行列式習(xí)題解答講解

1(1)108311(1)108311b211c2(3)a283a：ab(2)b(2)cax((4)x+yababyx+yxx+yxy01abc(2)(2)cab111(3)(3)a2b2c2axyx+y(4)yx+y(4)x+yxy=x(x+y)y+yx(x+y)+(x+y)yxy3(x+y)3x3=3xy(x+y)y33x2y3y2xx3y3x3=2(x3+y3)2(1)1234；(2)4132；(3)3421；(4)2413；nnnn(n1) 232 4…個(gè)32 2個(gè)…42(2n)(2n2)2個(gè)…12341p2p3p4p1231234123411233244112334423|4124||2141| (1)||；(2)||；L0117」L5062」LL00-1d」41244-12-101202c-c1202(1)2310520c-7c1032-144301170010=122(-1)4+3103-144-1109910c+cc+1c=12-22300-c+1c31717144022306402230621423002121c-c3-1412503212506250r-r3r-r3-122r-r3412301400ce=adfbebfcfbc4111111a001b001cdr+ar2000b0a0c10dc+dc32a1cd0ac0:a211bb21x=(a3+b3)yzyzxbb)2(b+2)2(bc2(c+1)2(c+2)2(c+3)2d2(d+1)2(d+2)2(d+3)2xy1aa41bb2b41cc2c415x00anx…x………0……0…0…a…aa…n1n2n1cc=21cc31a2100…xa200…=xn+axn1+…+ax+a.1n1n0b2a20(1)3+1(1)3+1a=(ba)(ba)b2a22xyzyzxxyzxyyzxyzxzaz+bxxax+byyay+bzxyzyzxxyb3yzxxyxyzxya3yzzxb=c2zxyx+b3yzyzxc+(2c+1)zy(b+2)2(b+3)2(c(b+2)2(b+3)2(c+2)2(c+3)2(d+2)2(d+3)26cccccca2ab2bc2a2b2c2abcd+a2a1b2b1c211c4ca2a49a214a6ac6cb2b49b214b6b42+=0c4cc2c49c214c6cc9cd2d49d214d6d421000abacadaa2b2a2c2a2d2a2a4b4a4c4a4d4a4=b2a2c2a2d2a2b2(b2a2)c2(c2a2)d2(d2a2)111=(ba)(ca)(da)b+ac+ad+ab2(b+a)c2(c+a)d2(d+a)=(ba)(ca)(da)100=(ba)(ca)(da)(cb)(db)117………aa………aax1當(dāng)n=2時(shí),D==x2+ax+a,命題成立.2ax+a1221D=xn1+axn2+…+ax+a,n1n2n1nxD=xD+a(1)n+1…nn1n…10…01…0………1…x.00n1nijD=1an1a…ann…aD=2a…a…annan13annan1n(n1)3ij1an1a……aaaan1a…a…a…a2naa21n1a………aa2n2na3n=…aan1…a…annn(n1)=(1)1+2+…+(n2)+(n1)D=(1)2D8n(n1)11n1n(n1)11n1n(n1)n(n1)同理可證D=(1)2::=(1)2DT=(1)2D2a…a1nnnn(n1)n(n1)n(n1)D=(1)2D=(1)2(1)2D=(1)n(n1)D=D32(1)D=n(2)D=na1xaax……aaanaan1a(3)(3)D=n+1a1k1,a…………aa;…xn…1…1……………(4)D=2nan0cn．．nijbn0．a(chǎn)b110cd00dna=ij;ij9(6)D=n11…112…1…………11…n12n解(1)D=na00…010a0…0000a…00………………000…a0100…0a0a…000a…00…00…00…0………0…a)1a0a…0(n1)(n1)a+an=anan2=an2(a21)(n2)(n2)xaxD=axn…axa0…0a…0…………0a00…．．．．D=n00…0a0…0a0…0…………0a00…1n(n(n+1)D…n+1aaana1………………1…n(n+1)n+1n(n+1)n+1n+1i>j1n+1i>j1=n(i一j)n+1i>j1n+1i>j1D=an0cn0abcd110bndn即而得即而得an1anancn1000+(1)2n+1bnncn．a(chǎn)n1an10ab11cd0…11c．．011d0bn10dn10．0:0dnbn1b0dn1d0nn2n2nn2n2D=(adbc)D2nnnnn2n2D=nn(adbc)D2niiii2i=2DD=2a1c1b1d1111D=nn(adbc)2niiiii=1(5)a=ijijD=det(a)=nij0123…1012…2101…3210……n2…n3…n4………0rr2rr,…23…1…022…1…10……11(6)D=n…n1a100…000a20…0000a3…00…11…0002…111…11cc+c,c+c2131cc+c,…41…0000=(1)n1(n1)2n20……………11…cc,cc1223cc,…340000…n10n0000…an11111…1(1+a)(aa…a)n12n1a100…000a20…00n00a3…00n…0000…n200000…an200000…0n1414a120+…00200…000a23…00a230…0000a3…000a34…00………………000…n10000…n10000…an1n000…an1n=(1+a)(aa…a)+aa…aaa+…+aa…an12n112n3n2n23n12naii=1i(x+x+x+x=5,|x1+2+x=2,1231234(5x+6x=1,xxx=0,xxx=0,51123123112=111101230530218151013015101300100010001100125138=10001100121013D=15220121=100051120214140=20151=9=51020319910=0011D=22310052021291=4601140=010052012313810=0011D=3235001231131251242512D=41123123112D1D56001560(2)D=0156001500015220Dx=2=2,2DDxDx=3=3,3Dx=4=14D560015605D=5D6D01500016=5(5D6D)6D=19D30DD=11000165100065100065100065D+6510651=D+64=19D30+64=1507D=251000100010651000651011600056001560015065=5106015006160560160