370 



HITCHCOCK. 



panding, first, into a sum of ordinary determinants or square bracket 

 expressions, the sign before every square bracket is changed because 

 the signs behave as if (51) were an ordinary determinant; but, since 

 the development follows the rows, the sign of every determinant (53) 

 will have been changed due to the interchange of rows; thus each 

 square bracket has its sign changed, we may say, both from within and 

 from without. Hence the whole expression is unaltered. On the 

 other hand, an interchange of two columns in (51) changes the sign 

 before every square bracket but leaves the brackets themselves un- 

 affected, for the development is according to rows; hence the sign of 

 the whole is changed. 



It is now not hard to see that the polyadic determinant (51) is the 

 same as ((<pi, <pz, • • •, <Pn))s- For we have seen that the first index is 

 non-signant, the second signant; the third is obviously signant since 

 it governs the columns of the ordinary determinants of the form (53) 

 but does not otherwise enter. Furthermore, on complete expansion 

 of (51) we shall have each term a product of n elements, one from each 

 of the double polyadics, one from each row of (51), and one from each 

 column of (51). But this is precisely the law of formation of a cubic 

 determinant having the elements of the polyadics as non-signant 

 layers. By Art. 6 this is the scalar in question and we may write the 

 identity 



<P\ • Ei, <p\ '. Eo, 



<p2 '• Ei, <po • E2, 



' , (pi ' E n 



^n^Ei, (Pn'^2, *•*, ^n"-E n 



((<Pl, <P2, "•» <Pn))s (54) 



where the polyadic determinant on the left is defined as above. 



To take the simplest possible illustration, let K = 1, N = 2, n = 2. 

 Our A"-adics become vectors in two dimensions. Take Ei = i, E 2 = j. 

 Let there be two dyadics <p\ = bni + bi?j = 6 m ii + ^mji + frmij + 

 fcmjj and similarly for ^2. The determinant on the left of (54)Jbe- 

 comes 



&mi + &112J, &mi + &122J 

 &2iii + &212J, &221I + &222J 

 whose elements are vectors. Expanding by the rule we have 

 [b m i + &112J, k>2ii + 6222J] — [&121I + &mj, &2ui + &212J] 



