第二讲补充题解答 (1)设矩阵 2-20 1-10 A 求A2-4B2-2BA+2AB 解: 原式=(A2-2BA)+(2AB-4B2)=(A-2B)A+2(A-2B)B=(A-2B)(A+2B) 000 4-40 000 8-44000 36-124 111 151 1,b=「bb2…bn17,计算A=ab3,B=b3a及A 解: a1b1 a1b2 b a2b1 a2b2 a2b ab= B=ba=b1 b2 1b1 alby agb1 a2b2 a2b k= abTabT.abT 4)“/ (3)设矩阵 A 2-24 求A(n为正整数) 解:因 所以 1 (4)设A2=A,B2=B,(A+B)2=A+B,证明:AB=0第二讲补充题解答 (1) 设矩阵 A =    2 −2 0 4 −2 2 1 3 1    , B =    1 −1 0 −2 1 −1 0 1 0    求A2 − 4B2 − 2BA + 2AB. 解: 原式 = ￾ A2 − 2BA
+ ￾ 2AB − 4B2
= (A − 2B) A + 2 (A − 2B) B = (A − 2B) (A + 2B) =    0 0 0 8 −4 4 1 1 1       4 −4 0 0 0 0 1 5 1    =    0 0 0 36 −12 4 5 1 1    (2) 设 a = h a1 a2 · · · an iT , b = h b1 b2 · · · bn iT , 计算 A = abT , B = b T a 及 Ak . 解: A = abT =       a1 a2 . . . an       h b1 b2 · · · bn i =       a1b1 a1b2 · · · a1bn a2b1 a2b2 · · · a2bn . . . . . . . . . anb1 anb2 · · · anbn       B = b T a = h b1 b2 · · · bn i       a1 a2 . . . an       = Xn i=1 aibi Ak = abT abT · · · abT =  b T a k−1 abT = nX−1 i=1 aibi !k−1       a1b1 a1b2 · · · a1bn a2b1 a2b2 · · · a2bn . . . . . . . . . anb1 anb2 · · · anbn       (3) 设矩阵 A =    1 −1 2 2 −2 4 −1 1 −2    求An (n为正整数). 解: 因 A =    1 2 −1    h 1 −1 2 i , 所以, An =   h 1 −1 2 i    1 2 −1      n−1 A = (−3)n−1 A (4) 设 A2 = A, B2 = B, (A + B) 2 = A + B, 证明: AB = 0