A 22 故A 263 8 24124 (5)4=1≠0故A存在 而 0A4=0 412=-2A2=5432=0A12=0 A13=0A23=043=2A43=-3 A14=0A24=0434=-5A4=8 200 从而1|-25 00 002-3 00 58 (6)A 由对角矩阵的性质知A-17 −  = A A A 1 1 故                   − − − − − = − 4 1 12 1 24 5 8 1 0 3 1 6 1 2 1 0 0 2 1 2 1 1 0 0 0 1 A (5) A = 1  0 故 −1 A 存在 而 A11 = 1 A21 = −2 A31 = 0 A41 = 0 A12 = −2 A22 = 5 A32 = 0 A42 = 0 A13 = 0 A23 = 0 A33 = 2 A43 = −3 A14 = 0 A24 = 0 A34 = −5 A44 = 8 从而               − − − − = − 0 0 5 8 0 0 2 3 2 5 0 0 1 2 0 0 1 A (6)             = an a a A  0 0 2 1 由对角矩阵的性质知                   = − an a a A 0 1 1 0 1 2 1 1 