000,所以 101 a=3I-aa 030 102 (b)因det(A)=9≠0,所以A可逆.A-1的计算如下: 201100 E 030010 05+B-B101/21/200 030 010 003/2-1/201 (1/3)*R2,(2/3)*R3, 101/21/200 01/30 001-1/302/3 (R1-(1/2)*F 1002/30-1/3 01/3 001-1/302/3 8)设AB是n×n矩阵,且A,B和A+B均可逆,证明:A-1+B-1也可逆,并求(A-1+B-1)”的 表达式 证:因A-1+B-1=A-1(A+B)B-1,且A,B,A+B均可逆,则有 A-1+B B(A+B)A (9)设n阶方阵A满足A3=2I(是单位阵),矩阵B=A2+2A+I,试证:B可逆,并求B-1的表达 式 证:BA2=(A2+2A+1A2=A4+2A3+A2,因A3=2L,所以 BA2=2A+4I+A2=B+3I →BA2-B=→B3(A2-I 因此B可逆,且 B 3 (10)设A是n阶方阵,且Ak=0(k≥2,求(I-A)-1解: (a) aaT =    −1 0 1    h −1 0 1 i =    1 0 −1 0 0 0 −1 0 1   , 所以 A = 3I − aaT =    2 0 1 0 3 0 1 0 2    (b) 因 det (A) = 9 6= 0, 所以 A 可逆. A−1 的计算如下: h A E i =    2 0 1 1 0 0 0 3 0 0 1 0 1 0 2 0 0 1    0.5∗R1,R3−R1 −−−−−−−−−→    1 0 1/2 1/2 0 0 0 3 0 0 1 0 0 0 3/2 −1/2 0 1    (1/3)∗R2,(2/3)∗R3, −−−−−−−−−−−−→    1 0 1/2 1/2 0 0 0 1 0 0 1/3 0 0 0 1 −1/3 0 2/3    (R1−(1/2)∗R3, −−−−−−−−−→    1 0 0 2/3 0 −1/3 0 1 0 0 1/3 0 0 0 1 −1/3 0 2/3    A−1 = 1 3    2 0 −1 0 1 0 −1 0 2    (8) 设 A,B 是n×n矩阵, 且A, B 和A+B 均可逆, 证明: A−1+B−1 也可逆, 并求 ￾ A−1 + B−1
−1的 表达式. 证: 因 A−1 + B−1 = A−1 (A + B) B−1 , 且 A, B, A + B 均可逆, 则有  A−1 + B −1 −1 = B (A + B) −1 A (9) 设 n阶方阵A满足A3 = 2I(I是单位阵), 矩阵 B = A2 + 2A + I, 试证: B可逆, 并求B−1的表达 式. 证: BA2 = ￾ A2 + 2A + I
A2 = A4 + 2A3 + A2 , 因 A3 = 2I, 所以 BA2 = 2A + 4I + A2 = B + 3I ⇒ BA2 − B = 3I ⇒ B  1 3  A2 − I   = I 因此 B 可逆, 且 B −1 = 1 3  A2 − I  (10) 设 A 是 n 阶方阵, 且 Ak = 0(k ≥ 2), 求 (I − A) −1