24 Proceedings of the Royal Society of Edinburgh. [Sess. 
is a factor of 
0*1 + b 4 
«2 + J 2 
a 3 + b 3 
cl 4 + b 4 
$5 + b. 
a 5 + b 2 
a i + h 
a 2 + b 4 
a 3 + b b 
a t + b- 
a 4 + b 3 
a b + i’i 
a 4 + b 5 
cl 2 F b 4 
a 3 + bt 
a 3 + b 4 
(*4 + &5 
a 5 F 6 j 
a 4 F b 2 
ol 2 + b,. 
a 2 + 65 
«3 + S l 
a 4 + b 2 
a 5 + b 3 
a 1 + b i 
F a G a - F & G & a factor of A 
5* 
Increasing the first column by e times the second, e 2 times the third, and so 
forth, we obtain for our new first column 
eF a +e 4 F 6 
e 2 F a + e 3F & 
e 3 F a + e 2 F & 
e 4 F a + eF& 
so that there thus results 
1 
a 2 F t >2 
a 5 + b 5 
+ F & . 
1 
a 2 + b 2 
a 5 F b h 
e 
a 1 + b 3 
«4 F b 4 
€ 4 
a ± + b 3 
. . a 4 F G 
e 2 
a 5 F b 4 . 
a 3 + b 2 
€ 3 
a 5 + b 4 
a 3 F b 2 
e 3 
a i + 65 
cc 2 + b 3 
0 
€ 
a i + b 5 
■ « 2 + h 
e 4 
a 3 + b 1 
. . a x F b 4 
e 
ol 3 F b 4 
a ± F b 4 
or, say, for shortness’ sake 
A 5 = F a .U + F 6 .V. 
If we write e _1 for e in this, F a , F & become G a , G 6 respectively, U and V 
are interchanged, .and of course A 5 is unaltered. We thus have the com- 
panion equality 
A 5 = G & .U + G a .V, 
and from the pair derive 
(F a - G & )A 5 = (F a G a - F 6 G 6 )V, 
which gives us what we desire. For, the linear expression F a — G& not 
being an exact divisor of F a G 0 — F&G 6 , the latter must be an exact divisor 
of A 5 . 
(4) From the same final equality there clearly comes the further result 
that F a — G& is an exact divisor of the determinant Y. Indeed, the equality 
leads us to all the divisors of Y as well as to those of A 5 . Thus, the factors 
of A 5 are seen to be three in number, two quadratics and one linear, one of 
the quadratics being F a G a — F&G& or Q x say, the other Q 2 got therefrom by 
changing e into e 2 , and the linear being evidently 2<x + 26. Substituting 
therefore 
CZl“ + 2l J )QlQ2 for A 5 
