Dr Muir on the Theory of Alternants. 
101 
and gives as his result 
( ( ( . h .h h h .h \(3) I «1 . a 2 . a s a 4 . a 5 \ 
Up A 2> A 3> A 4> \) • I A 1 A 2 A 3 A 4 A 5 ) 
II A 7l + a l A 7i+£l2 A h+tts A A a A 
= I A, A, A, a 4 a 5 ; 
[| ft+ai . /i+a2 . ct$ . 
+ A 1 A 2 A 3 A 4 A 5 / 
+ 
+ |'A» 3 + “ 3 A 4 +<l 4 A 5 +a 0 ’ 
wo h, h, h in fiinf Abtheilungen zu dreien vertheilt werden, 
namlich 
h + a 4 
li + a 2 
h + a 3 
°4 
a b 
h + a 1 
h + a 2 
a 3 
li + a 4 
a 5 
li + a j 
li + a 2 
a 3 
«4 
h + a b 
li + a i 
a 2 
h + a 3 
h + a 4 
ffljj 
h + a 1 
a -2 
h + a 3 
a 4 
h + a b 
e 
+ 
a 2 
a 3 
h + a 4 
h + a b 
a 1 
h + a 2 
li + a 3 
h + a 4 
a b 
a 1 
h + a. 2 
h + a 3 
a 4 
h + a b 
a i 
h + a. 2 
a s 
h + a 4 
h + a b 
a i 
a 2 
h + a 3 
li + a 4 
h -I - a b 
the table being intended to make clear the fact that the five indices 
of each of the ten alternants on the right of the" identity is got 
from the five 
CXgj ^ 3 > ^ 4 > ^5 
of the given alternant on the left by adding h to three of them. 
