110 
ME. A. CAYLEY ON THE EQUATION OE DEFEEEENCES 
The second equation and this equation may be written 
(3(z, 125 12c Is, 1)^=0, 
(35, 12c+4a^, 12d+%9js, 1)^=0, 
and the elimination of s from these equations gives the required equation in 6. The 
result may be obtained under either of the two fonns, 
+(15ac— 275^)^— 36(5(^— c^)} {a^^+3 («c— 5^)} + 3{2a5^+3(a(Z— 5c)}^=0 
and 
{4«TH-(24ac— 275^}^— 36(5(7— c^)} {a^^+12(«c— 5^)} + 3{ aM-\-Q{ad—hc)]-=0, 
the expansions of which respectively coincide with the before-mentioned result. 
In the case of the quartic equation <pv—(a, 5, c, djjv, 1)^=0, we have 
s^a+4s®254-6s^( 4c+«^)H-4s(8(Z+65^)+16c+24c^-l-c!^^=0, 
4 s^«-1-6sH5-4-4s (12c4-a^)+ 32(7-|-85^ =0, 
from which we derive another cubic equation ; and the two cubic equations are 
(4«, 245 , 48c + 4:a6, S2d+ m Is, 1)^=0. 
(45, 24c+10«^, 48(7-f445^, 32c +48c^+2a^^Is, 1)^=0, 
from which, if s be eliminated, we have the equation in 9. 
Similarly, for the quintic equation (pv=(a, 5, c, d, e,f'X'V, 1)®=0, we have 
s^a+ 5s^25+10s^( 4c+«^)+10s"( 86?+65^)+5s(16c+24c^-l-a4^)+32/+80<7^H-105^-=0. 
5s^«-fl05^45H-10s^(12c-f«^)+ 55(32(7+85^)+ (80c+40c^+((^^) =0. 
from which we derive another quartic equation ; the two equations are 
(5(?, 405 , 120c+ 10(^^, 160(7+ 405^ , 80c+ 40c^ + a^Is, 1)^=0, 
(55, 40c+20(?^, 120(7+1305^, 160c+280c^+12(^4^ 80/+200(7^ + 25 5^^Is, 1)^=0. 
from which, if 5 be eliminated, we have the equation in 0. 
But to apply Bezout’s method to the two equations each of the order w — 1, which 
result from the equation of the nih. order (^I'C, 1)”=0. The process is as follows : — 
Suppose, in general, that s is to be eliminated from the two equations 
Fs=0, Gs=0, 
each of the order n—\ -, it is only necessary to form the expression 
FsGs'— Fs'Gs 
S — ^ ’ 
which will be a function of s, s' of the form 
( ^0,0, ®0, 1 • • ®0, n-2l^, 1 )* (^ 5 1 )” 
