540 Proceedings of Royal Society of Edinburgh. [sess. 
2^ = 3100 
+ 3010 
+ 3001 
+ . . . 
Now the required multiplication is performed by adding each of 
these twelve sets of four indices to the indices 0 1 2 3 of the 
multiplicand, the first index to the first, the second to the second, 
and so on, the product being the sum of twelve alternants whose 
indices are the indices resulting from the twelve additions. It will 
he found that a considerable number of the twelve alternants 
vanish by reason of the equality of two of the four indices, and 
that, in fact, we have finally 
| a?b x c 2 d z |.2a 3 6 = | a% 1 c 2 d 4 \ + \a l b i c 2 d z \ + | a°b 2 c 5 d s \ 
+ |a°&W 4 | + |a°6Wj, 
= 2 | a Wd 4 1 - | a°Z> W j - 1 a°6 W | + 1 a°6 W 1 1 
(4) The nature of this process is seen to be such that if we were 
told beforehand that | a°b 2 c^d 5 | was one of the alternants appearing 
in the expression of the product, and were asked to find its co- 
efficient, we could do so only by finding all the other alternants as 
well. With a view to ascertaining whether there is no exception 
to this, let us not use the multiplier as it stands, but take instead 
the expression for it in terms of 2a, 3 ab , 2a&c, . . . . , viz., 
(2a) 2 2^ - 2 (2afr) 2 - 2a2a5c + 4 ^abcd* 
We then find 
| aPfrfid 3 1 . (' '2a) 2 l>,ab= I aPbWdf I . 'Za.'Zab = a°b l c 3 d 4 | + I cflfr&d 5 I } ,^ab 
= I a'bW 1 + 21 aWd 5 [ + I aWd 5 [ + I a»W | , 
- I aWd 3 1 . 2 (Jabf = - 2 | aWd 4 | 2 | aWd 5 I - 2 1 aWd 5 1 , 
- 1 a°b l c 2 d 3 I . '^a'Zabc = - | cdbWd 4 l - I a°b 2 c 3 d 5 I , 
+ 1 aW&d 3 1 . 4 'Zabcd = 4 I a}b 2 c 3 d 4 I , 
and thus by addition obtain 
I aWd 3 1 . 1d 3 b = 2 I bW(P t - I aWd 5 1 - I a°6 W | + I aWd 6 1 
as before. 
* See Hirsch’s Tables, given in Salmon’s Modern Higher Algebra, 4th ed., 
pp. 350-356. 
