线性代数课后题答案第五版全

1第一章行列式1第二章矩阵及其运算13第三章矩阵的初等变换与线性方程组30第四章向量组的线性相关性46第五章相似矩阵及二次型73第一章行列式1利用对角线法则计算下列三阶行列式1381141102解38114110224301111801321814124816442BACACBCBA解BACACBCBAACBBACCBABBBAAACCC3ABCA3B3C33222111CBACBA解222111CBACBABC2CA2AB2AC2BA2CB2ABBCCA4YXYXXYXYYXYX2解YXYXXYXYYXYXXXYYYXXYXYYXY3XY3X33XYXYY33X2YX3Y3X32X3Y32按自然数从小到大为标准次序求下列各排列的逆序数11234解逆序数为024132解逆序数为44143423233421解逆序数为532314241,2142413解逆序数为32141435132N1242N解逆序数为21NN321个52542个7274763个2N122N142N162N12N2N1个6132N12N2N22解逆序数为NN1321个52542个2N122N142N162N12N2N1个421个62642个2N22N42N62N2N2N1个3写出四阶行列式中含有因子A11A23的项解含因子A11A23的项的一般形式为1TA11A23A3RA4S其中RS是2和4构成的排列这种排列共有两个即24和423所以含因子A11A23的项分别是1TA11A23A32A4411A11A23A32A44A11A23A32A441TA11A23A34A4212A11A23A34A42A11A23A34A424计算下列各行列式171100251020214214解71100251020214214010014231020211021473234CCCC34114310221101414310221101401417172001099323211CCCC22605232112131412解2605232112131412260503212213041224CC041203212213041224RR0000003212213041214RR3EFCFBFDECDBDAEACAB解EFCFBFDECDBDAEACABECBECBECBADFABCDEFADFBCE411111111144DCBA100110011001解DCBA100110011001DCBAABARR10011001101021DCAAB101101111201011123CDCADAABDCCCDADAB1111123ABCDABCDAD15证明11112222BBAABABAAB3证明1112222BBAABABA00122222221213ABABAABAABACCCCABABABAAB2212221321ABAABABAB32YXZXZYZYXBABZAYBYAXBXAZBYAXBXAZBZAYBXAZBZAYBYAX33证明BZAYBYAXBXAZBYAXBXAZBZAYBXAZBZAYBYAXBZAYBYAXXBYAXBXAZZBXAZBZAYYBBZAYBYAXZBYAXBXAZYBXAZBZAYXA5BZAYYXBYAXXZBXAZZYBYBYAXZXBXAZYZBZAYXA22ZYXYXZXZYBYXZXZYZYXA33YXZXZYZYXBYXZXZYZYXA33YXZXZYZYXBA33303213213213212222222222222222DDDDCCCCBBBBAAAA证明2222222222222222321321321321DDDDCCCCBBBBAAAAC4C3C3C2C2C1得5232125232125232125232122222DDDDCCCCBBBBAAAAC4C3C3C2得022122212221222122222DDCCBBAA4444422221111DCBADCBADCBAABACADBCBDCDABCD证明6444422221111DCBADCBADCBA0001111222222222ADDACCABBADDACCABBADACAB111222ADDACCABBDCBADACAB00111ABDBDDABCBCCBDBCADACAB11ABDDABCCBDBCADACABABACADBCBDCDABCD51221100000100001AXAAAAXXXNNNXNA1XN1AN1XAN证明用数学归纳法证明当N2时2121221AXAXAXAXD命题成立假设对于N1阶行列式命题成立即DN1XN1A1XN2AN2XAN1则DN按第一列展开有1110010001111XXAXDDNNNNXDN1ANXNA1XN1AN1XAN因此对于N阶行列式命题成立6设N阶行列式DDETAIJ,把D上下翻转、或逆时针旋转90、或依副对角线翻转依次得7NNNNAAAAD1111111112NNNNAAAAD11113AAAADNNNN证明DDDNN21211D3D证明因为DDETAIJ所以NNNNNNNNNNAAAAAAAAAAD2211111111111111331122111121NNNNNNNNAAAAAAAADDNNNN21122111同理可证NNNNNNAAAAD11111212DDNNTNN212111DDDDDNNNNNNNN12121221311117计算下列各行列式DK为K阶行列式1AADN11,其中对角线上元素都是A未写出的元素都是0解AAAAADN00010000000000001000按第N行展开8111000000000000100001NNNAAA1121NNNAAANNNNNAAA22111ANAN2AN2A212XAAAXAAAXDN解将第一行乘1分别加到其余各行得AXXAAXXAAXXAAAAXDN0000000再将各列都加到第一列上得AXAXAXAAAANXDN00000000001XN1AXAN131111111111NAAANAAANAAADNNNNNNN解根据第6题结果有NNNNNNNNNNAAANAAANAAAD1111111111211此行列式为范德蒙德行列式911211111JINNNNJAIAD11211JINNNJI112112111JINNNNNJI11JINJI4NNNNNDCDCBABAD11112解NNNNNDCDCBABAD11112按第1行展开NNNNNNDDCDCBABAA000011111111100011111111112CDCDCBABABNNNNNNN再按最后一行展开得递推公式D2NANDND2N2BNCND2N2即D2NANDNBNCND2N2于是NIIIIINDCBDAD222而111111112CBDADCBAD所以NIIIIINCBDAD125DDETAIJ其中AIJ|IJ|解AIJ|IJ|0432140123310122210113210DETNNNNNNNNADIJN04321111111111111111111112132NNNNRRRR15242321022210022100021000011213NNNNNCCCC1N1N12N2116NNAAAD11111111121,其中A1A2AN0解NNAAAD11111111121NNNNAAAAAAAAACCCC1000010001000100010000113322121321111312112111000011000001100001100001NNNAAAAAAAANIINNAAAAAAAA1111131211211000001000000100000100000111121NIINAAAA8用克莱姆法则解下列方程组101123253224254321432143214321XXXXXXXXXXXXXXXX12解因为14211213513241211111D142112105132412211151D284112035122412111512D426110135232422115113D14202132132212151114D所以111DDX222DDX333DDX144DDX2150650650651655454343232121XXXXXXXXXXXXX解因为6655100065100065100065100065D150751001651000651000650000611D114551010651000650000601000152D70351100650000601000051001653D39551000601000051000651010654D1321211000051000651000651100655D所以66515071X66511452X6657033X6653954X6652124X9问取何值时齐次线性方程组0200321321321XXXXXXXXX有非零解解系数行列式为1211111D令D0得0或1于是当0或1时该齐次线性方程组有非零解10问取何值时齐次线性方程组010320421321321321XXXXXXXXX有非零解解系数行列式为101112431111132421D13341213132123令D0得02或3于是当02或3时该齐次线性方程组有非零解第二章矩阵及其运算1已知线性变换143213321232113235322YYYXYYYXYYYX求从变量X1X2X3到变量Y1Y2Y3的线性变换解由已知221321323513122YYYXXX故3211221323513122XXXYYY321423736947YYY321332123211423736947XXXYXXXYXXXY2已知两个线性变换32133212311542322YYYXYYYXYYX32331221133ZZYZZYZZY求从Z1Z2Z3到X1X2X3的线性变换解由已知221321514232102YYYXXX321310102013514232102ZZZ321161109412316ZZZ所以有3213321232111610941236ZZZXZZZXZZZX3设111111111A150421321B求3AB2A及ATB15解1111111112150421321111111111323AAB2294201722213211111111120926508503092650850150421321111111111BAT4计算下列乘积1127075321134解127075321134102775132271112374496352123321解12332113223110321312解2131223132111221263214242041312101314311041216解2041312101314311041265208765321332313232212131211321XXXAAAAAAAAAXXX解321332313232212131211321XXXAAAAAAAAAXXXA11X1A12X2A13X3A12X1A22X2A23X3A13X1A23X2A33X3321XXX322331132112233322222111222XXAXXAXXAXAXAXA5设3121A2101B问1ABBA吗解ABBA因为6443AB8321BA所以ABBA2AB2A22ABB2吗解AB2A22ABB2因为5222BA522252222BA2914148但43011288611483222BABA27151610所以AB2A22ABB23ABABA2B2吗解ABABA2B217因为5222BA1020BA906010205222BABA而718243011148322BA故ABABA2B26举反列说明下列命题是错误的1若A20则A0解取0010A则A20但A02若A2A则A0或AE解取0011A则A2A但A0且AE3若AXAY且A0则XY解取0001A1111X1011Y则AXAY且A0但XY7设101A求A2A3AK解12011011012A1301101120123AAA101KAK8设001001A求AK解首先观察180010010010012A2220020123232323003033AAA43423434004064AAA545345450050105AAAKAKKKKKKKKKK00021121用数学归纳法证明当K2时显然成立假设K时成立，则K1时，001001000211211KKKKKKKKKKKKAAA1111110010211KKKKKKKKKK由数学归纳法原理知19KKKKKKKKKKKA000211219设AB为N阶矩阵，且A为对称矩阵，证明BTAB也是对称矩阵证明因为ATA所以BTABTBTBTATBTATBBTAB从而BTAB是对称矩阵10设AB都是N阶对称矩阵，证明AB是对称矩阵的充分必要条件是ABBA证明充分性因为ATABTB且ABBA所以ABTBATATBTAB即AB是对称矩阵必要性因为ATABTB且ABTAB所以ABABTBTATBA11求下列矩阵的逆矩阵15221解5221A|A|1故A1存在因为122522122111AAAAA故|11AAA12252COSSINSINCOS解COSSINSINCOSA|A|10故A1存在因为COSSINSINCOS22122111AAAAA20所以|11AAACOSSINSINCOS3145243121解145243121A|A|20故A1存在因为214321613024332313322212312111AAAAAAAAAA所以|11AAA17162132130124NAAA0021A1A2AN0解NAAAA0021由对角矩阵的性质知NAAAA1001121112解下列矩阵方程112643152X解126431521X1264215380232212234311111012112X解1111012112234311X03323210123431131325381223101311022141X解11110210132141X21011013114212121010366121041114021102341010100001100001010X解11010100001021102341100001010X01010000102110234110000101020143101213利用逆矩阵解下列线性方程组13532522132321321321XXXXXXXXX22解方程组可表示为321153522321321XXX故0013211535223211321XXX从而有001321XXX205231322321321321XXXXXXXXX解方程组可表示为012523312111321XXX故3050125233121111321XXX故有305321XXX14设AKOK为正整数证明EA1EAA2AK1证明因为AKO所以EAKE又因为EAKEAEAA2AK1所以EAEAA2AK1E由定理2推论知EA可逆且EA1EAA2AK1证明一方面有EEA1EA另一方面由AKO有EEAAA2A2AK1AK1AK23EAA2AK1EA故EA1EAEAA2AK1EA两端同时右乘EA1就有EA1EAEAA2AK115设方阵A满足A2A2EO证明A及A2E都可逆并求A1及A2E1证明由A2A2EO得A2A2E即AAE2E或EEAA21由定理2推论知A可逆且211EAA由A2A2EO得A2A6E4E即A2EA3E4E或EAEEA3412由定理2推论知A2E可逆且34121AEEA证明由A2A2EO得A2A2E两端同时取行列式得|A2A|2即|A|AE|2故|A|0所以A可逆而A2EA2|A2E|A2|A|20故A2E也可逆由A2A2EOAAE2EA1AAE2A1E211EAA又由A2A2EOA2EA3A2E4EA2EA3E4E所以A2E1A2EA3E4A2E134121AEEA2416设A为3阶矩阵21|A求|2A15A|解因为|11AAA所以|521|52|111AAAAA|2521|11AA|2A1|23|A1|8|A|1821617设矩阵A可逆证明其伴随阵A也可逆且A1A1证明由|11AAA得A|A|A1所以当A可逆时有|A|A|N|A1|A|N10从而A也可逆因为A|A|A1所以A1|A|1A又|1111AAAAA所以A1|A|1A|A|1|A|A1A118设N阶矩阵A的伴随矩阵为A证明1若|A|0则|A|02|A|A|N1证明1用反证法证明假设|A|0则有AA1E由此得AAAA1|A|EA1O所以AO这与|A|0矛盾，故当|A|0时有|A|02由于|11AAA则AA|A|E取行列式得到|A|A|A|N若|A|0则|A|A|N1若|A|0由1知|A|0此时命题也成立因此|A|A|N119设321011330AABA2B求B25解由ABA2E可得A2EBA故321011330121011332211AEAB01132133020设101020101A且ABEA2B求B解由ABEA2B得AEBA2E即AEBAEAE因为01001010100|EA所以AE可逆从而201030102EAB21设ADIAG121ABA2BA8E求B解由ABA2BA8E得A2EBA8EB8A2E1A18AA2E18AA2A18|A|E2A182E2A14EA14DIAG212121,1,21DIAG42DIAG12122已知矩阵A的伴随阵8030010100100001A且ABA1BA13E求B解由|A|A|38得|A|2由ABA1BA13E得ABB3AB3AE1A3AEA11A261126213AEAE103006060060000660300101001000016123设P1AP其中1141P2001求A11解由P1AP得APP1所以A11AP11P1|P|31141P1141311P而11111120012001故31313431200111411111A6846832732273124设APP其中111201111P511求AA85E6AA2解85E62DIAG1158DIAG555DIAG6630DIAG1125DIAG1158DIAG120012DIAG100APP1|1PPP1213032220000000011112011112111111111425设矩阵A、B及AB都可逆证明A1B1也可逆并求其逆阵证明因为A1ABB1B1A1A1B127而A1ABB1是三个可逆矩阵的乘积所以A1ABB1可逆即A1B1可逆A1B11A1ABB11BAB1A26计算30003200121013013000120010100121解设10211A30122A12131B30322B则2121BOBEAOEA222111BAOBBAA而4225303212131021211BBA90343032301222BA所以2121BOBEAOEA222111BAOBBAA9000340042102521即3000320012101301300012001010012190340410252127取1001DCBA验证|DCBADCBA解41001200210100101002000021010010110100101DCBA而01111|DCBA故|DCBADCBA2828设22023443OOA求|A8|及A4解令34431A22022A则21AOOAA故8218AOOAA8281AOOA1682818281810|AAAAA464444241422025005OOAOOAA29设N阶矩阵A及S阶矩阵B都可逆求11OBAO解设43211CCCCOBAO则OBAO4321CCCCSNEOOEBCBCACAC2143由此得SNEBCOBCOACEAC2143121413BCOCOCAC所以OABOOBAO11121BCOA解设43211DDDDBCOA则29SNEOOEBDCDBDCDADADDDDDBCOA4231214321由此得SNEBDCDOBDCDOADEAD42312114113211BDCABDODAD所以11111BCABOABCOA30求下列矩阵的逆阵12500380000120025解设1225A2538B则5221122511A8532253811B于是850032000052002125003800001200251111BABA24121031200210001解设2101A4103B2112C则1111114121031200210001BCABOABCOA30411212458103161210021210001第三章矩阵的初等变换与线性方程组1把下列矩阵化为行最简形矩阵1340313021201解340313021201下一步R22R1R33R1020031001201下一步R21R32010031001201下一步R3R2300031001201下一步R33100031001201下一步R23R3100001001201下一步R12R2R1R3311000010000012174034301320解174034301320下一步R223R1R32R1310031001320下一步R3R2R13R20000310010020下一步R12000031005010312433023221453334311解12433023221453334311下一步R23R1R32R1R43R11010500663008840034311下一步R24R33R4522100221002210034311下一步R13R2R3R2R4R23200000000002210032011434732038234202173132解34732038234202173132下一步R12R2R33R2R42R21187701298804202111110下一步R22R1R38R1R47R141000410002020111110下一步R1R2R21R4R300000410001111020201下一步R2R3000004100030110202012设987654321100010101100001010A求A解100001010是初等矩阵E12其逆矩阵就是其本身33100010101是初等矩阵E121其逆矩阵是E121100010101100010101987654321100001010A2872212541000101019873216543试利用矩阵的初等变换求下列方阵的逆矩阵1323513123解1000100013235131231010110012004101231012002110102/102/30232/102/1102110102/922/70032/102/11002110102/33/26/7001故逆矩阵为210212112332672121023211220102334解100001000010000112102321122010230010030110000100122059401210232120104301100001001200110012102321106124301100001001000110012102321106126311101022111000010000100021106126311101042111000010000100001故逆矩阵为1061263111010421141设113122214A132231B求X使AXB解因为132231113122214,BA412315210100010001R35所以4123152101BAX2设433312120A132321B求X使XAB解考虑ATXTBT因为134313231221320,TTBA411007101042001R所以4171421TTTBAX从而4741121BAX5设101110011AAX2XA求X解原方程化为A2EXA因为101101110110011011,2AEA011100101010110001所以01110111021AEAX6在秩是R的矩阵中,有没有等于0的R1阶子式有没有等于0的R阶子式解在秩是R的矩阵中可能存在等于0的R1阶子式也可能存在等于0的R阶子式36例如010000100001ARA30000是等于0的2阶子式010001000是等于0的3阶子式7从矩阵A中划去一行得到矩阵B问AB的秩的关系怎样解RARB这是因为B的非零子式必是A的非零子式故A的秩不会小于B的秩8求作一个秩是4的方阵它的两个行向量是1010011000解用已知向量容易构成一个有4个非零行的5阶下三角矩阵0000001000001010001100001此矩阵的秩为4其第2行和第3行是已知向量9求下列矩阵的秩并求一个最高阶非零子式1443112112013解443112112013下一步R1R2443120131211下一步R23R1R3R1564056401211下一步R3R237000056401211矩阵的2秩为41113是一个最高阶非零子式2815073131213123解815073131223123下一步R1R2R22R1R37R115273321059117014431下一步R33R20000059117014431矩阵的秩是271223是一个最高阶非零子式302301085235703273812解02301085235703273812下一步R12R4R22R4R33R402301024205363071210下一步R23R1R32R10230114000016000071210下一步R216R4R316R2380230100000100007121000000100007121002301矩阵的秩为3070023085570是一个最高阶非零子式10设A、B都是MN矩阵证明AB的充分必要条件是RARB证明根据定理3必要性是成立的充分性设RARB则A与B的标准形是相同的设A与B的标准形为D则有ADDB由等价关系的传递性有AB11设32321321KKKA问K为何值可使1RA12RA23RA3解32321321KKKA210011011KKKKKR1当K1时RA12当K2且K1时RA23当K1且K2时RA312求解下列齐次线性方程组102220202432143214321XXXXXXXXXXXX解对系数矩阵A进行初等行变换有39A2122111212113/410013100101于是4443424