2 a.\}c*b - 1 6. 3 c] + *a.]}b 2 c - l c. 2 b~\‘ } 
