336 
Proceedings of Royal Society oj Edinburgh. [sess. 
Solving the 2nd and 3rd for y x , y 2 , and the 4th and 5th for S x , S 2 , 
we find on substitution in the 6 th 
KBV - (BEG + BFH)a x 3 /l x + 2(AEH - ABK + BFG)a x 2 /3 x 2 - 
(AHF + AEG )a x /3 x 3 + KA% 4 = 0 , 
which, when taken along with the remaining equation, gives 
KD 2 - 4ABK - DEG - DFH + 2AEH + 2BFG = 0 , 
or 
2 A D G 
D2B H 
F E K 
= 0 , 
as it should be. 
In the next place, however, it is important to notice that there 
are more than 10 such relations connecting the coefficients, A, B, 
C, L, D, etc. For, as the preceding mode of obtaining the rela- 
tions shows, we have oniy got to select a certain number of the 10 
equations cqa 2 = A , etc., from which it may he possible to eliminate 
the Greek letters, and one such relation must result. To take at 
once the most pertinent example, let us choose the eight equations 
oqGC-} — -A. , — D , 
@1@2 ~ B 5 filY2 "h &7l = ® J 
7i 72 = C > rA+ 72 8 i = K > 
3 X 3*, — Ij , S x (x<, — f- ^ 2^1 — G . 
From these, on eliminating a 2 , /3 2 , y 2 , S 2 , we have the four equa- 
tions 
Bcq 2 + A/3-g = D a x /? x , 
Cft 2 +By 1 2 =E/3 1 y 1 , 
Ly x 2 + CS 1 2 =Ky 1 S 1 , 
AS X 2 + Loq 2 = GS x a x , 
and there thus remains the problem of eliminating a 13 /3 1? y x , S 1? 
from these four. But this is the very problem we have been con- 
sidering as the extension of Sylvester’s, and it is its appearance in this 
connection which at once suggests the possibility of attacking it in 
the rear. To begin with, we infer that the new relation sought 
cannot be an independent relation, hut must be derivable from two 
or more of the initial ten, only three of which, by the way, are 
themselves independent. How, when we examine the ten care- 
fully, we find that there is a set of four which involve all the co- 
