(1)A= 25 41=5,A21=2×(-1),A12=2×(-1),42=1 AA 11 5-2 Au A 故A 2 (2)4=1≠0故A存在 Au= cos A2=sin 8 A12=-sin 8 A2= cos 从而 cos SIn n0 cose (3)4=2,故A存在 A21=2431=0 而 A 13 A2=6 22 210 故 167-1 1000 1200 (4)A 2130 1214 24A21=A31 A,,=24 42=12 33 84 44 6 10 120 A2=(-1)3230=-1213=(-1)4210=-12 120 00 A14=(-1)213=3 A23=(-1)2 0 12 12 100 00 A24=(-1)°213=-5A3=(-1)7120 26 (1)         = 2 5 1 2 A A = 1 A11 = 5, A21 = 2  (−1), A12 = 2  (−1), A22 = 1         − − =        =  2 1 5 2 12 22 11 21 A A A A A −  = A A A 1 1 故         − − = − 2 1 5 2 1 A (2) A = 1  0 故 −1 A 存在 A11 = cos A21 = sin A12 = −sin A22 = cos 从而         − = −     sin cos cos sin 1 A (3) A = 2 , 故 −1 A 存在 A11 = −4 A21 = 2 A31 = 0 而 A12 = −13 A22 = 6 A32 = −1 A13 = −32 A23 = 14 A33 = −2 故 −  = A A A 1 1           − − − − − = 16 7 1 2 1 3 2 13 2 1 0 (4)               = 1 2 1 4 2 1 3 0 1 2 0 0 1 0 0 0 A A = 24 A21 = A31 = A41 = A32 = A42 = A43 = 0 A11 = 24 A22 = 12 A33 = 8 A44 = 6 12 1 1 4 2 3 0 1 0 0 ( 1) 3 A12 = − = − 12 1 2 4 2 1 0 1 2 0 ( 1) 4 A13 = − = − 3 1 2 1 2 1 3 1 2 0 ( 1) 5 A14 = − = 4 1 2 4 2 1 0 1 0 0 ( 1) 5 A23 = − = − 5 1 2 1 2 1 3 1 0 0 ( 1) 6 A24 = − = − 2 1 2 1 1 2 0 1 0 0 ( 1) 7 A34 = − = −