391] SOLUTION OF A PROBLEM OF ELIMINATION. 41 
or, what is the same thing, 
X (a + kb) + X' (a + kb') + X" (a" + kb") = 0, 
X(b+kc) + \' (b' +kc')+ X" (b" + kc") = 0, 
X (c + kd) + X' (c' + kd') + X" (c" + kd") = 0, 
X (d + ke) + X' (d' + ke')+ X" (d" + ke") = 0 ; 
and representing the columns 
a 
b, 
a' 
V, 
a" 
b" : 
b 
b' 
c', 
b" 
c". 
c 
d, 
c 
d', 
c" 
d", 
d 
e> 
d! 
e, 
d" 
e", 
1, 
2, 
3, 
4, 
5, 
6, 
each equation is of the type 
X (1 + k2) + X / (3 + M) + X" (5 + k6) = 0. 
Multiplying the several equations by the minors of 135, each with its proper sign, 
and adding, the terms independent of k disappear, the equation divides by k, and 
we find 
X 2135 + X'4135 + X"6135 = 0; 
operating in a similar manner with the minors of 246, the terms in k disappear, and 
we find 
X 1246 + X' 3246 + X" 5246 = 0; 
again, operating with the minors of (146 + 236 + 245 + &246), we find 
X {1236 + 1245 + k (2146 + 1246)} 
+ X' {3146 + 3245 + k (4236 + 3246)} 
+ X" {5146 + 5236 + k (6245 + 5246)} = 0, 
where the terms in k disappear, and this is 
X (1236 + 1245) + X' (3146 + 3245) + X" (5146 + 5236) = 0. 
We have thus three linear equations, which written in a slightly different form are 
X 1235 +X' 3451 +X" 5613 = 0, 
X (1236 + 1245) + X' (3452 + 3461) + X" (5614 + 5623) = 0, 
X 1246 + X' 3462 + X" 5624 = 0, 
and thence eliminating X, X', X", we have 
1235, 1236 + 1245, 
3451, 3452 + 3461, 
5613, 5614 + 5623, 
1246 
3462 
= 0, 
5624 
C. VI. 
6