CONTINUANTS RESOLVABLE INTO LINEAR FACTORS. 347 
The performance of the first operation on the columns evolves the equations 
a+ 8B, = 5y,+a+p+3£,, 
6y,+a+ q+ 2B, , 
= Tyztat+rt+B,, 
= 8y,+a+s. 
ll 
Then the factor a@+8@, being removed, the cofactor becomes expressible as a 
determinant of the next lower order, viz. 
— 8p, : 8y, ats |; 
and as by hypothesis the performance of the operation 
col, + 4col, + 9col, + 16 col, 
enables us to remove the factor a+j—86,+128, we obtain three other conditional 
equations, to be followed at the next stage by two others, and last of all by one. Of 
these 4+ 3+2+1 equations the four first obtained are obviously best used to determine 
Pp, q, 7, s in terms of the /’s and y’s, the results being 
p = 8f, — 3B, - dy, 
q = 8B, - 2B, — 6y2; 
= 8p, r By = (ere 
s = 8B, =O ve: 
The remaining six form a very interesting set: after simplification they are 
128,-188,+ 528, == 105, 975 
648, — 816, + 76, = —45y, + 35y; 
156, — 188, = -10y, +77, 
456, -606,+ 148, = — 36y, + 35y, 
248, — 258, = — 157, +14y, 
c= [bee —YarecteYas 
Taking the first three and using with them the multipliers 7, —1, 1 respectively, we 
find, on adding, that 
5(B, +7) — 9(Be+ Ye) + 5(B3+ 7s) — Gt By) = 95 
similarly from the subset of two there results by subtraction 
3(B2 + y2) — 5(B3 +3) + 2(Bi+ ys) = 0; 
and the final set, of course, is 
(8, +3) — (Bytys) = 0. 
By means, therefore, of these three derived equations we arrive at the proposition that 
m the determinant under discussion the sum of any B and the corresponding y is 
constant. 
This being equivalent to only three equations, and other three being still un- 
accounted for, we put 
TAL) AD Vee ke = o— 6, o—f,, Cite o—f,, 
