工程數(shù)學線性代數(shù)課后習題

1、第一章行列式利用對角線法規(guī)計算以下三階行列式201(1)141183201解1411832(4)30(1)(1)1180132(1)81(4)(1)2481644abcbcacababc解bcacabacbbaccbabbbaaaccc3abca3b3c3111abca2b2c211解abca2b2c2bc2ca2ab2ac2ba2cb2(ab)(bc)(ca)(4)xyxyyxyxxyxy解xyxyyxyxxyxyx(xy)yyx(xy)(xy)yxy3(xy)3x33xy(xy)y33x2yx3y3x32(x3y3)2按自然數(shù)從小到大為標準次序求以下各擺列的逆序數(shù)(1)1234解逆序數(shù)為0

2、(2)4132解逆序數(shù)為441434232(3)3421解逆序數(shù)為532314241,21(4)2413解逆序數(shù)為3214143(5)13(2n1)24(2n)解逆序數(shù)為n(n1)232(1個)5254(2個)727476(3個)(2n1)2(2n1)4(2n1)6(2n1)(2n2)(n1個)(6)13(2n1)(2n)(2n2)2解逆序數(shù)為n(n1)32(1個)5254(2個)(2n1)2(2n1)4(2n1)6(2n1)(2n2)(n1個)42(1個)6264(2個)(2n)2(2n)4(2n)6(2n)(2n2)(n1個)寫出四階行列式中含有因子a11a23的項解含因子a11a23的項

3、的一般形式為(1)ta11a23a3ra4s此中rs是2和4構(gòu)成的擺列這類擺列共有兩個即24和42所以含因子a11a23的項分別是(1)ta11a23a32a44(1)1a11a23a32a44a11a23a32a44(1)ta11a23a34a42(1)2a11a23a34a42a11a23a34a42計算以下各行列式1241202105201174124cc412104110解120223120243122(1)10520c7c1032141031401174300104110c2c399100122c121c3002103141717142141312112325062214121402

4、140解3121c4c23122r4r23122123212301230506250622140rr2140413122012300000(3)abacaebdcddebfcfefabacaebce解bdcddeadfbcebfcfefbce1114abcdefadfbce111111a1001b1001c1001da100rar01aba0解1b10121b1001c101c1001d001d(1)(1aba0c3dc21abaad1)211c11c1cd01d010(1)(1)321abadabcdabcdad11cd證明:(1)a2abb23;2aab2b(ab)111證明a2abb2c

5、2c1a2aba2b2a22aab2bc3c12aba2b2a111100(1)31aba2b2a2(ba)(bababa2b2aa)12axbyaybzazbxxyzaybzazbxaxby(a3b3)yzx;azbxaxbyaybzzxy證明axbyaybzazbxaybzazbxaxbyazbxaxbyaybz1(ab)3xaybzazbxyaybzazbxayazbxaxbybzazbxaxbyzaxbyaybzxaxbyaybzxaybzzyzazbxa2yazbxxb2zxaxbyzaxbyyxyaybzxyzyzxa3yzxb3zxyzxyxyzxyzxyza3yzxb3yzxz

6、xyzxyxyz(a3b3)yzxzxya2(a1)2(a2)2(a3)2(3)b2(b1)2(b2)2(b3)20;c2(c1)2(c2)2(c3)2d2(d1)2(d2)2(d3)2證明a2(a1)2(a2)2(a3)2b2(b1)2(b2)2(b3)2(c4c3c3c2c2c1得)2(c2(c2)2(c3)2c1)d2(d1)2(d2)2(d3)2a22a12a32a5b22b12b32b5(c4c3c3c2得)c22c12c32c5d22d12d32d5a22a122b22b1220c22c122d22d1221111a2b2c2d2abcda4b4c4d4(ab)(ac)(ad)(b

7、c)(bd)(cd)(abcd);證明1111abcda2b2c2d2a4b4c4d411a1a1a0bcd0b(ba)c(ca)d(da)0b2(b2a2)c2(c2a2)d2(d2a2)(ba)(ca)(da)111bcda)b2(ba)c2(ca)d2(d(ba)(ca)(d1c1d1a)0ba)d(dbba)0c(cb)(cbb)(d(ba)(ca)(da)(cb)(db)c(c11a)ba)d(db=(ab)(ac)(ad)(bc)(bd)(cd)(abcd)x10000 x100a1xn1(5)00 xxn01anan1an2a2xa1an1xan證明用數(shù)學歸納法證明當n2時D2x

8、1x2a1xa2命題成立a2xa1假設(shè)對于(n1)階行列式命題成立即n1xn1a1xn2nnDa2xa1則Dn按第一列睜開有1000DnxDn1a(1)n1x100n11x1xDn1nna1xn1nnaxa1xa所以對于n階行列式命題成立6設(shè)n階行列式Ddet(aij),把D上下翻轉(zhuǎn)、或逆時針旋轉(zhuǎn)90、或依副對角線翻轉(zhuǎn)挨次得an1anna1nannanna1nD1a11a1nD2a11an1D3a11an1n(n1)D3D證明D1D2(1)2D證明因為Ddet(a)所以ijan1anna11a1nn1an1annD1a11(1)a1na21a2na11a1na21a2n(1)n1(1)n2an

9、1anna31a3nn(n1)(1)12(n2)(n1)D(1)2D同理可證n(n1)a11an1n(n1)n(n1)D2(1)2a1nann(1)2DT(1)2Dn(n1)n(n1)n(n1)D3(1)2D2(1)2(1)2D(1)n(n1)DD計算以下各行列式(Dk為k階行列式)(1)a1此中對角線上元素都是aDn1,未寫出的元a素都是0解a00010a000Dn00a00(按第n行睜開)000a01000a00001aa0000(1)2na(1)n10a000a(n1)000a0(n1)(n1)(n1)(1)n1(1)naananan2an2(a21)a(n2)(n2)xaa(2)Dna

10、xa;aax解將第一行乘(1)分別加到其他各行得axaaaxxa00Dnax0 xa0ax000 xa再將各列都加到第一列上得x(n1)aaaaDn0 xa00 x(n1)a(xa)n100 xa00000 xaan(a1)n(an)n(3)Dn1an1(a1)n1(an)n1;aa1an111解依據(jù)第6題結(jié)果有111n(n1)aa1anDn1(1)2(a1)n1(an)n1an1an(a1)n(an)n此行列式為范德蒙道德列式n(n1)Dn1(1)2(ai1)(aj1)n1ij1n(n1)(1)2(ij)n1ij1(n(n1)n(n1)1(ij)1)2(1)2n1ij1(ij)n1ij1an

11、bn(4)D2na1b1;c1d1cndn解anbnD2na1b1(按第1行睜開)c1d1cndnan1bn10ana1b1c1d1cn1dn1000dn0an1(1)2n1bna1b1c1d1cncn1再按最后一行睜開得遞推公式D2nandnD2n2bncnD2n2即D2n于是D2nn(aidibici)D2i2而D2a1b1cda1d1b1c111所以D2nn(aidibici)i1(5)Ddet(aij)此中aij|ij|;解aij|ij|01231012Dndet(aij)21013210n1n2n3n4bn1dn10(andnbncn)D2n21n2n3n40Dn解11111r1r2

12、1111111111r2r311111n1n2n3n4010000c2c11200012200c3c112220n12n32n42n5n1(1)n1(n1)2n21a11111a21,此中a1a2a0n111an1a111Dn11a21111ana100001c1c2a2a200010aa001c2c333000aa1n1n10000an1an10000a1110001a1011002a1a2a1an300011a10000n111a1n10000a1010001a1001002a1a1a2an30001a10n1n000001a1ii1(a1a2an)(1n1)i1ai用克萊姆法規(guī)解以下方程

13、組x1x2x3x45x12x2x34x422x13x2x35x423x1x22x311x40解因為1111D121414223153121151111511D1221414212142842315D2221501211302111151111512244261212142D32325D42312310113120所以x1D11x2D22x3D33x4D41DDDD5x16x26x31(2)x15x20 x25x36x40 x35x46x50 x45x51解因為56000D1560066501560001560001516000D056001507015601001561001556100D15

14、000703010603000560011556001D515600212015600015000011所以51000D210600005601145001560101556010D415600015003950010600015x11507x21145x3703x4395x42126656656656656659問取何值時齊次線性方程組x1x2x30 xxx0有非零解？123x12x2x30解系數(shù)行列式為D1111121令D0得0或1于是當0或1時該齊次線性方程組有非零解10問取何值時齊次線性方程組(1)x12x24x302x(3)xx0有非零解？123x1x2(1)x30解系數(shù)行列式為D1

15、32412342111111101(1)3(3)4(1)2(1)(3)(1)32(1)23令D0得02或3于是當02或3時該齊次線性方程組有非零解第二章矩陣及其運算已知線性變換x12y12y2y3x23y1y25y3x33y12y23y3求從變量x1x2x3到變量y1y2y3的線性變換解由已知x221y11y2x2315x3323y2y12211749y1xy21637y2故315x2y2323x3324y3y17x14x29x3y26x13x27x3y33x12x24x3已知兩個線性變換x12y1y3y13z1z2x22y13y22y3y22z1z3x34y1y25y3y3z23z3求從z1

16、z2z3到x1x2x3的線性變換解由已知x201y1201310z11y2z2x2232232201x3415y415013z23613z149z210116z3x16z1z23z3所以有x12z4z9z2123x310z1z216z33111123求3AB2A及設(shè)A111B124111051ATB111123111解3AB2A311112421111110511110581112132230562111217202901114292111123058ATB111124056111051290計算以下乘積(1)431712325701431747321135解123217(2)23165701

17、57720149(2)3(123)21(123)3(132231)(10)解212(3)1(12)322(1)2224解1(12)1(1)121233(1)3236131(4)21400121134131402131解214001267811341312056402a11a12a13x1(x1x2x3)a12a22a23x2a13a23a33x3解a11a12a13x1(x1x2x3)a12a22a23x2a13a23a33x3(a11x1a12x2a13x3a12x1a22x2a23x3x1a13x1a23x2a33x3)x2x3a11x12a22x22a33x322a12x1x22a13x

18、1x32a23x2x35設(shè)A12B10問1312ABBA嗎?解ABBA因為AB34BA12所以ABBA4638(2)(AB)2A22ABB2嗎?解(AB)2A22ABB2因為AB2225(AB)2222281425251429但A22ABB23868101016411812341527所以(AB)2A22ABB2(3)(AB)(AB)A2B2嗎?解(AB)(AB)A2B2因為AB22AB022501(AB)(AB)220206250109而A2B23810284113417故(AB)(AB)A2B2舉反列說明以下命題是錯誤的(1)若A20則A0解取A01則A20但A000(2)若A2A則A0或

19、AE解取A112A但A0且AE00則A(3)若AXAY且A0則XY解取A10X11Y11001101則AXAY且A0但XY7設(shè)A1023k1求AAA解A21010101121A3A2A10101021131Ak10k18設(shè)A10k01求A0解第一觀察A2101001010000A3A2A3323033200344362A4A3A0443004554103A5A4A0554005kkk1k(k1)k02Akkk100k用數(shù)學歸納法證明221022002k2當k2時明顯成立假設(shè)k時成立，則k1時，kkk1k(k1)k210Ak1AkA020kkk1100k00k1(k1)k1(k1)kk10k1(

20、k2k11)00k1由數(shù)學歸納法原理知kkk1k(k1)k2Ak02kkk100k9設(shè)AB為n階矩陣，且A為對稱矩陣，證明BTAB也是對稱矩陣證明因為ATA所以(BTAB)TBT(BTA)TBTATBBTAB從而BTAB是對稱矩陣10設(shè)AB都是n階對稱矩陣，證明AB是對稱矩陣的充分必需條件是ABBA證明充分性因為ATABTB且ABBA所以AB)T(BA)TATBTAB即AB是對稱矩陣必需性因為ATABTB且(AB)TAB所以AB(AB)TBTATBA求以下矩陣的逆矩陣1225解A12|A|1故A1存在因為25A*A11A2152A12A2221故A11A*52|A|21(2)cossinsin

21、cos解Acossin|A|101存在因為sincos故AA*A11A21cossinA12A22sincos所以A11A*cossin|A|sincos(3)121342541121|A|20故A1存在解A342因為541A11A21A31420A*A12A22A321361A13A23A3332142A11A*210所以1331|A|221671a1a0(4)2(a1a2an0)0ana10解Aa2由對角矩陣的性質(zhì)知0an110a1A1a201an解以下矩陣方程(1)25X461321解X2514635462231321122108(2)211113X2104321111132111解X2

22、104321111113101232343233022185233(3)14203112X110114131201解X1201111243110121101121661011101230124(4)010100143100X00120100101012001011431001解X100201001001120010010143100210100201001134001120010102利用逆矩陣解以下線性方程組x12x23x312x12x25x323x15x2x33解方程組可表示為123x11225x22351x33x1123111故x222520 x335130 x11從而有x20 x30

23、x1x2x322x1x23x313x12x25x30解方程組可表示為111x12213x21325x03x1111152故x221310 x332503x15故有x20 x3314設(shè)AkO(k為正整數(shù))證明(EA)1EAA2Ak1證明因為AkO所以EAkE又因為EAk(EA)(EAA2Ak1)所以(EA)(EAA2Ak1)E由定理2推論知(EA)可逆且(EA)1EAA2Ak1證明一方面有E(EA)1(EA)另一方面由AkO有E(EA)(AA2)A2Ak1(Ak1Ak)(EAA2Ak1)(EA)故(EA)1(EA)(EAA2Ak1)(EA)兩端同時右乘(EA)1就有(EA)1(EA)EAA2Ak

24、115設(shè)方陣A滿足A2A2EO證明A及A2E都可逆并求A1及(A2E)1證明由A2A2EO得A2A2E即A(AE)2E或A1(AE)E2由定理2推論知A可逆且A11(AE)2由A2A2EO得A2A6E4E即(A2E)(A3E)4E或(A2E)1(3EA)E4由定理2推論知(A2E)可逆且(A2E)11(3EA)4證明由A2A2EO得A2A2E兩端同時取行列式得|A2A|2即|A|AE|2故|A|0所以A可逆而A2EA2|A2E|A2|A|20故2E也可逆由A2A2EOA(AE)2EA1A(AE)2A1EA11(AE)2又由A2A2EO(A2E)A3(A2E)4E(A2E)(A3E)4E所以(A

25、2E)1(A2E)(A3E)4(A2E)1(A2E)11(3)EA416設(shè)A為3階矩陣|A|1求|(2A)15A*|2解因為A11A*所以|A|(2A)15A*|1A15|A|A1|1A15A1|2222A1|2)3|A1|A|117設(shè)矩陣A可逆證明其陪伴陣A*也可逆且(A*)1(A1)*證明由A11A*得A*|A|A1所以當A可逆時|A|有|A*|A|n|A1|A|n10從而A*也可逆因為A*|A|A1所以(A*)1|A|1A又A11(A1)*|A|(A1)*所以|A|(A*)1|A|1A|A|1|A|(A1)*(A1)*18設(shè)n階矩陣A的陪伴矩陣為A*證明(1)若|A|0則|A*|0(2)

26、|A*|A|n1證明(1)用反證法證明假設(shè)|A*|0則有A*(A*)1E由此得AAA*(A*)1|A|E(A*)1O所以A*O這與|A*|0矛盾，故當|A|0時有|A*|0(2)因為A11A*則AA*|A|E取行列式獲取|A|A|A*|A|n若|A|0則|A*|A|n1若|A|0由(1)知|A*|0此時命題也成立所以|A*|A|n119033ABA2B求B設(shè)A110123解由ABA2E可得(A2E)BA故2331330330B(A2E)1A11011012312112311020101且ABEA2B求B設(shè)A020101解由ABEA2B得(AE)BA2E即(AE)B(AE)(AE)00110所以

27、(AE)可逆因為|AE|010從而100201BAE03010221設(shè)Adiag(121)A*BA2BA8E求B解由A*BA2BA8E得(A*2E)BA8EB8(A*2E)1A18A(A*2E)18(AA*2A)18(|A|E2A)18(2E2A)14(EA)14diag(212)14diag(1,1,1)222diag(121)100022已知矩陣A的陪伴陣A*010010100308且ABA1BA13E求B解由|A*|A|38得|A|2由ABA1BA13E得ABB3AB3(AE)1A3A(EA1)1A3(E1A*)16(2EA*)121000106006010006001010606003

28、06030123設(shè)P1AP此中P1410求1102A11解由P1AP得APP|P|3P*14P111所以A11A=P11P1.1114311而1110111002021114101427312732故A1133110211116836843324設(shè)APP1111此中P10211115求(A)A8(5E6AA2)解()8(5E62)diag(1158)diag(555)diag(6630)diag(1125)diag(1158)diag(1200)12diag(100)(A)P()P1P()P*|P|1111002222102000303111000121111411111125設(shè)矩陣A、B及A

29、B都可逆證明A1B1也可逆并求其逆陣證明因為A1(AB)B1B1A1A1B1而A1(AB)B1是三個可逆矩陣的乘積所以A1(AB)B1可逆即A1B1可逆(A1B1)1A1(AB)B11B(AB)1A1210103126計算010101210021002300030003解設(shè)A12A21B31010321121B22303則AEEBAABB111112OA2OB2OA2B2而A1B1B21231235201210324A2B2212343030309A1EEB1A1A1B1B21252所以0124OA2OB2OA2B200430009121010311252即010101210124002100

30、23004300030003000927取ABCD10考據(jù)AB|A|B|01CD|C|D|10102000解AB010102002010CD101010100201401010101而|A|B|110|C|D|11故AB|A|B|CD|C|D|34O28設(shè)A4384O20求|A|及A22解令A34A20432212則AA1OOA2AO8A8O故8A11OA2OA28|A8|A8|A8|A|8|A|810161212A14O540O4054AOA4O2402262429設(shè)n階矩陣A及s階矩陣B都可逆求OA1BO解設(shè)OA1C1C2則BOCC43OAC1C2AC3AC4EnOBOC3C4BC1BC2

31、OEsACECA13n3由此得AC4OC4OBC1OC1OBC2EsC2B1所以O(shè)A1OB1BOA1OAO1CB1D1D2解設(shè)AO則CBD3D4AOD1D2AD1AD2CBD3D4CD1BD3CD2BD4ADEDA11n1由此得AD2OD2OCD1BD3OD3B1CA1CD2BD4EsD4B1AO1A1O所以CBB1CA1B1求以下矩陣的逆陣5200(1)210000830052解設(shè)A52B83則215211A15212B183212552EnOOEs358520011200A1于是2100A125000083BB10023005200581000(2)120021301214解設(shè)A10B30

32、C21則1214121000111200AOA1O2130CBB1CA1B11214100011002211102631151824124第三章矩陣的初等變換與線性方程組把以下矩陣化為行最簡形矩陣1021203130431021解2031(下一步r2(2)r1r3(3)r1)30431021r(1)r(2)0013(下一步23002010210013(下一步r3r2)0010102100130030210013001(下一步r33)(下一步r23r3)10210010(下一步r1(2)r2r1r3)0001100000100001(2)0231034304710231解0343(下一步r22(

33、3)r1r3(2)r1)047102310013(下一步r3r2r13r2)0013020100013(下一步r12)000001050013000011343(3)33541223203342111343解33541(下一步r23r1r32r1r43r1)22320334211134300488下一步r(4)r(3)(23003660051010r4(5)1134300122(下一步r13r2r3r2r4r2)00122001221102300122000000000023137(4)12024328302374323137解12024(下一步r12r2r33r2r42r2)32830237

34、430111112024(下一步r221r38r1r47r1)088912r0778110111110202rrr(1)rr)00014(下一步12243000141020201111r2r3)00014(下一步00000102020110300014000002010101123求A設(shè)100A010456001001789010解100是初等矩陣E(12)其逆矩陣就是其自己001101010是初等矩陣E(12(1)其逆矩陣是001E(12(1)101010001010123101A1004560100017890014561014521230101227890017823試利用矩陣的初等變換

35、求以下方陣的逆矩陣321315323321100321100解3150100141103230010021013203/201/23007/229/20101120101120021010011/201/21007/62/33/20101120011/201/2723632故逆矩陣為112101223201(2)02211232012132011000解02210100123200100121000112320010012100010495103002210100123200100121000100111034002101021232001001210001001110340001216101

36、200112201000101001011360001216101000112401000101001011360001216101124故逆矩陣為01011136216104(1)設(shè)A41213221B2231131解因為41213r100(A,B)2212201031131001A1B102所以X153124(2)021123設(shè)A213B231334解考慮ATXTBT因為求X使AXB234求X使XAB02312r10024(AT,BT)2132301017134310011424所以XT(AT)1BT1714從而XBA12114745110AX2XA求X設(shè)A011101解原方程化為(A2E

37、)XA因為(A2E,A)1101100110111011011000110101010011102E)1A011所以X(A1011106在秩是r的矩陣中,有沒有等于0的r1階子式?有沒有等于0的r階子式?解在秩是r的矩陣中可能存在等于0的r1階子式也可能存在等于0的r階子式1000R(A)3比方A0100001000000是等于0的2階子式100是等于0的3階子式000107從矩陣A中劃去一行獲取矩陣B問AB的秩的關(guān)系如何?解R(A)R(B)這是因為B的非零子式必是A的非零子式故A的秩不會小于B的秩8求作一個秩是4的方陣它的兩個行向量是(10100)(11000)解用已知向量簡單構(gòu)成一個有4個非零行的5階下三角矩陣0000100001000001000000此矩陣的秩為4其第2行和第3行是已知向量9求以下矩陣的秩并求一個最高階非