ELIMINANT OF A SET OF GENERAL TERNARY QUADRICS. 



681 



but, on performing on this the cyclical changes and adding, we obtain a cubic which, 

 like J and M 3 , is cyclically symmetrical, viz., 



- 2,4[1].^ + i:(3[4]-2[4']}.*% + Z{3[7] + 2[7']}-*2/ 2 + 3[0].^z. 

 The eight cubics of this kind are — 







X s 



f 



Z 3 



x 2 y 



2/ 2 z 



z 2 x 



xif- 



yz 2 







zar 



xyz 



-4[1] 



-4[2] 



-4[3] 



3[4]-2[4'] 



3[5]-2[5'] 



3[6]-2[6'] 



3[7] + 2[7'] 



3[8] + 2[8'] 



3[9] + 2[9'] 



3[0] 



-3[1] 



-3[2] 



-3[3] 



2[4]-5[4'] 



2[5]-5[5'] 



2[6]-5[6'] 



4[7]+ [7'] 



4[8]+ [8'] 



4[9]+ [9'] 



3[0] + 6[0'] 



-3[1] 



-3[2] 



-3[3] 



4[4]- [4'] 



4[5]- [5'] 



4[6]- [6'] 



2[7] + 5[7'] 



2[8] + 5[8'] 



2[9] + 5[9'] 



3[0] + 6[0'] 



-2[1] 



-2[2] 



-2[3] 



[4]-8[4] 



[5]-8[5'j 



[6]-8[6] 



5[7] 



5[8] 



5[9] 



3[0] + 12[0'] 



' -2[1] 



-2[2] 



-2[3] 



5[4] 



5[5] 



5[6] 



[7] + 8[7] 



[8] + 8[8'] 



[9] + 8[9'] 



3[0] + 12[0'J 



3 







-4[4'] 



-4[5'] 



-4[6'] 



2[7] 



2[8] 



2[9] 



[0] + 8[0'] 



3 







2[4] 



2[5] 



2[6] 



4[7'] 



4[8'] 



4[9] 



[0] + 8[0] 



" [1] 



" [2] 



- [3] 



w- m 



[5]- [5'1 



[6]- [6'] 



[7]+ [7'] 



[8]+ [8'] 



[9]+ [9'] 



[0] + 2[0'] 



On account of the cyclo-symmetry and the fact that the determinant in the first column 

 is always [l], in the second column always [2], and so on, the cubics may be denoted 

 shortly by the coefficients of the determinants only. The list would then stand as 

 follows : — 



A ( 



'-4, 



3-2, 



3 + 2, 



3 + 0) 



B 



1-3, 



2-5, 



4+1, 



3 + 6) 



B' < 



:-3, 



4-1, 



2 + 5, 



3 + 6) 



C < 



-2, 



1-8, 



5 + 0, 



3 + 12) 



C 



:-2, 



5 + 0, 



1 + 8, 



3 + 12) 



M, < 



: o, 



0-4, 



2 + 0, 



1 + 8) 



M' 8 ( 



o, 



2 + 0, 



0+4, 



1 + 8) 



J ( 



-i, 



1-1, 



1 + 1, 



1 + 2). 



(22) Of course, any seven of these are derivable from the remaining one with the 

 help of aggregates of multiples of the original quadrics. As a matter of fact, only two 

 such aggregates are necessary, viz., 



u$Ji 3 \-x + \u 1 h 2 f i [y + \uj % g z \z 



which will be found to be equal to 



