155 
1897-98.] Mr E. J. Nanson on a Problem of Sylvester’s. 
11. The elimination may readily be effected without any pre- 
liminary transformation as follows : — 
The equations (1) are 
bz 2 + cy 2 - 2xyz = 0 , ex 2 + az 2 - 2gxz - 0 , ay 2 + bx 2 - 2 hxy = 0, 
and from these we have 
8 fghx 2 y 2 z 2 — (bz 2 + cy 2 )(cx 2 + az 2 )(ay 2 + bx 2 ) 
= ax 2 (bz 2 + cy 2 ) 2 + by 2 {cx 2 + az 2 ) 2 + cz 2 (ay 2 + bx 2 ) - 4 abcx 2 y 2 z 2 
and therefore 
( abc + 2 fgh - af 2 - bg 2 - eh 2 )x 2 y 2 z 2 = 0. 
Hence, either A = 0 or one of the three, x, y, z, is zero. Now if 
x = 0 the second and third equations give 
ay 2 = 0 az 2 = 0, 
and therefore either a = 0 or y and z are both zero. Hence, unless 
x, y, z are all zero we must have 
A = 0 , or u = 0, or 5 = 0, or c= 0. 
Thus, the elimination of x, y, z from (1) leads to 
abc A = 0. 
Now the eliminant of three ternary quadrics is of order 12 in 
the coefficients. Hence, as all possible alternatives have been 
taken into account, it follows by symmetry that the characteristic 
of the true eliminant must be one of the three 
abc A 3 , a 2 b 2 c 2 A 2 , a 5 b 3 c 3 A. 
12. That the second of these three forms is the true eliminant 
is easily shown. For when the eliminant of a number of equations 
vanishes, the equations are satisfied by a common system of values 
of the variables. But in the present case there are eight such 
common systems. When a — 0 the equations are satisfied if 
x = 0 and bz 2 + cy 2 - 2xyz = 0 ; 
that is, there are two systems of values of x, y, z which satisfy (1), 
and these values are the tangential coordinates of the lines joining 
the point ^ = 0, £ = 0 to the intersections of ^ = 0 with the conic 
(5). Similarly, when 5 = 0 or c = 0 we get two systems. When 
