高等代数与解析几何习题解答2.pdf

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1、1Ù1ª§1NC1.OeN=´üN,=´÷N,=´_N?(1)f:C−→Ra7−→

2、a

3、−→(2)VAÛm,e½üþ,Nσ:V−→V−→−→−→−→−→−→a7−→σ(a)=a−2(a·e)e(3)f:N−→Nn7−→n+1):(1)ü,÷,Ø_.(2)ü,÷,_.(3)ü,÷,Ø_.2.f8ÜS8ÜS′N,g8ÜS′8ÜS′′N,y²:(1)XJgfüN,KfüN;(2)XJgf÷N,Kg÷N.y²:(1)é?¿s1,s2∈S,Xf(s1)=f(s2),Kgf(s1)=gf(s2).Ïg

4、f´üN,s1=s2,lf´üN.(2)é?¿s′′∈S′′,Ïgf´÷N,3s∈S,¦gf(s)=s′′,ls′=f(s)∈S′,¦g(s′)=s′′,g´÷N.3.f8ÜS8ÜS′_N,g8ÜS′8ÜS′′_N,Kgf8ÜS8ÜS′′_N,(gf)−1=f−1g−1.y²:ÏfgÑ_,¤±f−1g−1´8ÜS′′SN,−1−1−1−1−1−1(fg)(gf)=f(gg)f=f1S′f=ff=1S,−1−1−1−1−1−1(gf)(fg)=g(ff)g=g1S′g=gg=1S′′.§2Ûó5

5、·51·§2Ûó51.:!!12345671234567p=,q=.35476212517643¦pq,p−1qp,¿rp,q©OL«¤é¦È.!!12345671234567):pq=,p−1qp=,56312747541326p=(13)(34)(47)(25)(56),q=(12)(25)(56)(64)(47)(73).(¡üL«ªØ).2.Oe_Sê!,¿(½ÙÛó5:!12345614563287(1);(2);34561268547213!254398671(3);631279854!135···2n−1246

6、···2n(4).246···2n135···2n−1):(1)_Sê´8,ó.(2)_Sê´20,ó.(3)_Sê´11,Û.(4)_Sê´n,Ûó5ÓnÛó5.3.Oeü_Sê,¿(½ÙÛó5:(1)5317246;(2)384576192;(3)246813579;(4)987654321.):(1)_Sê´9,Û.(2)_Sê´18,ó.(3)_Sê´10,ó.(4)_Sê´36,ó.4.(½i9k,¦(1)237i864k5¤óü;(2)469k1i752¤Ûü.):(1)i=1,k=9.(2)i=8,k=3.5.

7、Oeü_Sê:·52·1Ù1ª(1)135···(2n−1)(2n)(2n−2)···642;(2)(2n+1)(2n)(2n−1)···321.):(1)n(n−1).(2)n(2n+1).6.®p_Sêa,¦p−1_Sê.):a.7.®üx1x2···xn_Sêa,¦xnxn−1···x2x1_Sê.n(n−1)):Ïx1x2···xn_Sê+xnxn−1···x2x1^Sê=,2xnxn−1···x2x1_Sê=x1x2···xn^Sê,¤±xnxn−1···x2x1_n(n−1)Sê=−a.2n(n

8、−1)∗8.y²:é?ÛØLêk,73_Sêkn2ü.y²:ék^êÆ8B{.Äk,k=1,213···n_Sê1;n(n−1)b½(Øék−1¤ák≤,=3nü2i1i2···in(1)Ù_Sêk−1,K73j

9、X,·Iéü(½A=.k=1,Ï?¿_Sê1üѱdü123···nüê,_Sê1üêuP1(n)=n−1.du?nüѱdn−1üVên.k=2,_Sê2üdeãª:(a)12···n−2n−1−→12···n−3nn−2n−1;(b)i1i2···in−1(_Sê1)−→i1i2···in−2nin−1;(c)i1i2···in−1(_Sê2)−→i1i2···in−1n.§3Ý·53·¤±_Sê2üêP2(n)=1+P1(n−1)+P2(n−1).ddP2(n−1)=1

10、+P1(n−2)+P2(n−2),.............................P2(3)=1+P1(2)+P2(2),(n−2)(n+1)¤±P2(n)=(n−2)+(n−1)+···+2=.