给青春涂上奋斗的底色 生命的底色

线性代数》同济大学版课后习题详解111abca2b2c2第一章行列式1利用对角线法则计算下列三阶行列式bccaabacbacb(ab)(bc)(ca)(1)014830418(4)013248abcbcacab(2)abcab3abc3216ccba33abc111abca2b2c2(3)(1)(1)814bbbaaa31)(14)18(1)cccxyxy(4)yxyxxyxy333x(xy)yyx(xy)(xy)yxy(xy)x323333xy(xy)y3xyxyx2(x3y3)2按自然数从小到大为次序求下列各排列的逆序数(1)1234解逆序数为0(2)4132解逆序数为441434232(3)3421解逆序数为532314241,21(4)2413解逆序数为3214143(5)13(2n1)24(2n)解逆序数为n(n21)32(1个)1211215254(2个)727476(3个)(2)n1)2(2n1)4(2n1)6(2n(6)13(2n1)(2n)(2n2)2解逆序数为n(n1)32(1个)5254(2个)(2n1)2(2n1)4(2n1)6(2n)42(1个)6264(2个)(2n)2(2n)4(2n)6(2n)(2n2)(n13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(1)a11a23a3ra4s其中rs是2和4构成的排列这种排列共有两个即24和42所以含因子a11a23的项分别是1)a11a23a32a44(1)1a11a23a32a44a11a23a32a441)a11a23a34a42(1)2a11a23a34a42a11a23a34a424计算下列各行列式4202021254101412120105201144207c2c4c37c341100411210310214(1)43411210310214c2c1c312c39017910021714011224236112023150200423411212312r2r40202423611202315c2c41122423611202315解bd231112423020cdaede112bb22aa0bbaa2a1c1c1c2c2b2b2b2a2a1bbaedeaccdbabdaba)aba)2a2b2abazxyyzxxyzbxbyzxyaaazxybbbyzxaaayzxbbbxyzaaa明xyzbbbzxyaaazxybbbyzxaaayzxbbbxyzaaa00101c1b1a10)001a1cbab110102ar100101c1b1a10解xyzbbbzxyaaabzbxbyyzaayzbbxbyzxyaaayzaaxyaadba11c2c301acba1()1xyzbbbzxyaaazxyyzx2bzxyzxybbbyzxaaaxyz2axyzb3yzxyzxzxyzxyxyzdaba()12bb22a2a1xx1dbb)(dba)241dd1c2c41bb241aa)((((()241dd1cc241bb24241aa111bacada0b(ba)c(ca)d(da)0b2(b2a2)c2(c2a2)d2(d2a2)111(ba)(ca)(da)b2(bba)c2(cca)d2(dda)11a)0cb0c(cb)(cba)d(d1a)d(dba)b3xyzyzxzxyxyzyzxzxy22223333abcd((((2222))))2222abcd((((111abcd((((2222abcd(ba)(ca)(d2222c3333555(a(b(cd(2a2b2c22222)2))22)333abcd((((222)))111abcd((((2222abcd(ba)(ca)(da)(cb)(db)c(c1b=(ab)(ac)(ad)(bc)(bd)(cd)(abcd)2a2b2c111abcd22222222abcd222222221111abcd22222222abcdx01x010000(5)nnxa1x000x1anan1an2a2xa1an1xann(n1)证明D1D2(1)2DD3D证明因为Ddet(aij)所以当n2时D2x1a2xa1x2a1xa2假设对于(n1)阶行列式命题成立即Dn1xn1na1x2an2xan1则Dn按第一列展开有证明用数学归纳法证明命题成立10DnxDn1an(1)n1x111x1nn1xDn1anxa1xan1xan因此对于n阶行列式命题成立6设n阶行列式Ddet(aij),把D上下翻转、或逆时针旋转90、或依副对角线翻转依次得an1anna1nannD1a11a1nD2a11an1anna1nan1a11D3D2an1annD1a11a1n(1)n1(1)n2(1)12(n2)同理可证n(n1)(1)2a11a1nn(n1)D3(1)2D2a11a1n(1)n1an1anna21a2na11a1na21a2nan1anna31a3nn(n1)(n1)D(1)2Dan1annn(n1)(1)2DTn(n1)(1)2Dn(n1)n(n1)(1)2(1)2D(1)n(n1)D7计算下列各行列式(Dk为k阶行列式)2)112)11Dn(1)2na11a(1)Dn其中对角线上元素都是a未写出的元素都是0（按第n行展开）xaaDnaxaaax将第一行乘(1）分别加到其余各行得xaaaaxxa00Dnax0xa0ax000xa(2)得再将各列都加到第一列上解00001x(n1)aaaa0000xa00a000Dn0xa000a0(n1)(n1)0000xa1aa(1)n[x(n1)a](xa)n1(n1)(n1)(3)(1)n11)nnnn2n22aaaa(a1)an(a1)n(an)nan1(a1)n1(an)n1a1an116题结果有Dn1根据第(n2)(nDn1n(n1)(1)2an1an(a(a1)n11)n(a(an)n1n)n此行列式为范德蒙德行列式(4)D2nann(n1)Dn1(1)2[(ai1)(aj1)]n1ij1n(n1)(1)2[(ij)]n1ij1n(n1)n(n1)1(1)2(1)2(ij)n1ij1(ij)n1ij1bna1b1c1d1an1bn10a1b1anc1d1cn1dn1000dn0an1bna1b1(1)2n1bnc1d1cn1dncn0再按最后一行展开得递推公式D2nandnD2n2bncnD2n2即D2n(andnbncn)D2n2ndn于是D2n(aidibici)D2i2anbnDa1b1D2c1d1a1d1b1c1cna1c1db11dn(按第行展开)n所以D2n(aidibici)i1(5)Ddet(aij)其中aij|ij|;解aij|ij|11000000a0a2aa1a0aanc2cc1c1234nnnn3210210110120123D1112131n1aaaa00010001001001101100nan00001a1ann11000001001001000nan0n10000n)1a1n8用克莱姆法则解下列方程组111111111111111100022111a111n(c1c1n)c2c1a11111a21111anDn200200465x25x1x13x3xx3x22x2x22x3xx1x12x3x14511112311256600060065065165105100为D因D527500060065065165101000D15221125410006500651065101000151000521123522112D3x3D2x21D1x1D4所以D30001000000607030561150006510001065106510051000D506516510510021210000065x1507x21145x3703x4395x1665665665665x4212665x1x2x309问取何值时齐次线性方程组x2x30有非零2x2x30解？解系数行列式为1D112于是当0或10非零解？1时该齐次线性方程组有非零解问取何值时解系数行列式为(1)x12x24x30齐次线性方程组2x1(3)x2x30有x1x2(1)x30124134D31111101(1(1)3()32(13)4(1)2(1)(3))23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已知线性变换x12y12y2y3x23y1y25y3x33y12y23y3求从变量x1x2x3到变量y1y2y3的线性变换解由已知x1x2x221y1315y2323y2y12211x1749y1故y2315x2637y2y2323x3324y3y17x14x29x3y26x13x27x3y33x12x24x32已知两个线性变换x12y1y3y13z1z2x22y13y22y3y22z1z3x34y1y25y3y3z23z3求从z1z2z3到x1x2x3的线性变换解由已知x1201y1201310z1x2232y2232201z2x3415y2415013z3613z11249z210116z3x16z1z23z3所以有x212z14z29z3x310z1z216z311123设A11B12求3AB2A及ATB11105111123111解3AB2A3111124211111105111105811113223056211117202901114292111123058ATB1111240561110512904计算下列乘积4317(1)12325701431747321135解123217(2)23165701577201493(2)(123)213解(123)2(132231)(10)12(3)1(12)322(1)2224解1(12)1(1)121233(1)3236131(4)21400121134131402131解214001267811341312056402a11a12a13x1(x1x2x3)a12a22a23x2a13a23a33x3x1(a11x1a12x2a13x3a12x1a22x2a23x3a13x1a23x2a33x3)x2x3a11x1a22x2a33x32a12x1x22a13x1x32a23x2x312105设AB问5设A13B12问(1)ABBA吗?解ABBA3412因为ABBA所以ABBA4638(2)(AB)2A22ABB2吗?解(AB)2A22ABB2因为AB2225a11a12a13x1(5)(x1x2x3)a12a22a23x2a13a23a33x3解(AB)2但A22ABB2222814525142938411681081234所以(AB)2A22ABB222(3)(AB)(AB)A2B2吗?解(AB)(AB)A2B2因为AB22250201(AB)(AB)25010609A3A2AAk10k11031而A2B2381041134故(AB)(AB)A2B26举反列说明下列命题是错误的2817108设A01求Ak001010221A201010220000002解首先观察(2)若A2A则A0或AE解取A1010则A2A但A(3)若AXAY且A0则XY解取0且AE100011111101则AXAY且A0但XY7设A101求A2A3Ak解A210101110213A4A5AA2A3A450236k1k(k1)k210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Akk1ABBA证明(即AB是对称矩阵充分性因为ATAAB)T(BA)TATBTABBTB且ABBA所以用数学归纳法证明当k2时显然成立假设k时成立，则k1时，k(k1)11必要性因为ATABTAB(AB)TBTATBA求下列矩阵的逆矩阵且(AB)TAB所以Ak1AkA2kk1k1010(1)k1(k1225|A|故A1存在因为1)k1k1(k2(k1)kk1A*A11A12A21A225221由数学归纳法原理知k1k(k1)1)k1k1k2(2)A1cossin|1A|A*sincosAk002kk1kcossinsincos|A|故A1存在因为A*9设AB为n阶矩阵，证明因为ATA所以(BTAB)TBT(BTA)TBTATB从而BTAB是对称矩阵A为对称矩阵，证明BTAB也是对称矩阵BTAB所以A1A11A12cossinsincos|1A|A*cossinsincos121(3)342541121解A342|A|20故A1存在因为541A112a1a2解下列矩阵方程an12A3A3A12A2A2A12A1A1A41326210所以A11A*1331|A|221671a1a20(4)(a1a2an0)0ana10解Aa2由对角矩阵的性质知0an(1)(2)25X13X142223082X2111011411203122528(1)(1)(3)解X12431101211011216610111123012104HYPERLINK\l"bookmark713"\o"CurrentDocument"010100143(4)100X001201HYPERLINK\l"bookmark717"\o"CurrentDocument"00101012010101143100解X100201001001120010123x11225x22351x331x1123111故x222520x335130x11从而有x20x30(2)2x1x23x12x2x323x315x301031401200100113利用逆矩阵解下列线性方程组x12x13x12x22x25x23x35x3x3解方程组可表示为解方程组可表示为HYPERLINK\l"bookmark331"\o"CurrentDocument"111x12HYPERLINK\l"bookmark368"\o"CurrentDocument"213x21HYPERLINK\l"bookmark280"\o"CurrentDocument"325x301x1111125故x221310x332503x15故有x20x3314设AkO(k为正整数)证明(EA)1EAA2Ak证明因为AkO所以EAkE又因为EAk(EA)(EAA2Ak1)所以(EA)(EAA2Ak1)E由定理2推论知(EA)可逆且(EA)1EAA2Ak1证明一方面有E(EA)1(EA)另一方面由AkO有E(EA)(AA2)A2Ak1(Ak1Ak)(EAA2Ak1)(EA)故(1EA)1(EA)(EAA2Ak1)(EA)两端同时右乘(EA)1就有(1EA)1(EA)EAA2Ak115设方阵A满足A2A2EO证明A及A2E都可逆并求A1及(A2E)1证明由A2A2EO得A2A2E即A(AE)2E或A12(AE)E由定理2推论知A可逆且A112(AE)由A2A2EO得A2A6E4E即(A2E)(A3E)4E或(A2E)41(3EA)E11由定理2推论知(A2E)可逆且(A2E)1(3EA)证明由2A2A2EO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A2A2EOA(AE)2EA1AAE)2A1EA11(A2E)又由A2A2EO(A2E)A3(A2E)4E(A2E)(A3E)4E所以(A2E)1(A2E)(A3E)4(A2E)1(A2E)141(3EA)解因为A11A*所以|A||(2A)15A*||21A15|A|A1||21A125A1|2A1|(2)3|A1|8|A|1821617设矩阵A可逆证明其伴随阵A*也可逆且(A*)1(A1)*证明由A11A*得A*|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(1A*)1|A|11A又A|A11|(A1)*|A|(A1)*所以|A|11111(A*)1|A|1A|A|1|A|(A1)*(A1)*18设n阶矩阵A的伴随矩阵为A*证明(1)若|A|0则|A*|0(2)|A*||A|n1证明(1)用反证法证明假设|A*|0则有A*(A*)1E由此得AAA*(A*)1|A|E(A*)1O所以A*O这与|A*|0矛盾，故当|A|0时有|A*|0(2)由于A11|A|A*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|n103319设A110ABA2B求B123解由ABA2E可得(A2E)BA故23311033033B(A2E)1A11011012312112311010120设A020且ABEA2B求B101解由ABEA2B得2(AE)BA2E即(AE)B(AE)(AE)001因为|AE|01010所以(AE)可逆从而100201BAE03010221设Adiag(121)A*BA2BA8E求B解由A*BA2BA8E得(A*2E)BA8EB8(A*2E)1A18[A(A*2E)]18(AA*2A)18(|A|E2A)8(2E2A)114(EA)14[diag(212)]4diag(12,1,11,22diag(121)100122已知矩阵A的伴随阵A*100308且ABA1BA13E求B解由|A*||A|38得|A|2由ABA1BA13E得ABB3A111B3(AE)1A3[A(EA1)]1A00003(E12A*)16(2EA*)100110600006006060030123设P1AP解由P1AP其中P得APP14101102所以A11A=P11P1.求A1114A11141033273127321102111168368433111124设APP其中P10211115求(A)A8(5E6AA2)解()8(5E62)diag(1158)[diag(555)diag(6630)diag(1125)]diag(1158)diag(1200)12diag(100)(A)P()P11P()P*|P|1111002222102000303111000121P|3P*1111P1114P3110211111411111125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A1(AB)B1B1A1A1B1而A1(AB)B1是三个可逆矩阵的乘积所以A1(AB)B1可逆即A1B1可逆(A1B1)1[A1(AB)B1]1B(AB)1A322010100011102210100算计11327A221A1设B22033则A1OEA2EOB1B2A1OA1B1B2A2B2而A1B1B21201312120331010AB0101CD10100101|A|B110|C||D110C取ABD解而ABCD|A||B|C||D10012验证|A||B|C||D0002000101010120A2B2201323034039A1OEA2EB1OB2A1OA1B1B2A2B210002100242435252428设A令A1113322010100011102210100243524210100A83443O3443A1OOA2A1OOA2A2AO18求|A8|及A42022OA28|A8||A18||A28||A1|8|A2|81016OEEnODD2BADCDD1BADCA1C1B1AOB123DDDDOED1D2D3D4AD1EnAD2OCD1BD3CD2BD4OB得AC此得由11A1003008210520以30(1CB求下列矩阵的逆阵OEEnO322851B1212525A1A设A解382532851B(1)OA1(1)BO解设OABOCC13CC42则A4A14O054OOA24O240O262429设n阶矩阵A及s阶矩阵B都可逆求4CCABC3CAB2CCC1CAOOB341CCC1C2DD1D1D由此得AC3AC4BC1BC2所以OBAO(2)AOCB1解设A设COBnEOOE0030082105200030022501201B1A1BA1把下列矩阵化为行最简形矩阵1021(1)20313043解(2)1000120021301214101221C12则11000111200AO1A1O2130CBB1CA1B11214112121801216524000131120014第三章矩阵的初等变换与线性方程组10212031(下一步r2(2)r1r3(3)r1)304313(下一步r2(1)r3(2))010210013（下一步001010210013（下一步000310210013（下一步000110210010（下一步0001r3r2)r33)r23r3)r1(2)r2r1r3)(3)23(3)23(2)100000001001103013(下一步1(下一步(下一步r2r32(3)r1r3(2)r1)r2r13r2)1323132335344422310113解23r4(5))1000(4)213221解31323100032233223353431113435108744224861042222200323432343101(下一步r23r138(下一步r2610r32r14)r33r1)(3)3222(下一步r13r2r3200740374(下一步r12r20r33r22r2)001001001001011104(下一步12(下一步0889077811r22r1r38r1r47r1)10112010001012(下一步r1r2r2(1)r4r3)44101E(12(1))010001HYPERLINK\l"bookmark380"\o"CurrentDocument"010123101A100456010001789001102020111(下一步r2r3)0014000010202011030001400000A求362514100110A001001设2010解100是初等矩阵E(12)其逆矩阵就是其本身101010是初等矩阵E(12(1))其逆矩阵是4561123078903试利用矩阵的初等变换321(1)315323321100解315010323001012100213001452HYPERLINK\l"bookmark192"\o"CurrentDocument"1012201782求下列方阵的逆矩阵1/23007/229/2HYPERLINK\l"bookmark370"\o"CurrentDocument"2~010112HYPERLINK\l"bookmark711"\o"CurrentDocument"10011/201/21007/62/33/2~0101120011/201/201410300000121132121010021620311`110100000121010010001124~01000101~00101136000121610124101故逆矩阵为06131610412134(1)设A221B2求X使AXB3113解因为41213r100102(A,B)22122~010153311300112432212112231002376112222000001010100112023222010103000001215329214014103000001211321210故逆矩阵为3010(2)3解0101~0~001~0~0011所以XA1B101512(2)设A求X使XA所以X(A2E)1A考虑ATXTBT因为所以从而(AT,BT)XT设A6式?解的r阶子式在秩是在秩是的矩阵中,有没有等于0的r的矩阵中1阶子式?有没有等于0的r阶子可能存在等于0的r1阶子式也可能存在等于0(AT)1BTBA1原方程化为(A2E)X(A2E,A)例如R(A)3是等于0的2阶子式是等于0的3阶子式AX2X求X从矩阵A中划去一行得到矩阵R(A)R(B)解这是因为B的非零子式必是因为8求作一个秩是4的方阵B问AB的秩的关系怎样?A的非零子式故A的秩不会小于B的秩它的两个行向量是0)(11000)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(10101000011000此矩阵的秩为4其第2行和第3行是已知向量310（1）11213443102解1121（下一步r1r2）13441121~3102（下一步r23r1r3r1）134411~45（下一步r3r2）459求下列矩阵的秩并求一个最高阶非零子式3231（2）2113701832132解21313（下一步r1r2r22r1r37r1）70518344~71195（下一步r33r2）21332715344~7119000矩阵的秩是2327是一个最高阶非零子式2121832307325810321121~0465000031矩阵的秩为24是一个最高阶非零子式1121837解23075（下一步r12r4r22r4r33r4）32580r1r4r2r4r3r4103200121703635(下一步r23r1r32r1)0242010320012000000103012170000100000103201032001217000010000017016(下一步r216r4r316r2)01420075580320矩阵的秩为3700是一个最高阶非零子式10设A、B都是mn矩阵证明A~B的充分必要条件是R(A)R(B)证明根据定理3必要性是成立的充分性设R(A)R(B)则A与B的标准形是相同的设A与B的标准形为D则有由等价关系的传递性A~DD~B有A~B123k11设A12k3问k为何值可使k23(1)R(A)1(2)R(A)2(3)R(A)3kk当当当r~)A)(R(k3331时时2k2)A122k(RAk11k且时2且12求解下列齐次线性方程组(1)于是x12x1x22x3x2x3x40x40102x12x2x32x40解对系数矩阵A进行初等行变换有kk1(k1)(k2)11211010A2111~013122120014/3x1x2x3x4故方程组的解为x43x4x4x443343kx1x2x3x4（k为任意常数）5x110x2x35x40解对系数矩阵A进行初等行变换有12111201A3613~0010510150000x12x2x4于是x2x2x30x4x4故方程组的解为x12x2x3x4(2)3x16x2x33x402100k1x1x2x3x100k2为任意常数）于是2x13x2x35x403x1x22x37x404x1x23x36x40x12x24x37x40对系数矩阵A进行初等行变换有2315100A3127~~010AA41360011247000(3)解0001x10x20x30x40故方程组的解为00004x25x37x403x23x32x4011x213x316x402x2x33x40解对系数矩阵A进行初等行变换有010100~72631533114312132477171701701422338B3110~01011341138006于是R(A)2而R(B)3故方程组无解x1313137x31137x4于是1920x2x31731270x4x3x3x4x4故方程组的解为313x11717x2k11920k(k1k2为任意常数)x3117217x41013求解下列非齐次线性方程组:4x12x2x32(1)3x11x22x31011x13x28解对增广矩阵B进行初等行变换有2x3yz4x2y4z53x8y2z134xy9z6解对增广矩阵B进行初等行变换有2314102112450112B~B38213~000041960000于是yz2zzx2即yk12(k为任意常数)z12xyzw1(3)4x2y2zw22xyzw1x2z1解对增广矩阵B进行初等行变换有21111B422122111112z12y12yzxyz11/21/201/2~00010000006757ww1797zz1757zxyz120121210k1xyzw1200k2为任意常数）2xyzw1(4)3x2yz3w4x4y3z5w2解对增广矩阵B进行初等行变换有21111/71/76/7B3134~5/79/75/713520000k217571k1xyzw1797067570（k1k2为任意常数）14写出一个以22xc131c20401为通解的齐次线性方程组解根据已知可得x122x234x32c131c204HYPERLINK\l"bookmark463"\o"CurrentDocument"x3401与此等价地可以写成x12c1c2x2x3x43c14c2c1c2或x12x3x4x23x34x4或x12x3x40x23x34x40(3)要使方程组有有无穷多个解必须R(A)R(B)3故)(1)20(1因此当)(2)0(11时方程组有无穷多个解16非齐次线性方程组2x1x2x32x12x2x3x1x22x32因此2时方程组无解这就是一个满足题目要求的齐次线性方程组x1x2x31x1x2x3x1x2x32(1)有唯一解(2)无解(3)有无穷多个解?111解B11112r112~011(1)00(1)(2)(1)(1)215取何值时非齐次线性方程组2112121解B121~0112(1)11223000(1)(2)要使方程组有解必须(1)(2)0即12当1时当取何值时有解？并求出它的解10110110~21111221B方程组解为(1)要使方程组有唯一解必须R(A)3因此当1且2时方程组有唯一解.(2)要使方程组无解必须R(A)R(B)故(1)(2)0(1)(1)20x1x31或x2x3x1x2x3x31x3x3x111即x2k10(k为任意常数)x310当2时2112101B1212~01111240000方程组解为x1x2x3x32或2或x1x2x3x32x32x3x112即x2k12(k为任意常数)x310(2)x12x22x3117设2x1(5)x24x322x14x2(5)x31问为何值时此方程组有唯一解、无解或有无穷多解?并在有无穷多解时求解2221解B254224512542~011100(1)(10)(1)(4)要使方程组有唯一解必须R(A)R(B)3即必须(1)(10)0所以当1且10时方程组有唯一解.要使方程组无解必须R(A)R(B)即必须(1)(10)0且(1)(4)0所以当10时方程组无解.要使方程组有无穷多解必须R(A)R(B)3即必须(1)(10)0且(1)(4)0所以当1时方程组有无穷多解此时，增广矩阵为1221B~00000000方程组的解为x1x2x3x1x2x2x3x3121或x2k1k200(k1k2为任意常数)x31018证明R(A)1的充分必要条件是存在非零列向量a及非零行向量bT使AabT证明必要性由R(A)1知A的标准形为10(1,0,,0)0000即存在可逆矩阵P和Q使1PAQ0(1,0,,0)01AP10(1,0,,0)Q101令aP10bT(100)Q1则a是非零列向量0是非零行向量且AabT充分性因为a与bT是都是非零向量所以A是非零矩阵从而R(A)因为1R(A)R(abT)min{R(a)R(bT)}min{11}1所以R(A)119设A为mn矩阵证明(1)方程AXEm有解的充分必要条件是R(A)m程AXEm有解的充分必要条件是R(A)m(2)方程YAEn有解的充分必要条件是R(A)n证明注意方程YAEn有解的充分必要条件是ATYTEn有解由(1)ATYTEn有解的充分必要条件是R(AT)n因此，方程YAEn有解的充分必要条件是R(A)R(AT)n20设A为mn矩阵证明若AXAY且R(A)n则XY证明由AXAY得A(XY)O因为R(A)n由定理9方程A(XY)O只有零解即XYO也就是XYbT1设v1(110)Tv2(0及3v12v2v3解v1v2(110)T(01(101101)T(101)T3v12v2v33(110)T(31203(012)T2设3(a1a)2(a2a)5(a3a2(101510)TTa3(41解由3(a1a)2(a2a)5(a3a16(3a162a25a3)11)T1)Tv3(340)T求v1v22(011)T(340)T3121430210)Ta)求a11)a)整理得其中a1T(2513)T第四章向量组的线性相关性证明由定理7方程AXEm有解的充分必要条件是R(A)R(AEm)而|Em|是矩阵(AEm)的最高阶非零子式故R(A)R(AEm)m因此方16[3(2,5,1,3)T2(10,1,5,10)T5(4,1,1,1)T](14)3已知向量组Aa1(012Bb1(2113)T证明B组能由A组线性表示证明由3)Ta2(32)Tb2(0012)Ta3(2301)T211)Tb3(441但A组不能由B组线性表示11301r11301r11301(B,A)02211~02211~02211111100221100000知R(B)R(BA)2显然在A中有二阶非零子式故R(A)2又R(A)R(BA)2所以R(A)2从而R(A)R(B)R(AB)因此A组与B组等价03(A,B)210132041024~r03110113023124220461578179475225511153602010100031241615704135知R(A)R(AB)3由所以B组能由A组线性表示2112414~r0103021022~r0111~0001000知R(B)2因为R(B)R(BA)所以A组不能由B组线性表示已知向量组Aa1(011)Ta2(10)Bb1(101)b2(121)Tb3(321)T证明A组与B组等价证明由5已知R(a1a2a3)2R(a2(1)a1能由a2a3线性表示(2)a4不能由a1a2a3线性表示证明(1)由R(a2a3a4)3知a2无关又由R(a1a2a3)2知a1a2表示(2)假如a4能由a1a2a3线性表示a4能由a2a3线性表示a2a3线性表示a3a4)3证明a3a4线性无关故a2a3也线性a3线性相关故a1能由a2a3线性则因为a1能由a2a3线性表示故从而a2a3a4线性相关矛盾因此a4不能由a16判定下列向量组是线性相关还是线性无关(1)((2)(2131)T(210)T(130)T(140)T(01)2)解(1)以所给向量为列向量的矩阵记为A因为HYPERLINK\l"bookmark376"\o"CurrentDocument"121r121r121HYPERLINK\l"bookmark378"\o"CurrentDocument"A314~077~011HYPERLINK\l"bookmark239"\o"CurrentDocument"101022000所以R(A)2小于向量的个数从而所给向量组线性相关(2)以所给向量为列向量的矩阵记为B因为210|B|340220002所以R(B)3等于向量的个数从而所给向量组线性相无关问a取什么值时下列向量组线性相关？a1(a11)Ta2(1以所给向量为列向量的矩阵记为1)Ta3(1由a)T有a1b1(12)Tb1(01)T,a2b2(24)T(00)T(24)而a1b1a2b2的对应分量不成比例是线性无关的10举例说明下列各命题是错误的(1)若向量组a1a2am是线性相关的则a1可由a2am线性表示解设a1e1(1000)a2a3am0则a1a2am线性相关但a1不能由a2am线性表示|A|a111a1aa(a1)(a1)知当a1、0、1时R(A)3此时向量组线性相关8设a1a2线性无关a1ba2b线性相关求向量表示的表示式解因为a1ba2b线性相关故存在不全为零的数11(a1b)2(a2b)0解Ab用1由此a1a2线性2使a112a112(1)a22设cca1(1c)a2cR设a1线性相关？试举例说明之解不一定例如当a1(12)T,a2线性相关b1a2(2b2也线性相关4)T,b1(1问a1b11)T,a2b2是否一定b2(00)T时(2)若有不全为0的数12m使1a1mam1b1mbm0成立则a1a2am线性相关,b1b2bm亦线性相关解有不全为零的数12m使1a1mam1b1mbm0原式可化为1(a1b1)m(ambm)0取a1e1b1a2e2b2amembm其中e1e2em为单位坐标向量则上式成立而a1a2am和b1bm均线性无关(3)若只有当121a1才能成立则a1a2性无关解由于只有当12由1a1成立所以只有当121(a1b1)2(a2成立因此a1b1a2b2mam1b1mbm0am线性无关,b1b2m全为0时等式mam1b1mbm0m全为0时等式b2)m(ambm)0ambm线性无关m全为0时等式bm亦线取a1a2们满足以上条件但a1am0取b1bm为线性无关组则它a2am线性相关(4)若a1a2am线性相关,b1b2bm亦线性相关则有不全为0的数12m使1a1mam01b1mbm0同时成立解a1(10)Ta2(20)Tb1(03)Tb2(04)T1a12a201221b12b201(3/4)2120与题设矛盾11设b1a1a2b2a2a3b3a3a4b4a4a1证明向量组b1b2b3b4线性相关证明由已知条件得a1b1a2a2b2a3a3b3a4a4b4a1于是a1b1b2a3b1b2b3a4b1b2b3b4a1从而b1b2b3b40这说明向量组b1b2b3b4线性相关12设b1a1b2a1a2bra1a2ar且向量组a1a2ar线性无关证明向量组b1b2br线性无关证明已知的r个等式可以写成111(b1,b2,,br)(a1,a2,,ar)011上式记为b1b2BAK因为|K|10br线性无关K可逆所以R(B)R(A)r从而向量组13求下列向量组的秩,并求一个最大无关组(1)a1(1214)Ta2(9100104)Ta3(2428)T解由1921921221004r0820r00(a1,a2,a3)110201900004480320000知R(a1a2a3)2因为向量a1与a2的分量不成比例故