1887.] Mr A. H. Anglin on Summation of Alternants. 199 
1. Employing n letters a, b, c, .. . h, k, l — starting with, the case 
of two general indices y and z, we have 
| a°Zdc 2 . . . h n ~ 3 k y l z | = £}(abc . . . I) | (y - n + 2), (z - n + 1) | , 
which, for shortness, may he written in the form 
[y, z] = I 2, 1 | £- (A). 
Then 
[y+1, z] + [y, z+l] = 
3 2 
2 1 
t' + 
2 1 
3 2 
= 
3 1 
3 1 
3, 1 
which result is the only Extension of equation (A). 
Again, in the case of three general indices x, y, z, writing the 
equation 
| a% 1 r 2 . . . g n ~ A h x k y l z | = tf(abc ... T) | (x - n + 3), (y - n + 2), (z - n + 1) | 
in the form 
we have 
where 
0, V, s] H 3, 2, 1 | £ ( b ) 5 
[x+l, y, z ] + [x, y+l,z] + [x, y,z+ 1] = 
4 3 2 
3 2 1 
3 2 1 
3 2 1 
+ 
4 3 2 
+ 
3 2 1 
3 2 1 
3 2 1 
4 3 2 
Expanding each determinant in terms of the elements of the first 
column, the coefficient of x - n + 4 is | 2, 1 | , while that of x - n + 3 
consists of the sum of two determinants which by (1) is equal to 
1 3, 1 | ; and similarly for the coefficients of the elements involving 
the other indices. 
Hence 
4 2 1 
4 2 1 
4 2 1 
+ 
3 3 1 
3 3 1 
3 3 1, 
and thus 
2 [> + i, y, «] = | 4 , 2 , i \c h .... ( 2 ). 
Further, we have 
0+i, v + 1’ z 1 + \. x + 1> y> z + 1] + 0> y + 1> s + 1] = s 2 £ > 
where 
4 3 2 
4 3 2 
3 2 1 
4 3 2 
+ 
3 2 1 
+ 
4 3 2 
3 2 1 
4 3 2 
4 3 2 
j 
