11.设A,B都是n阶方阵,求证 (1)(A+E)2=A2+2A+E,(A+E)(A-E)=A2-E (2)当AB≠BA时 (A+B)2≠A2+2AB+B2;(A+B)(A-B)≠A2-B2 证明:(1) (A+E)2=A(A+E)+E(+E)=42+A+A+E=A2+2A+E (A+E)(A-E)=(A+E)A-(A+E)E=42+A-A-E=A2-E (2)当AB≠BA时, (A+B)=(A+B)(A+B)=(A+B)A+(A+B)B=A+BA+AB+B-A+2AB+B (A+B)(A-B)=(A+B)A-(A+B)B=A2+BA-AB+B2+A2-B2 证毕。 12.若A=(B+E),证明A2=A当且仅当B2=E 证明 A2=A 1(B+E)2=(B+E) B2+2B+E=2B+2E分 B=E 18.设n阶方阵A满足:A2-A+E=0,证明A为可逆阵,并求A 证明:由A2-A+E=0可以知道A-A2=E,即A(E-A)=E。所以A为可逆阵 且A-1=(E-A) 19.设A为方阵,某正整数k>1,A=0,证明: (E-A)-1=E+A+42+…+A-11 11. A, B—¥nê ߶yµ (1)(A + E) 2 = A2 + 2A + Eß(A + E)(A − E) = A2 − E (2)AB 6= BAû (A + B) 2 6= A 2 + 2AB + B 2 ; (A + B)(A − B) 6= A 2 − B 2 y²µ(1) (A + E) 2 = A(A + E) + E(A + E) = A 2 + A + A + E = A 2 + 2A + E (A + E)(A − E) = (A + E)A − (A + E)E = A 2 + A − A − E = A 2 − E (2)AB 6= BAûß (A+B) 2 = (A+B)(A+B) = (A+B)A+(A+B)B = A 2+BA+AB+B 2 6= A 2+2AB+B 2 (A + B)(A − B) = (A + B)A − (A + B)B = A 2 + BA − AB + B 2 6= A 2 − B 2 y." 12. e A = 1 2 (B + E) ßy²A2 = A Ö= B2 = E" y²µ A2 = A ↔ 1 4 (B + E) 2 = 1 2 (B + E) ↔ B2 + 2B + E = 2B + 2E ↔ B2 = E 18.  n ê A ˜vµA2 − A + E = 0ßy² A èå_ ßø¶A−1" y²µdA2 − A + E = 0å± A − A2 = Eß=A(E − A) = E"§±A èå_ ß ÖA−1 = (E − A)" 19.  A èê ß,Í k > 1ßAk = 0ßy²µ (E − A) −1 = E + A + A 2 + · · · + A k−1