22 



A IM AK N.-KSS. 



M.-N. Kl. 



(6) <r a, a. 



We will expand this determinant according to the /»-rowed determinants 

 of the last p rows. Let p columns be determined l)y l\, /-j . . . . kp, such 

 that /-J <C ^2 "^ • • • • "^ ^P- '^''^^' r>l''''T coiiiplcmcnt of the /»-rowed deter- 

 minant in question containing these columns is obtained by striking out 

 from the set of unit vectors all the C/.-^ and forming the (;/ — /»l-rowed de- 

 terminant of die rest. With regard to the sign of the algebraic complement 

 we observe that the sum of the indices of the last p rows in (6) is 



{n — p + 1 + ti)p 



= np 



p\p 



According to Laplace's theorem, the space complement of ^p a^ a., . . . . <Xp 

 is equal to the sum of all y j terms of the form 



(A) (- 1) 



1 



e^t,- 



C;, 



0» 





^\ k^ 



ap k- 



each being a (special) polyadic of the (;/ — /))''^ order. 



In order to obtain the left member of the .équation 151 we take the 

 indeterminate product of t» by each of these terms iAi and then deriving 

 the space complement of each of the polyadics, obtained in this way, of 

 order h — p + 1 . But eacli of these polyadics can be expanded into a sum 

 of;/ others by putting ü = ZjVj and then multiplying distributi\-ely. Neglecting 



the scalar factor we thus all together get [ ] '^ terms of the following form : 



(B) 



.^ k; 



c„ 



Ci 



e,; 



and our final task is to derive the space complement of each of these, i. e. 



(C) 



e,X" 



ei 



.e>t, 



e„ 



^1 c„ 



then multiply by the corresponding scalar and sum. 



