392 HITCHCOCK. 



m, = AhB" S - 2A"sB" s + A" s Bh (129) 



Since the matrices are of the second order, A" s and B"s are the de- 

 terminants of their respective matrices; hence on substituting, 



w . 2 = (a n + o 22 ) 2 (611622 — &12621) — 2(an022 — 012021) (611622 — 6)2621) 



+ («ll022 — 0] 2 2 l) (6ll + 6 2 2) 2 



= (a 2 n+ a 2 22 ) (611622- 612621) + (6 2 n+ 6V) {a n a- 22 - a n a n ) 



-\- 2011022611622 — 2012021612621 (130) 



Again 



6m3 = ^ 3 sB 3 s - 3^ S B 5 (.4 2 ) S (5 2 ) S + 2(A*) S (B 3 ) S (131) 



by the rule. But A 3 = A s A 2 - A" S A hence 



(A*)s = A s s - 3A S A" S (132) 



by (128), and similarly for B. Substituting values in (131) from (128) 

 and (132) to get rid of scalars of powers of matrices we find 



m 3 = A s B s A"sB"s (133) 



= (On -f" 22 ) (611 + 622) («lia22 — ai202l) (611622 — 612621) 



and finally 24m 4 = A* S B* S - 6A* S B* S (A>) S (B% + 3U 2 ) 2 5 (5 2 ) 2 S 



+ SAsB s (A s ) s (B*)s - Q(A*) s (B*)s (134) 



But A 4 = /1^4 3 - A" S A- hence 



(^4) g = ^ _ iA - s A"s + 2.4'V (135) 



so that by eliminating the scalars of powers from (134) 



mi = A"s>B" s - (140) 



Now in the present example d-C = A-CB\ 3 C = A S CB\ etc. Hence 



x = O-'C = 4 „l B „ 2 [AsBsA" 8 B" 8 C - (A*bB"b- 2A" 8 B" s 



S + A" s B-s)ACB + AsBsA-CB- - A S CB*] (141) 



Simplifying by the aid of the Hamilton-Cayley equations for A and B, 



—^—rlAsBsC - B S AC - A S BC + ACB] 



A S B s 



.v = 



= -TT^r^sI - A)C[B S I -B} = A-*CB~\ 

 A sB s 



checking with the known solution. 



