《线性代数》(第二版)智能教学系统 习题解答 第五章A组题

次型二习题五〔〕(x,x,x)2xx4xx2xx〔〕f〔〕f=；21221231323(x,x,x,x)2xx2xx2xx2xx=。123412131434202x1f(x,x,x)=(x,x,x)011x,1231232210x3202011210(x,x,x)f。123〔0111100010011010x1xf(x,x,x,x)=(x,x,x,x)2，12341234x3x401111000(x,x,x,x)f1001。1234101012010112202112112211〔〕；〔〕。12121012221120122X(x,x,x)〕设T12311221x121f(x,x,x)(x,x,x)02x=X=T1231232x31222x2xxxxx4xx=2123。121323X(x,x,x,x)〔T123412010x12112211xf(x,x,x,x)(x,x,x,x)=X=2T121x123412341032x1140122xxxx2xxxxxxxx=2224。1213232434f(x,x,x)2xx4xx4xx〔〕〔〕〔〕=；21221231223f(x,x,x)2xx2xx=；1231223f(x,x,x)x2x3x4xx4xx=。2222123131223f(x,x,x)123220212020=。A220E)212(2)(=，=20221，4A，。1232(2E)X01。T11(E)X02(2)。T24(4E)X03(2,2。T3,,将1231122(,,)T，333123111212(,,)T，333221221(,,)T，33333令122333212,,)==，333123232133200010004=diag(-2,1,4)=。122xyyy323132112333yyyx，323231212xyyy333312(x,x,x)f1232yy4y21222。3〔111212120000002==，220021112222y2y222。3〔231233222000501=，=，3330012213332y5yy21222。3(x,x,x)x2x5x2xx2xx6xx〔〕f〔〕f〔〕f=；212223123121323(x,x,x)2xx4xx=；1231213(x,x,x)4xx2xx2xx=。123121323x1f(x,x,x)=x2+2x(xx)+(xx)2－(xx)2+2x6xx5x++22231231123232323(xxx)x4xx4x=2+2++2，3123223x2f(x,x,x)=(xxx)2+x2+4xx+4x2=(xxx)2+(x2x)2。123123223312323111(y,y,y)(x,x,x)012000设=T=T=，123123y令Y=BXy2。221〔〕令110yxyyx112211yy110yxx=。2122xy3x001y333f(x,x,x)=2(yy)(yy)+4(yy)y12312121232y2y4yy4yy=－++21221323yy)yy)=2－2，1323101011000(y,y,y)(z,z,z)设=T=T=，1231232z2z令Z=BY2。221〔4z4zz21222。3(x,x,x)x2x4x2xx4xx〔〕f〔〕f〔〕f=；2122231231223(x,x,x)x3xx2xx2xx6xx=；212223123121323(x,x,x)4xx2xx6xx=。123121323(x,x,x)f123110122A=。024100012024110010001100100000110110012024122024A=。100100112012E010010001001001令112012=，001f(x,x,x)yy=212。2123〔f(x,x,x)y4yy=。212223123〔1f(x,x,x)2yy6y=。2122232123f(x,x,x)5x5x2xx6xx6xx=212223123121323cf(x,x,x)12351315333c=。A。f(x,x,x)5x5x3x2xx6xx6xx即=212223123121323513153333=，4y9y222。3nf(x,x,,x)xxxxxx=122n12n22n1nn1xyy10010110112nxyy222n111yyxn1，=，nnPxyy11n1nn1xyy01102n122n1100xyy2n12n(x,x,,x)f122nyy212y2n1y2。n2n(x,x,x)2x3x3x2x8f=(a>0)21222312323fy2y5y2a21223y2y5yf2和。21222320003a0a3=。故A200E)03(2)(69a)a=。=220a35。1(E)X01(0。T12(2E)X02。T25(5E)X03。T3,,将123111,)T，22123111T，22111(0,,)T，223301011,,)0==。2211231022(x,x,x)5x6x4x4xx4xx〔〕f〔〕f〔〕f=；2122231231223(x,x,x)10x2xx8xx24xx28xx=；212223123121323(x,x,x,x)xx4x7x6xx4xx4xx=2122232412341314232xx4xx。2434(x,x,x)f123520262024。52026202452，26即A(x,x,x)f123442。1442，A1f(x,x,x,x)〔1234103201223242。2227103012324，tf(x,x,x)x4xxx10xx6xx〔〕〔〕〔〕=；212223123121323f(x,x,x)xx5xx2xx4xx=；212223123121323f(x,x,x)2xxx2xxx=。2122231231223f(x,x,x)1231t5t43=。5311t5t43tt，1t4t=2，=2t45314t02tt02tf(x,x,x)〔1231t1t12125=。1t=1t2tt，2t14t0。5f(x,x,x)〔123210t11=。2t002由2111t2A|=1，22t2。B为nBA及B)ABTTTTBCC12CCCC，TT1122CCCCCCCCC)CCC，TTTTTT221122122122即AAATAATAATT又AAATB为nBA及BfXAXgXBX，T，Tx,x,,x12nX0X0T，T。X(AB)XXAXX0，故=T=T+TX(AB)X=TAA和AA为A1-**AA(A)(A)AA1即11TT1(A)(A)AA*即**TT*ACCT。ACC)CC)QQ1T111T=T，11A|A|A|A|CC)C[C]PP1T=T，*=111T=1AA1C)(C)1T=1T=A故A和A1*A为mAA为mT〔为nTA为mAT为n=mn组O,XO,有OXA〕0。TTTAAT〔A为mAT为nnAO,XO,有=OXO,有T00XA〕A0。TTTTTAATf(x,x,,x