151 
1S97-9S.] Mr E. J. Kan son on a Problem of Sylvester’s. 
where 
A = 
a h g 
h b f 
g f c 
so that “ rejecting the special ( N.B . , not irrelevant) factor abc, we 
obtain ” 
A =0. 
2. Kow, as will be shown presently, abc is a factor of the true 
eliminant of (1). But abc is not a factor of (3). For, if we 
multiply columns 4, 5, 6 by be, ca, ab ; then multiply columns 
1, 2, 3 by - af, bf, cf, and add to column 4 ; multiply by ag, - bg, 
eg, and add to column 5 ; multiply by ah, bh, - ch , and add to 
column 6, we find the value of the determinant in (3) to be 
1 
aW 
. c b 
c a 
b a . 
. . aA! 
all’ 
aG’ 
g . bW 
bB’ 
hr 
. h cG' 
cV 
cC’ 
where A', B', C', F', G', H' are the co-factors of a, b, c, f, g, h in A . 
But this is 
abc' 
. c b 
c . a 
b a . 
A' 
H' 
G' 
H' 
B' 
F # 
C' 
F' 
C 
that is 2 A 2 . 
3. Kext, Cayley points out, first, that the equations (1), (2), are 
found by expressing that an arbitrary point on the line 
V , s) (f, V, 0 = 0 • • • ( 4 ) 
is on the conic 
(«, £, «,/, g, (f> *i, f) 2 =o . . (5) 
and, second, that in virtue of the identical equations 
Aa? + Hy-f Gz = 0 v 
ILr + By + Fz = 0 > . . . (6) 
G^+Fy+Cz = o) 
the equations (1), (2) are equivalent to only three independent 
