339 
1896 - 97 .] 
Dr Muir on Quaternary Quadrics. 
treating similarly the 3rd term we have . 
(BG 2 + D 2 L-4ABL>y% 2 
= B?j 2 (Giv) 2 + Lw 2 (Dy) 2 - 4ABLy% 2 , 
= W ^ 2 + L*y + + - 4ABL yV 
= BLW + BA =^ 4 + LB% 2 ;r 2 + LA 2 ^ , 
X L 
-% 2 
(L^ + Bw^^BLa^ + A 2 ^ 
and, finally, from the two results by subtraction there emerges the 
desired equation. 
Using 0 and <£> as before, and taking advantage of the cyclical 
substitution, we thus have 
AO 2 - DG<9 + (BG 2 + D 2 L - 4ABL) = 0 , 
B<£ 2 - ED<£ + (CD 2 + E 2 A - 4BCA) = 0 , 
C<9 2 - KE<9 + (LE 2 + I\ 2 B - 4CLB) = 0 , 
L<f) 2 - GK cj> + (AK 2 + G 2 C - 4LAC) = 0 . 
From the 1st and 3rd of these there results the eliminant 
A 
BG 2 + D 2 L - 4ABL 
2 
A 
DG 
DG 
BG 2 + D 2 L - 4ABL 
C 
LE 2 + K 2 B - 4BCL 
c 
KE 
• 
KE 
LE 2 + K 2 B - 4CLB 
or 
(ALE 2 + ABIC 2 - BCG 2 - CLD 2 ) 2 - 4BL(AEIv - CDG) 2 
- (AEK - CDG)(EG - DK)(DEL - BGK) , 
or 
2A 2 B 2 K 4 - 22ACB 2 K 2 G 2 - 2ABCL2 G 2 E 2 + 8ABCL • DEKG 
- DEKG 2 ABK 2 + 2ABE 2 K 2 G 2 , 
which agrees completely with the result of § 6. 
From the 2nd and 4th, or simply by cyclical substitution, we 
obtain the eliminant also in the form 
( - ALE 2 + ABK 2 + BCG 2 - CLD 2 ) 2 - 4CA(BKG - DEL) 2 
- (AEK - CDG)(EG - DK)(DEL - BGK) 
where the last term, being invariant to the cyclical substitution, is 
the same as before, and where therefore the two other terms, not 
being individually invariant, must be so when taken together. 
