304 Proceedings of Royal Society of Edinburgh. 
A(BB'-A'C>;2 + B(AA'-B'C>z + - K^)xy 0 \ 
A(A'2-BC)?/^ + B(CC' - B'A>^z; + C(BB' - C'A>y = 0 I (e) 
A(CC'-A'BV + B(B'2-CA>a; + C(AA' - C'B>?/ = 0 j 
the resultant of which is immediately seen to he 
ABC . A2 = 0. 
7. Again, taking the same two equations viz. (yg), and now 
eliminating xy, we obtain an equation involving xz^ yz, z ^ — that is 
to say, involving the simple unknowns x, y^ z, tlie full set being 
(BB'-A'C> + (AA'-B'C')y + (C'^ -AB)^ =0*^ 
(A'2 -BC) a: + (CC'-B'A')y + (BB' - C'A> = 0 I (^). 
(CC'-A'B> + (B'2 -CA)2/ + (AA'-C'B> = 0 j 
Here the resultant is in a still simpler form, viz , 
A2 = 0. 
8. We have thus arrived at the resultant in three different ways 
by means of a determinant of the third order, viz., 
in § 5, where the unknowns are x^, z^ ; 
in § 6, where the unknowns are yz, zx, xy \ 
and in § 7, where the unknowns are x^ y, z . 
There are other ways, however ] and though none of them is any 
simpler than that of the preceding paragraph, it is nevertheless 
interesting to see them, especially when they are brought into 
comparison with those already obtained. 
Thus, taking (^j), (^ 2)5 eliminating x^ and z‘^ from (ag), we 
have. 
A(CC' - A!^')yz + (2A'B'C' - AA'^ - CCy.x + C(AA' - B'C')xy = 0 
A(BB' - CA:)yz + B(AA' - B'C' )zx + (2 A'B'C' - BB'2 - KM‘^)xy = 0 
(2A'B'C' - CC'2 - BB'2)y^ + B(CC' - A'B>x + C(BB' - C'A>y = 0 
where it would be easy to show that the resultant takes the form 
2A'B'C'*A2 = 0; 
indeed the determinant here and the determinant of § 5 are trans- 
formable into one another by changing rows into columns. It is 
much more important, however, having now got two sets of equations 
in yz^ zx, xy, to compare the one with the other. Doing so, we find 
