372] 
ON THE RECIPROCATION OF A QUARTIC DEVELOPABLE. 
507 
or we have 
(ad — be) 2 — 4 (ac — b 2 ) (bd — c 2 ) = 0, 
which is in fact the equation 
□ = a 2 d 2 — 6abcd + 4ac 3 + 4<b 3 d — 3 b 2 c 2 = 0, (1) 
and thus the reciprocal is in fact the quartic torse, given by the equation (1). 
The equations (3), (4), (5) lead to 
x8a + y8b + z8c + w8d = 0 ; 
but (a, b, c, d) are in consequence of these equations connected by a single equation, 
viz. the equation (1); the equation just obtained is thus the only relation between 
the differentials (8a, 8b, 8c, 8d): it is clear from the equations (2) that this is in fact 
the equation 
8D = 8 a □ 8a + 8 b □ 8b + 8 C □ 8c + 8 d □ 8d = 0. 
The equations (1), (2), (3) lead in like manner to 
a8x + b8y + c8z + d8iv = 0, 
which in virtue of the equations (5) is equivalent to 
p8 (3xz — y 2 ) + q8 (yz — 9xw) 4- r8 (3yw — z 2 ) = 0, 
viz. it is a consequence of 
8 (3xz — y 2 ) = 0, 8 (yz — 9xw) = 0, 8 (3yw — z 2 ) = 0. 
But inasmuch as (x, y, z, w) are connected by a two-fold relation, the equations 
(1), (2), (3) must lead to one other linear relation between the differentials (8x, 8y, 8z, 8w): 
and I proceed to show that this is so. 
Differentiating we have 
dr 8a — 3cd 8b + (— 3bd + 6c 2 ) 8c + ( 2ad — 3be) 8d + Adx + x8\ — 0, 
— 3cd! 8a + ( 12bd — 3c 2 ) 8b + (— 3ad — 6bc) 8c + (—Sac — 6b 2 ) 8d + \dy + ydX = 0, 
(— 3bd + 6c 2 ) 8a + (— Sad — 6bc) 8b + ( 12ac — 3b 2 ) 8c— Sab 8d + \dz + zd\ — 0, 
( ‘lad — 3be) 8a + (— 3ac + 6b 2 ) 8b — Sab 8c + a 2 8d+ Xdiv + wd\ — 0. 
Consider the matrix 
— 3 cd , 
— 3 bd + 6c 2 , 
lad — 3 be, 
— 3 cd , 
12 bd — 3c 2 , 
— Sad — 6bc, 
— Sac + 6b 2 , 
— 3 bd + 6c 2 , 
— Sad — 6bc, 
12ac — 3b 2 , 
— Sab , 
lad — 3 be 
— 3ac + 6b 2 
— Sab 
a 2 
64—2