MAJOR P. A. MACMAHON: MEMOIR ON THE 



The denominator determinant has the value (1) (2) 2 (3) ; Dividing numerator and 

 denominator by (1)', and then putting x equal to unity, we find 



(a+l)(a + 2)(a+3), 6 (6+1) (6 + 2), (0-1)0(0+,!), (d-2) (d-l) (d) 



(6+1) (6+2), c(c+l) , (d-l)d 



7> j_ O C. 4- 1 d 



a + o 



and herein putting a-6+1, a-o + 2, a-d+3, 6-e+l, 6-rf + 2, c-d+1 separately 

 equal to zero, we in each case find two columns becoming identical and the deter- 

 minant vanishing. Hence the sum of the coefficients has the proper numerical value. 



Art. 18. As a second test I will show that the quotient of determinants becomes 

 unity on putting a = 6 = c = d. 



The numerator determinant becomes 



(a 



( 



x(a+2)(a+3) 



x>+3) 



of 



x (a+2) 



(a) (a+1) 

 (a+1) 



x 





(a) 

 1 



Transform this by taking 

 For New First Row- 



1" Row+x a (l+x+x 8 )x2 nd Row+x 2 " +1 ( 1+x + x2 ) x3rd Row + x^x 4 th Row, 



For New Second Row 



2 nd Row + x" (1 + x) x 3 rrt Row+ x 2 "" 1 " 1 x 4 th Row, 



3 rd Row+x a x4 th Row, 



For New Third Row- 

 and it becomes 



i 1 r r 3 



i, i, .1, & 



x, 1, 1, x 



X 3 , X, 1, 1 



X 6 , X 3 , X, 1 



and thus the quotient of determinants is unity. 

 This verifies numerous particular cases. 

 Art. 19. A third test is to show that the quotient of determinants has the value 



when 6 = c = a, </ = -!. 



The proof is too long to find a place here. 



