Linear Algebra Solutions 50

# Linear Algebra Solutions 50 - 18 Suppose that At = −A...

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Unformatted text preview: 18. Suppose that At = −A, that A ∈ Mn×n (F ), where n is odd. Then det At = det(−A) det A = (−1)n det A = − det A. Hence (1 + 1) det A = 0 and consequently det A = 0 if 1 + 1 = 0 in F . 19. 1 r r r 1 1 r r 1 1 1 r 1 1 1 1 C4 → C 4 − C 3 C → C3 − C2 =3 C2 → C 2 − C 1 = 1 0 0 0 r 1−r 0 0 r 0 1−r 0 r 0 0 1−r = (1 − r)3 . 20. 1 a2 − bc a4 1 b2 − ca b4 1 c2 − ab c4 = = = R2 → R2 − R1 R3 → R3 − R1 = 1 a2 − bc a4 2 − ca − a2 + bc b4 − a4 0b 0 c2 − ab − a2 + bc c4 − a4 b2 − ca − a2 + bc b4 − a4 c2 − ab − a2 + bc c4 − a4 (b − a)(b + a) + c(b − a) (b − a)(b + a)(b2 + a2 ) (c − a)(c + a) + b(c − a) (c − a)(c + a)(c2 + a2 ) (b − a)(b + a + c) (b − a)(b + a)(b2 + a2 ) (c − a)(c + a + b) (c − a)(c + a)(c2 + a2 ) = (b − a)(c − a) b + a + c (b + a)(b2 + a2 ) c + a + b (c + a)(c2 + a2 ) = (b − a)(c − a)(a + b + c) 1 (b + a)(b2 + a2 ) . 1 (c + a)(c2 + a2 ) Finally 1 (b + a)(b2 + a2 ) 1 (c + a)(c2 + a2 ) = (c3 + ac2 + ca2 + a3 ) − (b3 + ab2 + ba2 + a3 ) = (c3 − b3 ) + a(c2 − b2 ) + a2 (c − b) = (c − b)(c2 + cb + b2 + a(c + b) + a2 ) = (c − b)(c2 + cb + b2 + ac + ab + a2 ). 53 ...
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