ON DOUBLE POLYADICS — THE LINEAR MATRIX EQUATION. 385 



If we assume A' = 1 we return to the case of Art. 13 and 14 and have 



(ij 1 ji, ji | ij)*s = ij*jiji*ij - ij*ijji*ji = +1 (104) 



and finally if we had merely four dyads ij, ji, ji, and ij, the scalar 



((ij, ji, Ji, ij))s = (105) 



since all such scalars vanish unless the antecedents are distinct. 



Again, we may form double and mixed products of double dyadics. 

 Assuming K = 1 the star product written singly will refer to the 

 dyadic factors. A double star will refer to antecedents and conse- 

 quents as in the usual double product, thus if A, M, B, N are dyadics 



AM**BN = A*BM*N (106) 



and we may have 



AM*: BN = A*BM: N (107) 



and 



AM:*BN = A: BM*N (108) 



while 



AM: :BN= A: BM:N (109) 



It is evident that, regarding the double dyadic as an element, we 

 could start a fresh investigation of each of these types of product with 

 respect to orthogonality, idemfactors, and so on ad inf. But we have 

 now ample tools for the applications it is desired to make. 



PART II. THE SOLUTION OF THE LINEAR MATRIX 



EQUATION. 



16. Transformation from Matrices to Double Dyadics. 



Any matrix A of order N is defined as a binary assemblage of ele- 

 ments dik where subscripts run from 1 to N. We add two matrices 

 by adding corresponding elements. We multiply by the rule 



(AB) ik = Xa i8 b 8k (s = 1, 2,- • •, N) (110) 



If Ai, Ai, • - •, Ah and B\, B 2 - • •, Bh and C are known matrices 

 while a; is a required matrix, the linear matrix equation may be written 



B{x) = A^Bi + A 2 xB 2 -\ h A h xB h = C. (Ill) 



