1887.] Dr T. Muir on the Theory of Determinants. 
457 
b, c 
t -it t 
a , u , c 
on peut etre assure qu’on aura toujours 
( ab' - ab)c - (ac — a'c)b + (be - b'c)a = 0 . 
“ Et si au lieu de joindre la premiere equation, c’eftt ete la 
seconde, nous aurions trouve de meme 
(ab' - a'b)c - (ac - a'c)b' + (be' - b'c)a = 0.” 
Similarly in regard to the quantities 
a , b, c, cl 
a\ b', c\ d' 
„tl iff ft itt 
a , o , c y a 
the identity 
\(ab' — ab) c" - (ac — ac) b" + (be' - b'c) a nl \d 
- [(ab' - a'b)d" - (ad' - a'd)b" + (bd' - b'd)a"]c 
+ [(ac - a'c) d" — (ad' - a'd)c" + (cd' — cd) a"~\b 
- [(be - b'c) d" - (bd' - b'd) c" + (cd' - c'd) b"]a = 0 
and two others are established, the general theorem of course being 
merely referred to as easily obtainable. 
Thus far there is in substance nothing new. What we have 
obtained is simply a different aspect of Vandermonde’s theorem, 
that when two indices of either set are alike the function vanishes , or, 
as we should now say, a determinant with two rows identical is 
equal to zero. Indeed the identities are used by Vandermonde in 
Bezout’s form when solving a set of simultaneous equations. But 
what follows is important. 
By taking two of these identities 
(ab' - a'b)c - (ac' — ac)b + (be - b'c)a = 0 
(ab' - ab)c - (ac - a'c)b' + (be' - b'c)a = 0 , 
multiplying both sides of the first by d\ both sides of the second 
by d, and subtracting, there is obtained in regard to the quantities 
a, b, c, d 
a\ b\ c , d' 
the identity 
(ah' - a'b)(ed' - c'd) - (ac - a'c)(bd' - b'd) + (be - b'c)(ad' - a' cl) — 0 . 
