同济大学线性代数第六版全

同济大学线性代数第六版 答案 八年级地理上册填图题岩土工程勘察试题省略号的作用及举例应急救援安全知识车间5s试题及答案 (全)第一章行列式利用对角线法规计算下列三阶行列式201(1)141183201解1411832(4)30(1)(1)1180132(1)81(4)(1)2481644abcbcacababc解bcacabacbbaccbabbbaaaccc3abca3b3?c3111abca2b2c2111解abca2b2c2bc2ca2ab2?ac2ba2cb2(ab)(bc)(ca)xyxy(4)yxyxxyxyxyxy解yxyxxyxyx(xy)yyx(xy)(xy)yxy3(xy)3x33xy(xy)y33x2yx3y3x32(x3y3)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)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(n1)232(1个)5254(2个)727476(3个)(2n1)2(2n1)4(2n1)6(2n1)(2n2)(n1个)(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(2n1)(2n2)(n1个)42(1个)6264(2个)(2n)2(2n)4(2n)6(2n)(2n2)(n1个)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)ta11a23a3ra4s其中rs是2和4组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)ta11a23a32a44(1)1a11a23a32a44?a11a23a32a44(1)ta11a23a34a42(1)2a11a23a34a42?a11a23a34a42计算下列各行列式12412021052001174124cc412104110解1202232021122(1)4310520c41032141031401177c300104110c2c399100122c112c30021031417171421413121123250622141cc2140rr2140解3121423122423122123212301230506250622140rr2140413122012300000(3)abacaebdcddebfcfef解abacaebcebdcddeadfbcebfcfefbce1114abcdefadfbce111111a100(4)1b1001c1001da100rar01aba0解1b10121b1001c101c1001d001d(1)(1aba0c3dc21abaad1)211c11c1cd01d010(1)(1)321abadabcdabcdad111cd5证明:a2abb22aab2b(ab)3;111证明a2abb2c2c1a2aba2b2a22aab2b2aba2b2a111c3c1100(1)31aba2b2a2(ba)(ba)aba(ab)3ba2b2a12axbyaybzazbxxyz(2)aybzazbxaxby(a3b3)yzx;azbxaxbyaybzzxy证明axbyaybzazbxaybzazbxaxbyazbxaxbyaybzxaybzazbxyaybzazbxayazbxaxbybzazbxaxbyzaxbyaybzxaxbyaybzxaybzzyzazbxa2yazbxxb2zxaxbyzaxbyyxyaybzxyzyzxa3yzxb3zxyzxyxyzxyzxyza3yzxb3yzxzxyzxyxyz(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证明a2(a1)2(a2)2(a3)2b2(b1)2(b2)2(b3)2(c4?c3?c3?c2?c2?c1得)c2(c1)2(c2)2(c3)2d2(d1)2(d2)2(d3)2a22a12a32a5b22b12b32b5(c4?c3?c3?c2得)c22c12c32c5d22d12d32d5a22a122b22b1220c22c122d22d1221111abcd(4)a2b2c2d2a4b4c4d4(ab)(ac)(ad)(bc)(bd)(cd)(abcd);证明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)x10000x100xna1xn1an1xan(5)00x10anan1an2a2xa1证明用数学归纳法证明当n2时D2x1x2a1xa2命 题 快递公司问题件快递公司问题件货款处理关于圆的周长面积重点题型关于解方程组的题及答案关于南海问题 成立a2xa1假设关于(n1)阶行列式命题成立即Dn1xn1?a1xn2an2xan1则Dn按第一列展开有1000DnxDn1a(1)n1x100n11x1xDn1anxna1xn1an1xan因此关于n阶行列式命题成立设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an1anna1nannanna1nD1aaD2aaD3aa111n11n1n111n(n1)证明D1D2(1)2DD3D证明因为Ddet(aij)所以an1anna11a1n1an1annD1a11(1)na1na21a2na11a1na21a2n(1)n1(1)n2an1anna31a3n(1)12(n2)(n1)D(1)n(n1)2D同理可证n(n1)a11an1n(n1)n(n1)D(1)2(1)2DT(1)2D2a1nannn(n1)n(n1)n(n1)(1)n(n1)DDD3(1)2D2(1)2(1)2D7计算下列各行列式(Dk为k阶行列式)a1(1)Dn,其中对角线上元素都是a未写出的元素都1a是0解a000a000aDn000100000(a001)n10a0000(1)n1(1)na100(按第n行展开)00a01a0000(1)2naa0(n1)a(n1)(n1)(n1)ananan2an2(a21)a(n2)(n2)xaa(2)Dnaxa;aax解将第一行乘(1)分别加到其余各行得axaaaDnxxa00ax0xa0ax000xa再将各列都加到第一列上得x(n1)aaaaDn0xa00[x(n1)a](xa)n100xa00000xaan(a1)n(an)n(3)Dn1an1(a1)n1(an)n1;aa1an111解根据第6题结果有111n(n1)aa1anDn1(1)2an1(a1)n1(an)n1an(a1)n(an)n此行列式为范德蒙道德列式n(n1)Dn1(1)2[(ai1)(aj1)]n1ij1n(n1)(1)2[(ij)]n1ij1n(n1)(1)2(1)(ij)n(n1)1j)2(in1ij1n1ij1anbn(4)D2na1b1;c1d1cndn解anbnD2na1b1(按第1行展开)c1d1cndnan1bn10a1b1anc1d1cn1dn1000dn0an1bn1(1)2n1bna1b1c1d1cn1dn1cn0再按最后一行展开得递推公式D2nandnD2n2bncnD2n2即D2n(andnbncn)D2n2n于是D2n(aidibici)D2i2而a1b1a1d1b1c1D2cd11n所以D2n(aidibici)1Ddet(aij)其中aij|ij|;解aij|ij|0123n11012n2Dndet(aij)2101n33210n4n1n2n3n4011111r1r21111111111r2r311111n1n2n3n4010000c2c11200012200c3c112220n12n32n42n5n1(1)n1(n1)2n21a11111a21,其中a1a2an0(6)Dn111an解1a111Dn11a21111ana100c1c2a2a200aa33c2c3000000100110aaa01112n000000100010a1a2an001000000n(a1a2an)(11)i1ai001001001an1an110an1an00a1001a1002a1311a10n1111an00a1001a1002a1301a1n1na1001i1i用克莱姆法规解下列方程组x1x2x3x45x12x2x34x422x13x2x35x423x1x22x311x40解因为1111D12141422315312115111151122141214284D12315142D22215012113021111511115D3122442612121422325D42312310113120所以xD11xD22xD33xD411D2D3D4D5x16x26x31x15x20(2)x25x36x40x35x46x50x45x51解因为56000D1560066501560001560001516000510000560010600D1015601507D20056011450015600156100150101556100560101500015600D301060703D4015003950005600106001150001556001D515600015602120015000011所以x11507x21145x3703x4395x42126656656656656659问取何值时齐次线性方程组x1x2x30有非零解xxx0123x12x2x30解系数行列式为D1111121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(1)x12x24x30齐次线性方程组2x(3)xx0有非1x223x1(1)x30零解解系数行列式为D132412342111111101(1)3(3)4(1)2(1)(3)(1)32(1)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x221y11x2315y2x3323y2y12211x1749y1故y2315x2637y2y323x3324y32y17x14x29x3y26x13x27x3y33x12x24x3已知两个线性变换x12y1y3y13z1z2x22y13y22y3y22z1z3x34y1y25y3y3z23z3求从z1z2z3到x1x2x3的线性变换解由已知x201y1201310z11232y2232201z2x2x3415y415013z23613z11249z210116z3x16z1z23z3所以有x12z4z9z2123x310z1z216z33设A111B123求3AB2A及ATB111124111051111123111解3AB2A311112421111110511110581112132230562111217202901114292111123058ATB111124056111051290计算下列乘积431712325701431747321135解123217(2)23165701577201493(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)a12a22a23x2a13a23a33x3解a11a12a13x1(x1x2x3)a12a22a23x2a13a23a33x3x1(a11x1?a12x2?a13x3a12x1?a22x2?a23x3a13x1?a23x2?a33x3)x2x3a11x12a22x22a33x322a12x1x22a13x1x32a23x2x35设A12B10问1312(1)ABBA吗解ABBA因为AB34BA124638(2)(AB)2?A22ABB2吗解(AB)2?A22ABB2因为AB2225(AB)222222525但A22ABB238411所以(AB)2?A22ABB2(3)(AB)(AB)A2B2吗解(AB)(AB)A2B2所以ABBA14142968101016812341527因为AB22AB022501(AB)(AB)220206250109而A2B23810284113417故(AB)(AB)A2B2举反列 说明 关于失联党员情况说明岗位说明总经理岗位说明书会计岗位说明书行政主管岗位说明书 下列命题是错误的(1)若A20则A0解取A01则A20但A000(2)若A2?A则A0或AE解取A11则A2?A但A0且AE00(3)若AXAY且A0则XY解取A10X11Y11001101则AXAY且A0但XY7设A10求A2?A3Ak1解A21010101121A3A2A10101021131Ak10k110求Ak8设A010解首先察看1010221A2010102200000023323A3A2A033200344362A4A3A0443004A5A4554103A0554005kkk1k(k1)k2k02Akkk100k用数学归纳法证明当k2时显然成立假设k时成立，则k1时，kkk1k(k1)k210Ak1AkA02kkk10100k00k1(k1)k1(k1)kk102k1(k1)k100k1由数学归纳法原理知kkk1k(k1)k2Ak02kkk100k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(BTAB)TBT(BTA)TBTATBBTAB进而BTAB是对称矩阵设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(AB)T(BA)TATBTAB即AB是对称矩阵必要性因为ATABTB且(AB)TAB所以AB(AB)TBTATBA求下列矩阵的逆矩阵1225解A12|A|1故A1存在因为25A*A11A2152A12A2221故A11A*52|A|21(2)cossinsincos解Acossin|A|10故A1存在因为sincosA*A11A21cossinA12A22sincos所以A11A*cossin|A|sincos121(3)342541121|A|20故A1存在因为解A342541A11A21A31420A*A12A22A321361A13A23A3332142A11A*210所以1331|A|221671a1a02(a1a2an0)(4)0ana10Aa2解由对角矩阵的性质知0an110a1A1a201an解下列矩阵方程(1)25X461321解X2514635462231321122108211113(2)X2104321111132111解X2104321111113101232343233022185233(3)14X2031121101解X141312011201111243110121101121661011101230124010100143(4)100X00120100101012014310010101解X100201001001120010010143100210100201001134001120010102利用逆矩阵解下列线性方程组x12x23x312x12x25x323x15x2x33解方程组可 表 关于同志近三年现实表现材料材料类招标技术评分表图表与交易pdf视力表打印pdf用图表说话 pdf 示为123x11225x22351x33x1123111故x222520x335130x11进而有x20x30x1x2x322x1x23x313x12x25x30解方程组可表示为111x12213x21325x03x1111152故x221310x332503x15故有x20x3314设AkO(k为正整数)证明(EA)1EAA2Ak1证明因为AkO所以EAkE又因为E?Ak(EA)(EAA2Ak1)所以2k1(EA)(EAAA)E由定理2推论知(EA)可逆且12k1(EA)EAAA证明一方面有E(EA)1(EA)另一方面由AkO有E(EA)(AA2)A2Ak1(Ak1Ak)2k1(EAA?A)(EA)故(EA)1(EA)(EAA2Ak1)(EA)两端同时右乘(EA)1就有(EA)1(EA)EAA2Ak1设方阵A知足A2?A2EO证明A及A2E都可逆并求A1及(A2E)1证明由A2?A2EO得A2?A2E即A(AE)2E或A1(AE)E2由定理2推论知A可逆且A11(AE)2由A2?A2EO得A2?A6E4E即(A2E)(A3E)4E或(A2E)1(3EA)E4由定理2推论知(A2E)可逆且(A2E)11(3)EA4证明由A2?A2EO得A2?A2E两端同时取行列式得|A2?A|2即|A||AE|2故所以A可逆而A2EA2|A2E||A2||A|20故A2E也可逆由A2?A2EOA(AE)2EA1A(AE)2A1EA112(AE)又由A2?A2EO(A2E)A3(A2E)4E(A2E)(A3E)4E所以(A2E)1(A2E)(A3E)4(A2E)1(A2E)11(3E4A)16设A为3阶矩阵|A|12求|(2A)15A*|解因为A11|A|A*所以|(2A)15A*||12A15|A|A1||1A215A21||2A1|(2)3|A1|8|A|1821617设矩阵A可逆证明其陪同阵A*也可逆且(A*)1(A1)*证明由A11A*得A*|A|A1所以当A可逆时有|A||A*||A|n|A1||A|n10进而A*也可逆因为A*|A|A1所以(A*)1|A|1A又A1|A1|(A1)*|A|(A1)*所以(A*)1|A|1A|A|1|A|(A1)*(A1)*18设n阶矩阵A的陪同矩阵为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A11A*则AA*|A|E取行列式获得|A||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A033ABA2B求B110123解由ABA2E可得(A2E)BA故2310330332E)1A3B(A110110123121123110101220设A020且ABEAB求B1012解由ABEAB得(AE)BA2E即(AE)B(AE)(AE)001因为|AE|01010所以(AE)可逆进而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)18(2E2A)14(EA)14[diag(212)]14diag(1,1,1)22diag(121)100022已知矩阵A的陪同阵A*010010100308且ABA1BA13E求B解由|A*||A|38得|A|2由ABA1BA13E得ABB3AB3(AE)1A3[A(EA1)]1A3(E1A*)16(2EA*)12100016000601000600101060600306030123设P1AP其中P1410求A111102解1111111由PAP得APP所以A?A=PP.|P|3P*14P111411311而1110111002021114101427312732故1133A1102111168368433111124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()P1P()P*|P|111100222210200030311100012111111111125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)B1B1?A1?A1B1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)1A1210103126计算010101210021002300030003解设A112A221B131B22301032103则A1EEB1A1A1B1B2OA2OB2OA2B2而A1B1B21231235201210324A2B2212343030309A1EEB1A1A1B1B21252所以0124OA2OB2OA2B200430009121010311252即01010121012400210023004300030003000927取ABCD10考据AB|A||B|01CD|C||D|10102000解AB0101020020104CD10101010020101010101而|A||B|110|C||D|11故AB|A||B|CD|C||D|34O28设A4384O20求|A|及A22解令A134A2204322则AA1OOA2AO8A8O故8A11OA2OA28|A8||A8||A8||A|8|A|8101612124A14O540O054AOA4O2402262429设n阶矩阵A及s阶矩阵B都可逆求OA1(1)BO解设OA1C1C2则BOC3C4OAC1C2AC3AC4EnOBOC3C4BC1BC2OEsACECA13n3由此得AC4OC4OBC1OC1OBC2EsC2B1OA1OB1所以BOA1O(2)AO1CBAO1D1D2解设则CBD3D4AOD1D2AD1AD2EnOCBD3D4CD1BD3CD2BD4OEsADEDA11n1由此得AD2OOD2OCDBDDB1CA113Es3CD2BD4D4B1AO1A1O所以CBB1CA1B1求下列矩阵的逆阵5200(1)210000830052解设A52B83则215211A15212B18323212552585200112001A1于是2100A25000083BB10023005200581000(2)120021301214解设A10B30C21则1214121000111200AOOA12130CBB1CA1B11214100011002211102631151824124第三章矩阵的初等变换与线性方程组把下列矩阵化为行最简形矩阵(1)1021203130431021解2031(下一步r2(2)r1r3(3)r1)3043~1021(下一步r(1)r(2))001323002010210013(下一步r3r2)00101021001300302100130010210010001(下一步r33)(下一步r23r3)(下一步r1(2)r2r1r3)~100000100001(2)0231034304710231解034304710231~00130013(下一步r22(3)r1r3(2)r1)(下一步r3r2r13r2)02010~0013(下一步r12)00000105~0013000011343(3)33541223203342111343解33541(下一步r3r1r32r1r3r1)22320243342111343~0048823400366(下一步r(4)r(3)r(5))005101011343~00122(下一步r13r2r3r2r4r2)001220012211023~00122000000000023137(4)12024328302374323137解12024328302374301111~12024088912077811(下一步r2r2r3r2r2r2)134(下一步r2r1r8r1r7r1)234111110202~00014(下一步rrr(1)rr)122430001410202~01111(下一步rr3)0001420000010202~0110300014000000101011232设100A010456求A001001789010解100是初等矩阵E(12)其逆矩阵就是其本身001101010是初等矩阵E(12(1))其逆矩阵是001E(12(1))101010001010123101A1004560100017890014561014521230101227890017823试利用矩阵的初等变换求下列方阵的逆矩阵321315323321100321100解315010~0141103230010021013203/201/23007/229/2~010112~0101120021010011/201/2~1007/62/33/20101120011/201/2723故逆矩阵为632112101223201(2)02211232012132011000解02210100123200100121000112320010~01210001049510300221010012320010~01210001001110340021010212320010~012100010011103400012161012001122~010001