362 
Proceedings of Royal Society of Edinburgh. [sess. 
Diminishing (or increasing) each element of each row by the cor- 
responding element of the next row, we change these into 
and 
a 
b - a 
c-b 
d - c 
e-d- g 
9 
a 
b - a 
c-b-g 
d+g-c-f 
9 
a-g 
b+g-a-f 
c+f- b-e 
9 
-9 
g-f 
a + f-e 
b + e - a - d 
9 
g-f • 
f~e 
e - d 
a + d - c 
a 
b + a 
c + b 
d + c 
e + d-g 
. 
a 
b + a 
c + b - g 
d-g+e-f 
• 
• 
a-g 
b-g+a-f 
c - f+ b-e 
• 
-9 
-g-f 
a- f-e 
b - e + a - d 
9 
~9~f 
-f-e 
-e-d 
a- d- c 
and now increasing (or diminishing) each element of the 2nd 
column by the corresponding element of the 1st, each element of 
the 3rd by the corresponding element of the new 2nd, and so on, 
we have, finally, two identical determinants, 
a 
b 
c 
d 
e-g 
a b 
c 
d 
e-g 
• 
a 
b 
i-g 
d-f 
a 
b 
e-g 
d-f 
• 
0 
a-g 
*-f 
c-e 
and 
a 
-9 b-f 
e-e 
• 
-9 
-f 
a-e 
b-d 
-9 
~f 
a-e 
b-d 
-9 
-/ 
- e 
-d 
a-e 
-9 ~f 
- e 
-d 
a-e 
so that the eliminant is equal to 
(a + b + c + d + e +f+ g)(a -b + c-d + e-f+g) 
x 
a b c d e-g 
. a b c- g d-f 
. . a- g b -f c- e 
. - g -f a-e b-d 
- g - f - e - d a- c 
as was to be shown. 
3. The essence of the proof is seen to be the resolution of the 
eliminant into two determinants closely resembling each other, the 
removal of a linear factor from each, and the reduction of the two 
co-factors into one and the same form. 
The first of the two determinants is got from the eliminant by. 
