36 



DR THOMAS MUIR ON THE 



Setting the factor 00 aside we consequently obtain the eliminant in the form 



5 



9 + 9' 

 4 7 + 7' 



8 + 8' 

 6 

 



+ Zi 







6 



8 



7 + 7' 







5 



5 



8 + 8' 



11 



- 24 







4 



5 



+ 8' 







11 



6 



9 + 9' 



8 



+ 12(03 -6'8-6'8'+ 68') - 15( 56'- 37 -37'- 08') 

 + 45(34 + 8'9 + 8'9' - 06') + (123 - 456)0', 



which, on the development of the determinants, and substitution of 



-±0456 +±049 IT for -±0048', 



-±45512 +±4589 for -±0456', 



±446 IT -±47IIT2 for ±049'II , 



±4678' -±77812 for ±078'9', 



±16711 -±1788' for ±1268, 



±16711 -±1268' for ±17'88', 



±44611 -±1556 for ±4488', 



gives the result of § 31. Denoting, therefore, the eliminant of § 31 by E we have 



7l = 00- E = y 5 . 



In the same way it can be shown that 



y 2 = 0'0'-E, 



y 3 = (2 12 + 6'7)-E, 



y 4 = (-3 10 + 67')-E. 



(39) The extraneous factors in the cases of y 3 and y 4 are in appearance a 

 little peculiar, as they give no evidence of being symmetrical with respect to the 

 circular substitution. That they really possess this symmetry is, however, readily 



be changed into 



and thus into 



I a ihf 3 



{| \m 2 c. 

 {I h m 2 C i 

 {l \™*h 



IffiV^I + I W m 2>h\ I C l92 J h\} 

 if 1 n 2 h 3 \ + | b 1 m li n 3 \ | cj 2 h 3 \} 

 I f x n 2 g 3 1 + | bjm.fig | ! Cl f 2 g 3 | } 



I bjm 2 c 3 | • {I a x l 2 f 3 1 ! g 1 7t^ z 

 + I b 1 m 2 n 3 \ ■ {| a x l 2 f 3 | | Cl g 2 h 3 



so that by a second use of the said theorem we have 



and finally 



or by a third use of the same theorem 



- I 6,m 2 c 3 

 + | b l m. i n 3 



\fi9zh\ 



I ftfjh 



I a ih9 s I ( AnJh 

 I a ik9z 1 I cJJh I 



I a ih n 3 I I fi9ih I 

 I «A C 3 I I fi9A I 



+ I «i^3 



I J \n>29 3 1 } 



6 1 m 2 w 3 1 

 b x m 2 c 3 1 



a x l 2 n 3 



ft 1^2 C 3 



I b{m.,a 3 1 I c x n 2 a 3 



