A RIGID BODY IN ELLIPTIC SPACE. 
309 
Now let S imply summation from 1 to n —1 inclusive. Then writing the equations 
of condition in the form 
£'(na)= — 12 0 a 0 , &c., 
we can show, as before, that 
£ , (tat) 2 , (nf)+ =lfl 0 2 (00), 
i.e., 
ZX'n&jdj) + %%'n = ifi 0 2 (oo). 
Subtract this from the former equation, then 
S'n 1 o 0 (o,i)+n 0 2 (o,o)=o. 
In this way we deduce the following equations 
n 0 (o,o)+ft 1 (o,i)+n 3 (o,2)+ ... =o 
n 0 (i,o)+o 1 (M)+o*(i,2)+ ... =o 
fl 0 (2,0) + n 1 (2,l) + O 8 (2,2)+ ... =o 
The condition that these should be simultaneously satisfied is that the determinant 
should vanish ; that is, 
( 0 , 0 ) ( 0 , 1 ) ( 0 , 2 ) 
( 1 , 0 ) ( 1 , 1 ) ( 1 , 2 ) 
( 2 , 0 ) ( 2 , 1 ) ( 2 , 2 ) 
= 0 . 
39. Since (0,1) = (1,0), &c., this determinant is symmetrical. Let [0,0], [0,1], . . . 
denote the coefficients of (0,0), (0,1), ... in the expansion of the determinant. Then 
[0,1]=[1,0]. 
Then we can solve any (n— 1) of the equations to find the ratios O 0 : n 1 : . . . 
Suppressing the first equation we get 
fl O :[ 0 , 0 ]=(- l )*“ 1 « 1 : [ 0 , 1 ] 
Similarly, by suppressing the second equation, we find that 
n o :[l, 0 ]=(-l)”- 1 n 1 :[l,l] 
and so on. Hence finally 
Xl 0 2 : XV : . . . =[ 0 , 0 ] : [ 1 , 1 ] : [2,2] . . . 
