1896 - 97 .] Dr Muir on Quaternary Quadrics. 
333 
D, E, K, G their values as found from the original equations. Or, 
we may proceed synthetically, thus : — 
From (P) AKB xyzw = - CDL x 2 y 2 + LEA^y/% + BGC x 2 yw y 
and from (a) AKB xyzw = ABG yz 2 w - ABDz% 2 + ABE xzw 2 ; 
.'. by addition 
4AKB xyzw= - 2CDLr 2 y 2 - 2ABD^% 2 
+ 2ABGyz% + 2LEA*y% + 2BGC xhyw + 2 ABEa-^o 2 . 
Again, from ( 1 ), (3) Y)~K.xyzw = BLx 2 z 2 + AL y 2 z 2 + BCxhv 2 + AC y 2 w 2 , 
and from (2), (4) EG xyzw = BLLY 2 + CL.r 2 y 2 + ABzho 2 + AC yhc 2 ; 
by addition 
D(EG - DK )xyzw = D(CLx 2 y 2 + AB zV - AL y% 2 - BGxho 2 ) . 
From this and the previous result it follows that 
(4 AKB + DEG - D 2 K \xyzw = - D(CLr 2 y 2 + ABzV + AL y 2 z 2 + BCx% 2 ) 
+ 2ABGy^%' + .... 
= -D(C* 2 + A* 2 ) (Ly 2 + Bio 2 ) 
+ 2BG(Gr 2 + Az 2 )yw + 2 AE(L// 2 + B w 2 )xz , 
which, on division by xyzw, gives as desired 
D-<f>6 - 2BG</> - 2 AE- 6 + (4 AKB + DEG - D 2 K) = 0 . 
This is not invariable to the cyclical substitution, but leads at 
once to the three others 
E-0tj> - 2CD -6 - 2BIv<£ + (4BGC + EKD - E‘-G) = 0 
K-<t>0 - 2LE .0 - 2CG-6* + (4CDL + KGE - K 2 D) = 0 
G-6<t> - 2AK-0 - 2LD-0 + (4LEA + GDK - G 2 E) = 0 , 
whence, on the elimination of <f>$, <£, 0, we have 
D 
BG 
AE 
4AKB + D(EG - DK) 
E 
BK 
CD 
4BGC + E(DK - EG) 
K 
LE 
CG 
4CDL + K(EG - DK) 
G 
LD 
AK 
4LEA + G(DK - EG) 
D 
BG 
AE 
K(2AB - D 2 ) 
E 
BK 
CD 
G(2BC - E 2 ) 
= 0 , 
K 
LE 
CG 
D(2CL - K 2 ) 
G 
LD 
AK 
E(2LA - G 2 ) 
