线性代数(王定江)第1章答案.pdf
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线性代数(王定江)第1章答案.pdf
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-1-习习题题一一1利用对角线法则计算下列行列式:
(1)21231
(2)8.23=
(2)22.ababba=(3)112214141264215.311=+=(4)333(a)(ab)(ab)(ab)ababbabababababababab+=+33332233(333()2()ababababababab=+=+.2按自然数从小到大为标准次序,求下列各排列的逆序数:
(1)2467315
(2)7426315(3)n(n1)21(4)246(2n)135(2n1)解:
(1)000035210t=+=
(2)012135214t=+=
(1)(3)01
(1)2nntn=+=L
(1)(4)12Lnntn+=+=3写出5阶行列式含有因子412213aaa的项.解:
5534412213aaaaa和5435412213aaaaa.4计算下列各行列式:
(1)012312034720
(1)32
(2)11252001000+=.-2-
(2)01110001000
(1)12
(1)
(1)!
02000010nnnnnn+=LLLLLMMMML(3)21122121412412021202-4120241240724-101052010520015220-4011701170117rrrrrrrr32324120212021501170117174598500152200017857072400945rrrrrr+=+.(4)1001011011011010110101001abbacccddd=()
(1)1.abcdbdcdabcdabadcd=+=+5求解下列方程:
(1)0913251323221321122=xx解:
因为22222211231123100122301001312315013113323190133xxxxxx=22222231
(1)
(1)(933)3
(1)(4)33xxxxxx=+=,所以原方程的解为1x=,或2x=.-3-
(2)0111132323232=cccbbbaaaxxx解:
因为2323232311()()()()()()11xxxaaaxaxbxcabbcbcbbbccc=,所以原方程的解为xa=或xb=或xc=.6证明:
(1)322)(11122yxyyxxyxyx=+证明:
2222222222()22()22111xxyyxxyyxxyxyxyyxyxyxy+=+3223333()xxyxyyxy=+=.
(2)yxzxzyzyxbabzaybyaxbxazbyaxbxazbzaybxazbzaybyax)(33+=+证明:
axbyaybzazbxaybzazbxaxbyazbxaxbyaybz+axayazaxaybxaxbzazaxbzbxayazaxayazbyaybxaxaybxbyazaxayazaxbzazbyayazbybz=+-4-byayazbyaybxbybzazbybzbxbzazaxbzazbybzbxaxaybxbybxaxaybxaxbzbxbyaybxbybz+33axayazbybzbxxyzyzxayazaxaybxbyayzxbyxyazaxaybxbybzzxyxyz=+=+3333()xyzxyzxyzayzxbyzxabyzxzxyzxyzxy=+=+.(3))()()(222abbcaccbacbacbabaaccb+=+证明:
222222bccaababcbcacababcabcabcabc+=222111()()()()()abcabcabccacbbaabc=+=+.(4)11100121100001000001nnnnnnxxaxaxaxaxaaaaa=+LLMMMOMMLLL证明:
012101211000100001000100000100001nnnnnxxxxxaaaaaaaaaaxa=+LLLLMMLLLL-5-120121010000010000001nnnnnnnnnaaxaaxaaxaaxa=+LLMLLLLL
(1)1011()
(1)1nnnaax+=+LO
(1)11110
(1)()nnnnnnaxaxaxa+=+L1110.nnnnaxaxaxa=+L7计算下列行列式(3n):
(1)aaaaDn001000000100LLMMMMLL=解:
11100100000
(1)00000100100nnnnnaaaaaaDaaaaa+=+LLMMMMOOOLOL1
(1)122
(1)nnnnnnaaaaaa+=+=O.
(2)nDnLMMMMLLL333333333233331=解:
-6-3113331333333332332100100
(1)23333200000133320033nDnnn+=LLLLLLMMMMMMMOMOLL333301026(3)!
0023nn=O.(3))1(100000220000111321=nnnnDnLMMMMMLLL解:
123112311110001100002200022000001
(1)00010nnnnnDnnn+=LLLLLLLMMMMMMMMMMLL1110002200
(1)
(1)030020001nnnn+=LLLMMMMML11
(1)
(1)!
(1)
(1)!
(1)22nnnnnn+=.(4)1111)()1()()1(1111LLMMMLLnaaanaaanaaaDnnnnnnn=+解:
-7-11111111111
(1)()1
(1)()
(1)
(1)()1
(1)()111nnnnnnnnnnnnnnaaanaaanaaanDaaanaaanaaan+=LLLLLMMMMMMLLLL
(1)
(1)2201111111
(1)
(1)()
(1)()
(1)()nnnnijnnnnnnnaaanijaaanaaan+2)阶行列式中,值必为零的有(BD).(A)行列式主对角线上的元素全为零(B)上(或下)三角形行列式主对角线上有一个元素为零(C)行列式零元素个数多于n个(D)行列式非零元素个数小于n个7四阶行列式1001002200330044ababbaba的值等于(D).(A)12341234aaaabbbb(B)12341234aaaabbbb+(C))()12123434(aabbaabb(D))()23231414(aabbaabb解:
110010010000112200220022230000033300233140000044414000411ababababababbbabaababbababbaaa=()()231434232314141221bbbbaaaaaabbaabbaa=.8若347534453542333322212223212)(=xxxxxxxxxxxxxxxxxf,则方程0)(=xf的根的个数为(B).(A)1(B)2(C)3(D)4解:
212321012221222322101()333245353312244357434373xxxxxxxxxxfxxxxxxxxxxxxx=2100221002121(29)33121221714370xxxxxxxxxxxx=.-13-9如果122211211=aaaa,则方程组=+=+0,022221211212111bxaxabxaxa的解是(B).(A)2221211ababx=,2211112babax=(B)2221211ababx=,2211112babax=(C)2221211ababx=,2211112babax=(D)2221211ababx=,2211112babax=解:
11211211222222211112111222221222122bababababaxaaaabaaaaa=,11111211121222221112111221221222122abbaababbaxaaaaabaaaa=.二填空题:
1222cbacbabaaccb+=)()()(acbcabcba+解:
222222bccaabbcacababcabcabcabcabc+=222111()()()()().abcabcabcbacbcaabc=+=+2xxxxxxaaaxa+0000004321=31234()xxaaaa+解:
-14-1234123423434431234000000000000000().axaaaaaaaxaaaaaaxxxxxxxxxxxaaaa+=+34001030100211111=2解:
1111111111123412001110200
(1)2342.1030234003010040004=4若311151113=0,则=2,3,6解:
因为311311111151351(3)151113313113=111(3)062(3)(6)
(2)002=,所以2,3,6=.5设5213301211114321=D,则41424344AAAA+=0解:
41424344AAAA+-15-12341234123411110125012502103036500010111101230002=.6多项式333231232221131211)(axaxaxaxaxaxaxaxaxxf+=的次数最多是1次解:
111213212223313233()xaxaxafxxaxaxaxaxaxa+=+131212132322222333323233111113111211121321212321222122233131333132313233xxxxxaxaxxaaxxxxxaxaxxaaxxxxxaxaxxaaaxxaxaaaxaaaaxxaxaaaxaaaaxxaxaaaxaaa=+121311131112111213222321232122212223323331333132313233xaaaxaaaxaaaxaaaxaaaxaaaxaaaxaaaxaaa=+121311131112111213222321232122212223323331333132313233111111111aaaaaaaaaaaaaaaxaaaaaaaaaaaa=+.7若方程组123111111112axaxax=有无穷多个解,则a=2解:
因为方程组有无穷多个解,所以2112111111111121=
(2)11
(2)010112111001
(2)
(1)0aaaaaaaaaaaaaaaa+=+=+=+=求解得2a=或1a=.又因为当1a=时方程组无解,所以2a=.
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