ELIMINANT OF A SET OF GENERAL TERNARY QUADRICS. 671 



gates of multiples of the given quadrics : consequently they can be used along with the 

 latter for the purposes of dialytic elimination. The simplified eliminant of the 6th order 

 thus obtained is therefore 



a, \ e, 2/i 2 9l 2\ 



a, \ c, 2/ 2 2g 2 2\ 



a 3 b 3 c 3 2/ 3 2g 3 2h 3 



2 I \c 2 h 3 I - 4 1 cj 2 g 3 \ 2 \ \c 2 g 3 | + 4 1 c^, 2 / 3 1 2 1 c x a 2 f 3 \ \ a x i 2 c 3 | 



- 4 1 a x g 2 h 3 1 • 2 I c x aj 3 \ \ a x l 2 c 3 \ 2 | c x a 2 \ 3 | + 4 1 aj 2 g 3 \ 2 \ a x \g 3 | 



2 1 a x b 2 g 3 ! - 4 1 6^2/3 1 . 2 I \c 2 h 3 I I a x \c 3 I 2 I ^^2/3 1 + 4 I 5^2^ 



(9) The cubic required, in place of the Jacobian, for the simplification of Sylves- 

 ter's eliminant of the 10th order, has been obtained as follows : — 



Multiplying the three given quadrics by \f 2 g 3 j , - \f 1 g B | , \f x g 2 | respectively, and 

 adding, we have 



k/^si-* 2 - i^/siy + K/^N 2 + 2 i/i^A \- xt J- 



Comparing this with the first of the three quadrics derived in the preceding section, viz., 



2 1 VA iy ~ 4 1 c i/ 2 & I"' + ( 2 1 Ws l+ 4 l c A/ 3 1 } -V z + 2 1 c i a 2 /s \' zx + I <* A c s l-«y > 



we readily see that if we multiply the latter by z and the former by 42, and add, we 

 shall obtain a cubic free of terms in x s , y d , z 3 , x 2 y, xy 2 , viz., the cubic 



{ 2 I \c 2 h 3 1 - 4 1 \gj 3 1 } -y 2 z + 2 | c x a 2 f 3 \-z 2 x 

 + { 2 I Ws I + 4 1 c A/s I } -^ 2 + 4 1 aj 2 g 3 [zx 2 + { | a x h 2 c 3 \ + 8 |/^ s \}-xyz . 



This is exceedingly simple as compared with the Jacobian, but it fails in the matter of 

 symmetry. Proceeding therefore again to the original triad of quadrics, and multiply- 

 ing them by | b 2 g% \, — | b 1 g 3 1 , | b x g 2 | respectively, we obtain after addition a quadric 

 with no term in y 2 , viz., 



loA&l-* 2 - IWsN 2 + ^WsI-f + 2 I^Al^y- 



The multiple of this by 2y, if added to the cubic just obtained, gives us all that can be 

 desired, viz., the cubic 



2\a x b 2 g 3 \-x 2 y + 2\b x c 2 h 3 \-y 2 z + 2\c x a 2 f 3 \-z 2 x 

 + 4 1 b x g 2 h 3 \-xy 2 + 4:\c 1 h 2 f s [ys 2 + 4:\a x f 2 g 3 \-zx 2 



+ { I «i ^2 c s I + 8 l/i# A I } - X V Z • 



From the mode of obtaining it we see that if we call it M 3 , and denote the first of 

 the three derived quadrics of § 8 by M 2 , we have 



M 2 -z + 2\uJ) 2 g z \'y + 4\u x f 2 g 3 \-z = M 



