211 
1888-89.] Dr T. Muir on the Theory of Determinants. 
To this proof the following note is appended (p. 207) 
“Cette demonstration quoiqu’assez simple semble reposer 
cependant sur un artifice de calcul : mais en cherchant nne 
demonstration direde , j’ai rencontre une difficulte d’un genre 
particulier. En effet, on trouve facilement que l mQ terme de 
l’une des fonctions en question est aussi egal oil au meme 
terme de l’autre, ou generalement au m me , et que, dans le 
dernier cas, le rn me terme de la premiere est aussi egal au 
Z me de la seconde, abstraction faite des signes. (ix. 5) 
Mais l’identite de ces derniers (qui est de rigueur) exige des 
explications tres-longues et beaucoup moins elementaires que 
la demonstration que je viens de donner.” 
The remaining six or seven pages of the paper are more interest- 
ing, and concern the subject of vanishing aggregates of products 
of pairs of determinants. The theorems were suggested by taking, 
as we now say, a determinant of even order having its last n rows 
identical with its first n rows, e.g., the determinant 
(abab , 1234) , 
and using theorem (3) to expand it in terms of minors formed from 
the first n rows and their complementary minors. When n is even, 
a proof is thus obtained, as we have seen in the footnote to the 
account of Bezout’s paper of 1779, that the first half of the expan- 
sion is equal to zero. When n is odd, the method fails, although 
the proposition is still true.* Keiss’s enunciation is as follows 
(p. 209) 
* It is worthy of note in passing, that a common method does exist for 
establishing the two cases, — a method quite analogous to Reiss’s, but difficult 
of suggestion to one who used his notation, or indeed to any one who had no 
notation suitable for determinants whose elements had special numerical 
values. All the change necessary is to make the last n elements of the first 
column each equal to zero. This causes no difference in the result when n 
is even, e.g., from the identity 
a 4 a 2 a 3 a 4 
\ b 2 b 3 b 4 _ ^ 
• 6^2 ^3 $4 
• b 2 b 3 b 4 
we have, as before, 
\(hb 2 \.\a 3 b 4 \ - \a 1 b 3 \.\a 2 b 4 \ + | a x b A [.| a 2 b 3 | = 0 ; 
and when n is odd, the second half of the terms which previously gave trouble 
do not occur. 
