BOUNDARY PROBLEMS AND DEVELOPMENTS. 57 



Proof: Let Y(x) be any solution of equation (1) whatever. Then 

 the relation Y(x) = Y(x)$(x) defines a matrix <£(#), and substitut- 

 ing from this expression in equation (1) we have 



— F$ = A Y<t>, 

 dx 



d$ dY 

 i.e. Y— + —* = AY$. 



dY 

 But by hypothesis — = A 1 . 



d<$ 

 Hence Y-- + AY* = AY$, 



d$> 

 i.e. Y— = 0. 



dx dx 



Y_ 

 dx 



dx 



Y 



dx 



d<£> 

 Inasmuch as I Y(x) I =)= 0, it follows that — = 0, and $(x) = C. 



dx 



Q. E. D. 



A mere interchange of the roles played by the rows and the columns 

 of the matrices involved transforms the discussion carried out thus far 

 for equation (1) to the corresponding discussion for the equation 

 Y'= YA. The facts in the two cases may, therefore, be established 

 by precisely the same methods. 



The pair of related equations 



(1) Y' = AY, 



(2) Z' = - ZA, 



are said to be adjoint, either being the adjoint of the other. 



Theorem: If Y(x) is any matrix solution of equation (1), and Z(x) 

 is any matrix solution of equation (2), then 



Z(x) Y(x) = C. 

 Proof: From the relation 



A V-—Y 7— 



dx dx dx 



