版权说明:本文档由用户提供并上传,收益归属内容提供方,若内容存在侵权,请进行举报或认领
文档简介
习题解答习题一(A)1.用消元法解以下线性方程组:x12x23x34,(1)3x15x27x39,2x13x24x35.解由原方程组得同解方程组x12x23x34,x22x33,得方程组的解为x1x32,令x3c,得方程组的通解为x22x33.x1c2,x22c3,x3c,此中c为随意常数.x12x2x3x41,(2)x12x2x3x41,x12x2x35x45.解由原方程组得同解方程组x12x2x3x41,4x44,02,所以方程组无解.x1x22x31,x12x2x32,(3)3x12x25x33,x15x30.解由原方程组得同解方程组x1x22x31,x2x30,4x31,得方程组的解为x15,x21,x31.4442x12x2x43,2x13x2x33x46,(4)3x14x2x32x40,x13x2x3x42.解由原方程组得同解方程组x13x2x3x42,3x2x3x410,3x3x49,x43,得方程组的解为x12,x21,x34,x43.2.用初等行变换将以下矩阵化成行阶梯形矩阵和行最简形矩阵:122(1)212.221122122100解212r012r010,得221001001122100行阶梯形:012(不独一);行最简形:010.0010013211(2)1232.4423102132111232255解1232r04105r01,得244423000000001021123225行阶梯形:04105(不独一);行最简形:0152.00000040023(3)11.12231110解11r01r01,得1200001110行阶梯形:01(不独一);行最简形:01.000011111(4)203211361.242643111111111110011220321r02501r01001解361200700,得1242643000000010000000111111001120250101001行阶梯形:070(不独一);行最简形:.0000102000000000003.用初等行变换解以下线性方程组:x13x23x35,(1)2x1x24x311,x2x33.100M2133M537,解B214M11r010M011M39001M29得方程组的解为x127,x320,x299.3x1x24x33x41,(2)2x1x26x35x42,x12x22x32x42.1143M11143M1解B2165M2r0321M0,1222M20000M1得方程组无解.x12x23x34x44,x13x23x41,(3)x2x3x44,7x23x3x418.1000M471234M42151303M1r0101M解B2,0111M40012M230731M182M00000x147,2得方程组的解为x2x415,令x4c,得方程组的通解为2x32x423.2x147c152c23,x4c,此中c为随意常数.,x2,x32222x1x2x32x43x52,6x13x22x34x45x53,(4)6x13x24x38x413x59,4x12x2x3x42x51.21123M21100112M263245M32解Br00104M3,634813M900010M042112M100000M0x11x21x51,222得方程组的解为x34x53,令x2c1,x5c2,得方程组的通解为x40.x1c1c21c1,x34c23,x40,x5c2,此中c1,c2为随意常数.22,x22(B)x1x2x31,1.当为什么值时,线性方程组x1x2x3,有无量多解,并求解.x1x2x32111M1111M1解B11Mr010M1.11M2001M21111M1当1时,Br000M0,方程组有无量多解,且解为000M0x1x2x31.令x2c1,x3c2,得方程组的通解为x1c1c21,x2c1,x3c2,此中c1,c2为随意常数.3.(结合收入问题)已知三家企业A、B、C拥有以以下图所示的股份关系,即A企业掌握C企业50%的股份,C企业掌握A企业30%的股份,而A企业70%的股份不受此外两家企业控制等等.ABC3现设A、B和C企业各自的营业净收入分别是12万元、10万元、8万元,每家企业的联合收入是其净收入加上其他企业的股份按比率的提成收入.试确立各企业的结合收入及实质收入.解A企业的结合收入为元,实质收入为元;企业的结合收入为元,实质收入为元;C企业的结合收入为元,实质收入为元.习题二A)1.利用对角线法例计算以下队列式:cossin(1).sincos解原式1.xy(2)x2y2.解原式xy(yx).123(3)312.231解原式18.abc(4)0ab.00a解原式a3.00a(5)0ab.abc解原式a3.2.按定义计算以下队列式:00a0(1)b000.f00c0d0eb000c解原式a(13f0cab(111)1)dabcd.0dee010L0002L0(2)LLLLL.000Ln1n00L010L0解原式n(1)n102L0(1)n1n!.MMM00Ln13.利用队列式的性质,计算以下队列式:abacae(1)bdcdde.bfcfef111111解原式abcdef111abcdef0224abcdef.11100211112222(2).333344441111解原式0444192.00660008axaaa(3)aaxaaaaax.aaaaax11111000解原式(4ax)aaxaa(4ax)ax00(4ax)x3.aaaxaa0x0aaaaxa00x23100(4)1201.03518510154120112011201解原式23100011020110203518035180035125101540015100151123512215.01151111L11a10L0(5)10a2L0,此中a0,i1,2,L,n.iMMMLM100Lanr1r1aii2in解原式
n10L010i1ai1a10L0n1n)ai.10a2L(1i1ai0i1MMMLM100Lan4.利用队列式睁开定理,计算以下队列式:12140121(1)01.13013102010100101222121323解原式01117.1331131301311139365827(2)53.4278450030323014153223141531443144解原式1434318.343066663443a100L010a20L0000a3L00(3)MMMMM.000Lan10100L0an00L01a10L0a20L000a2L0解原式n1(1)0a3L00anMMMMMMM00Lan100Lan01a20L0n11(n1)0a3L0a1a2Lan(1)(1)MMM00Lan1a2a3Lan1a1a2Lana2a3Lan1(a1an1).210L00121L00(4)Dn012L00MMM.MM000L21000L12解将队列式按第一行睁开,得Dn2Dn1Dn2,则DnDn1Dn1Dn2LD2D121121,2所以DnDn11Dn22LD1(n1)n1.5.利用队列式睁开定理证明:当时,有0L001L0001L00n1n1DnMMOMM.M000L000L1证将队列式按第一行睁开,得Dn()Dn1Dn2,则DnDn1(Dn1Dn2)2(Dn2Dn3)Ln2(D2D1)n2[()2()]n,所以DnDn1n.(1)由Dn对于与对称,得DnDn1n.(2)n1n1由(1)与(2)解得Dn(类比于高中学过的由数列an与an-1的关系推导通项公式)abc6.利用范德蒙品德列式计算队列式a2b2c2.bcacababc111解原式(abc)a2b2c2(abc)abc111a2b2c2(abc)(ba)(ca)(cb).214211257.设D,试求A14+A24+A34+A44和M11+M12M13M14.31335111解A14+A24+A34+A440;1111M11+M12M13M14A11A12A131125A14133351110100346502132654422422142284.22062062612608.利用克拉默法例解以下线性方程组:x1x2x3x45,x12x2x34x42,(1)3x2x35x42,2x13x1x22x311x40.解经计算,得D142,D1142,D2284,D3426,D4142,所以方程组的解为x11,x22,x33,x41.x12x23x34x411,(2)x2x3x43,x13x2x40,7x23x3x45.解经计算,得D16,D116,D20,D332,D416,所以方程组的解为x11,x20,x32,x41.2x1x23x30,9.试问取何值时,齐次线性方程组3x14x27x30,有非零解.x12x2x30解方程组有非零解,则D0.又213D3475(3),12所以3.x1x2x30,10.试问、取何值时,齐次线性方程组x1x2x30,有非零解.x12x2x30解方程组有非零解,则D0.又11D11(1),121所以1或0.(B)1.选择题:2a111a135a123a12a11a12a1331(1)设a21a22a23a0,则2a21a235a223a22().3a31a32a332a311a335a323a323(A)2a(B)2a(C)3a(D)3a1a11a135a12a123解原式c22(3)a211a235a22a22c5c3c3(3)c23a3115a32a32c2c3a33
61a11a12a13(a21a22a23)2a.3a31a32a333选(A).a100b1(2)四阶队列式0a2b20的值等于().0b3a30b400a4(A)a1a2a3a4bb12b3b4(B)a1a2a3a4bb12b3b4(C)a1a2bb12a3a4b3b4(D)a2a3b2b3a1a4bb14解将队列式的第4行挨次与第3行、第2行互换,再将队列式的第4列挨次与第3列、第2列互换,得a100b1a1b1000a2b20b4a400a2a3b2b3a1a4b1b4.0b3a3000a2b2b400a400b3a3选(D).(3)设线性方程组a11x1a12x2b10,a11a121,则方程组的解为a21x1a22x2b2若a21a220.().(A)xb1a12,x2a11b1(B)xb1a12,x2a11b11b2a22a21b21b2a22a21b2(C)x1b1a12,x2a11b1(D)x1b1a12,x2a11b1b2a22a21b2b2a22a21b2解将方程组写成标准形式:a11x1a12x2b1,有a21x1a22x2b2.Da11a121,D1b1a12b1a12,D2a11b1a11b1,a21a22b2a22b2a22a21b2a21b2所以方程组的解为D1b1a12D2a11b1x1b2a22,x2a21.DDb2选(C).1111xabc0的根的个数为((4)方程f(x)=a2b2c2).x2x3a3b3c3(A)1(B)2(C)3(D)4解方法一:将f(x)按第1列睁开,知f(x)为3次多项式,所以有3个根.选(C).方法二:f(x)(ax)(bx)(cx)(ba)(ca)(cb)有3个根x1a,x2b,x3c.选(C).a10a202.计算四阶队列式D40b10b2.c10c200d10d2a10a20a1a200解D4c10c20c1c2000b10b200b1b20d10d200d1d2a1a2b1b2(a1c2a2c1)(b1d2b2d1).c1c2d1d2111x13.计算四阶队列式D411x11.1x111x1111x11x1111x1D4x1x1111x11解xx111xx1111x1111111100xx10x043x(1)2xxxx4.1x0010004.计算n阶队列式rnrn1rn1rn2LLLr2r1解Dn
123212Dn321LLLnn1n2123Ln1111L1111L1MMMM111L1
LnLn1Ln2.LL1n134Lnn11100L001cj120L00Mc1MM2jnMMM1122L20100L120L1n(n1)(1)122LMMM122L
000(1)1n(n1)2n2.M2n12aa200012aa2005.计算五阶队列式D5012aa20.0012aa200012a解方法一:一般地,对于此类n阶队列式,将其按第一行睁开,得Dn2Dn12Dn2,则DnDn1(Dn1Dn2)2(Dn2Dn3)Ln2(D2D1)n2[(2)222]n,有DnDn1n(Dn2n1)n2Dn22nLn1D1(n1)nn12(n1)n(n1)n,所以D56a5.方法二:由习题二(A)的第5题,适当时,有n1n1n(n1)n,Dnlim(n1)lim所以D56a5.x00L0a01x0L0a16.计算n阶队列式Dn01xL0a2.MMMMM000Lxan2000L1xan1解将队列式按第一行睁开,得DnxDn1a0,则Dnx(xDn2a1)a0x2Dn2a1xa0Lxn1D1an2xn2La1xa0xn1(xan1)an2xn2La1xa0xnan1xn1La1xa0.13267.已知1326、2743、5005、38742743都能被13整除,不计算队列式的值,证明00553874能被13整除.13262743证00553874
c41000c1c4100c2c410c3
13213262742743500.50053873874由已知,得后队列式的第4列拥有公因子13,所以原队列式能被13整除.1111abcd(ab)(ac)(ad)(bc)(bd)(cd)(abcd).8.证明:b2c2d2a2a4b4c4d4证结构5阶队列式11111abcdxD5a2b2c2d2x2,a3b3c3d3x3a4b4c4d4x4则D5(ba)(ca)(da)(cb)(db)(dc)(xa)(xb)(xc)(xd).(1)将D5按第5列睁开,得11111111D5abcd4(abcd3(2)a2b2c2d2xa2b2c2d2)xL.a3b3c3d3a4b4c4d4比较(1)与(2)右侧x3的系数,知结论建立.x1x2x3ax40,9.证明:当(a1)24b时,齐次线性方程组x12x2x3x40,有非零解.x1x23x3x40,x1x2ax3(ab)x40证方程组的系数队列式111aD1211(a1)24b,113111aab当D0,即(a1)24b时,方程组有非零解.10.应用题:(1)1;(2)xy10.习题三A)1.以下矩阵中,哪些是对角矩阵、三角矩阵、数目矩阵、单位矩阵.11000100300200,C420,D030A,B01.030100530030解D是数目矩阵,也是对角矩阵;A、C是三角矩阵;B都不是.1121232.设矩阵A111,B122.211031(1)计算2AB;(2)若X知足3AX2B,求X.347解(1)2AB100;411110(2)X2B3A577.695a1c1d1b1c1d13.设有3阶方阵Aa2c2d2,Bb2c2d2,且A1,B2,求A2B.a3c3d3b3c3d3a12b13c13d1解A2Ba22b23c23d2a32b33c33d3a1c1d1b1c1d19(a2c2d22b2c2d2)9(A2B)45.a3c3d3b3c3d34.计算以下矩阵的乘积:2396.(1)6614解原式09.0181311(2)0422.70118解原式6.61(3)3212.3解原式10.1(4)2321.3321解原式642.9632001002(5)010010.0030013解原式E3.a11a12a13x1(6)x1x2x3a12a22a23x2.a13a23a33x3解原式a11x12a22x22a33x322a12x1x22a13x1x32a23x2x3.1031005.已知矩阵A021,B021.求:001301(1)AB与BA;(2)(AB)(AB)与A2B2.1003103解(1)AB343,BA043;3013010906006(2)(AB)(AB)600,A2B2300.6096001a0)可互换的全部矩阵.6.求与矩阵A(a01解设与A可互换的矩阵Bx1x2.由ABBA,得x3x4x1ax3x1,x1x4,x2ax4ax1x2,x2x2,x3x3,x30,x4ax3x4,x4x4.令x2c,x4b,得Bbc,此中b,c为随意常数.0b7.利用概括法,计算以下矩阵的k次幂,此中k为正整数:cossin.(1)cossin解令Acossin,有sincosA2cos2sin2,A3cos3sin3,Lsin2cos2sin3cos3则Akcosksink.sinkcosk12.(2)10解令A12,有A21416,A418010,A3010,L,则11Ak12k.011103)011.001110解令A011,有0011211331461510A2012,A3013,A4014,A5015,L0010010010011kCk2则Ak01k.0018.已知矩阵123,111,令AT,求An,此中n为正整数.23解AnT(T)n1(T)n1(T)3n1(T)11n1133n133n2223n121223n123n1.3n133n3n3n1312329.若A为n阶对称矩阵,P为n阶矩阵,证明PTAP为对称矩阵.由于(PTAP)TATA证PTAT(PT)TPTAP,所以PTAP为对称矩阵.10.利用公式法求以下矩阵的逆矩阵:(1)A34.21141A*14解A50,又A*,所以A155.23A2355100(2)A210.3311001A*100解A10,又A*210,所以A1210.331A331122(3)A212.221解A270,又A*3A,所以A11A*1A.A91111(4)A1111111.11111解A160,又A*4A,所以A11A*1A.A411.解以下矩阵方程:(1)12X021.2321312102132021483X解232132121325.10101(2)设XAXB,此中A111,B1.1011解由XAXB,得(EA)XB.又110EA101,EA30,102则EA可逆,且XEA1B.经计算,得()11021(EA)1E(EA)*321.A3011102111所以XEA1B.()33211001110100001321(3)001X010987.01010065410110000110100解001001,010010,则01001010010010012100112331X001987010456.01065410078912.设Adiag(1,2,1),且矩阵B知足A*BA2BA8E,求矩阵B.解等式A*BA2BA8E两边左乘以A,得ABA2ABA8A.又A20,上式两边右乘以A1,得2B2AB8E,即(EA)B4E,所以B4(EA)14diag(1,1,1)2A.2213.设A,B,C都是n阶矩阵,证明:ABC可逆的充分必需条件是A,B,C都可逆.证ABC可逆ABC0ABC0A0,B0,C0A,B,C都可逆.2A1.设n阶方阵A知足A3AO,证明A2E可逆,并求2E.14证由A23AO,得(A2E)(AE)2E,即(A2E)AEE,2所以A2EA1AE.可逆,且2E215A为n阶矩阵,且A3O,证明EA及EA都是可逆矩阵..设证由A2O,得(EA)(EAA2)E及(EA)(EAA2)E,所以EA及EA都是可逆矩阵.16.已知A为三阶方阵,且A2,求:(1)2A1;(2)A*;(3)A*1A1.2解(1)原式1A1(1)311.22A16(2)原式24.A(3)A*1A1AA11A15A1,有222原式5A1(5)31125.22A16123117.设A231,求A*.312解A18,则A*1AA.A1818.(1)设P1APB,证明BkP1AkP.100100(2)设APPB,且P210,B000,求A与A2011.211001证(1)Bk(P1AP)kP1A(PP1)A(PP1)L(PP1)APP1AkP.(2)由APPB,得APBP1,且A2011PB2011P1.又100100P1210,B2011000B,411001100所以A200,A2011PBP1A.61119.利用分块矩阵计算以下矩阵的乘积:1210103001010121(1)021002.0300030003解将矩阵进行以下分块:12M1010M3001M01A1E01M21EB1LLLLL,LLLLL,OA2OB200M2100M2300M0300M03则原式A1EEB1A1A1B1B2.又OA2OB2OA2B2A1B1B2123023512123490121032,A2B203030,291251所以原式0122004.90009a0100c(2)0a01c010b0d.0010b0d解将矩阵进行以下分块:a0M100c0aM01aEEc0CLLLLL,LL,EbEdE10Mb0d001M0b0ddac则原式aEECaCdEacd.EbEdECbdEbdccbd20.利用分块矩阵求以下矩阵的逆矩阵:130(1)120.005解将矩阵进行以下分块:13M012M0A1OA,LLLLOA200M51111231A1则AO355OA21.又A121115523055A1110551005
,A21511,所以5.21001300(2)03.030042解将矩阵进行以下分块:21M0013M00A1O,ALLLLL00M33OA200M423111A11O211331则A1155132,OA2.又A11312,A242211553231050512050所以A15.1010230210232000001200(3)01300.0002500021解将矩阵进行以下分块:2M00M0LLLLLL0M12M0A0M13M0LLLLLL0M00M20M00M2则A1diag(A11,A21,A31).又111321112A122,A2131110000203200所以A10110000015.880001144110021.设矩阵A0100,利用分块矩阵计算00120021
0L0A10=A2,LA35115251,A3188,211144A4.解将矩阵进行以下分块:11M0001M00ALLLLLdiag(A1,A2),00M1200M21则A4diag(A4,A4).又A4144140,A412101240411400A40100004140004041
,所以.2500130022.设矩阵A02,利用分块矩阵计算A2012.0100122解将矩阵进行以下分块:25M0013M00ALLLLLdiag(A1,A2),00M2100M122则AA1A21(8)8,所以A2012201282012.AOBC123.(1)设A.C,且m阶矩阵B和n阶矩阵C均可逆,试证明A1OOB1O0a10L000a2L0(2)设矩阵AMMMM,此中a,a,L,a1.12n为非零常数,求000Lan1an00L0证OBOC1BB1OEO(1)由于OB1OOCC1OE,所以A可逆,CE且A1OC1B1O.(2)将矩阵进行以下分块:0Ma10L00M0a2L00MMMMOBAM00Lan1C,0OLLLLLLanM00L0则A1OC1.又BB1OA
1diag(a11,a21,L,an11),C1(an1),所以000an1a1100010a2100.00an11024.利用矩阵的初等行变换判断以下矩阵能否可逆;如可逆,求其逆矩阵.130(1)312.433103M130130M100510101M31解AE312M010r010.433M00151010000M131221035由于011E,所以A不行逆.5000122(2)212.221100M122122M100999010M212,解AE212M010r221M001999001M221999122999所以A可逆,且A1212.9992219993201(3)0221.123201213201M1000解AE0221M01001232M00100121M00011000M1124r0100M0101,0010M11360001M216101124所以A可逆,且A10101.11362161011111111(4)A.111111111111M10001111M0100解AE111M001011111M00011000M101022r0100M011022,0011M1100220000M0011所以A不行逆.25.利用矩阵的初等行变换解以下矩阵方程:123130(1)324X1027.2101078123M130100M645解324M1027r010M212EX,210M1078001M333645所以X212.333531830(2)X132590.5212150515852解将方程两边转置,得332XT3915.由121000515M100100M147332M010r010M258EXT,121M001001M369123得X456.78926.求以下矩阵的秩:1562(1)2132.143015621562解A2132r0992,所以R(A)2.1430000021324(2)42517.211822132421324解A42517r00151R(A)2.211820000013122123(3)21.31143513121312解A2123r0747R(A)2.321100001435000031325(4)53234.13507751413132513507解A53234r049113135070000R(A)3.1751410000012127.设矩阵A251,且R(A)3,求的值.1161012111610解A251r01510.116100033(3)由R(A)3,得3.123k28.设矩阵A12k3,问k取何值时,使得k23(1)R(A)1;(2)R(A)2;(3)R(A)3.123k123k解A12k3r02(k1)3(k1),有k23003(k1)(k2)当k1且k2时,R(A)3;当k1时,R(A)1;当k2时,R(A)2.10129.设A是43矩阵,且A的秩为2,而B111,求R(AB).123解B20,则R(AB)R(A)2.30.设A为n阶矩阵,知足A25A6EO,证明:R(A2E)R(A3E)n.证由A25A6EO,得(A2E)(A3E)O,所以R(A2E)R(A3E)n.又R(A2E)R(A3E)R(A2E)R(A3E)R(E)n,所以R(A2E)R(A3E)n.11031.设三阶矩阵A212,试求R(A)与R(A*).122110110解A212r012R(A)2.122000由于R(A)231R(A*)1.32.求解以下线性方程组:x1x24x30,(1)2x19x26x30,3x15x22x30.解方程组的系数矩阵114114A296r012.352001由于R(A)3,所以方程组只有零解.x12x2x31,(2)2x1x2x33,x12x23x37.解方程组的增广矩阵121M1100M1BA211M3r010M1,123M7001M2所以方程组的解为x1,x21,x32.12x13x2x37x40,3x1x22x37x40,(3)4x1x23x36x40,x12x25x35x40.解方程组的系数矩阵10012317273127r010A2,41360015125520000得方程组的解为x11x4,2x27x4,2x35x4.2令x42c,得方程组的通解Xc(1,7,5,2)T,此中c为随意常数.x1x2x3x40,(4)2x13x2x3x42,3x12x2x3x45,3x16x2x3x44.解方程组的增广矩阵1111M01111M0BA2311M2r0133M23211M5.0055M73611M40000M5由于R(A)3R(B)4,所以方程组无解.2x1x25x315,(5)x13x2x34,x14x26x311,3x12x24x319.解方程组的增广矩阵215M15102M7BA131M4r011M1146M11000M,0324M19000M0得方程组的解为x12x37,x2x31.令x3c,得方程组的通解Xc(2,1,1)T(7,1,0)T,此中c为随意常数.x1x23x32x41,(6)x1x22x3x42,2x12x27x37x41,2x12x28x310x40.解方程组的增广矩阵1132M11107M4BA1121M2r0013M12277M10000M,022810M00000M0得方程组的解为x1x27x44,x33x41.令x2c1,x4c2,得方程组的通解为Xc1(1,1,0,0)Tc2(7,0,3,1)T(4,0,1,0)T,此中c1,c2为随意常数.33.试问取何值时,以下非齐次线性方程组无解、有独一解、有无量多解.(1)x1x2x31,(1)x1(1)x2x3,x1x2(1)x31.解方程组的系数队列式111A111(3)2.111当A0,即0且3时,方程组有独一解.当0时,111M1111M1BA111M0r000M1.111M1000M0由于R(A)1R(B)2,所以方程组无解.当3时,211M1112M2BA121M3r033M5.112M2000M0由于R(A)R(B)23,所以方程组有无量多解.(2)x12x22x31,(2)2x1(5)x24x32,2x14x2(5)x31.解方程组的系数队列式222222r3r2)(1)2.A254254(10245011当A0,即1且10时,方程组有独一解.当10时,822M1254M2BA254M2r011M1.245M11000M1由于R(A)2R(B)3,所以方程组无解.当1时,122M1122M1BA244M2r000M0.244M2000M0由于R(A)R(B)13,所以方程组有无量多解.x1x3,34.试问取何值时,非齐次线性方程组4x1x22x32,有解,并求解.6x1x24x323解方程组的增广矩阵101M101MBA412M2r012M23.614M23000M1当1时,BA
101M1r012M1,000M0有R(A)R(B)23,则方程组有无量多解,且解为x1x31,x22x31.令x3c,得方程组的通解为Xc(1,2,1)T(1,1,0)T,此中c为随意常数.35.求平面上三点(x1,y1),(x2,y2),(x3,y3)共线的充分必需条件.解设直线方程为axbyc0.则x1ay2bc0,平面上三点(x1,y1),(x2,y2),(x3,y3)共线xaybc0,有非零解22x3ay3bc0x1y11111x2y210,即x1x2x30.x3y31y1y2y3(B)1.选择题:(1)设A,B为n阶矩阵,以下结论正确的选项是().(A)若A、B是对称矩阵,则AB也是对称矩阵.(B)ABABA2B2.(C)若ABO,且A可逆,则BO.(D)若A与B等价,则A与B相等.解选(C).2nn矩阵,则必有().()设A和B均为(A)AB=A+B.(B)ABBA.(C)AB=BA.(D)AB1A1B1.解选(C).(3)设A为n(n2)阶矩阵,A*是A的陪伴矩阵,k为常数,则(kA)*().(A)A*.(B)kA*.(C)kn1A*.(D)knA*.解由陪伴矩阵的定义,知选(C).(4)设A和B均为n阶非零矩阵,且ABO,则A和B的秩().(A)必有一个等于零.(B)一个等于n,一个小于n.(C)都等于n.(D)都小于n.解由ABO,得R(A)R(B)n.又AO,BO,知R(A)1,R(B)1.所以R(A)n,R(B)n,应选(D).(5)对于非齐次线性方程组AmnXn1m1,若R(A)r,则().(A)当rm时,AmnXn1m1有解.(B)当rn时,AmnXn1m1有独一解.(C)当mn时,AmnXn1m1有独一解.(D)当rn时,AmnXn1m1有无量多解.解当rm时,mR(B)R(A)rmR(A)R(B)r,应选(A).213421262.设矩阵A02130334,试求(EA1B)TAT002,B0045.100020005解(EA1B)TAT(A(EA1B))T(AB)T,则(EA1B)TAT(AB)TAB0.1013.设矩阵A020,且A2BABE,试求B.301解由A2BABE,得(AE)(AE)BAE.又AE30,有(AE)BE,两边取队列式,得AEB1,所以11B.AE31004.设矩阵A230,且B(EA)1(EA),试求(EB)1.045解EB(EA)1(EA)(EA)1(EA)2(EA)1,则100(EB)11(EA)120.22300105.设矩阵A100,BP1AP,试求B2012A2.001解A2diag(1,1,1)A4E,所以200B2012A2(B4)503A2EA2020.0001116.设矩阵A111,矩阵X知足A*XA12X,试求矩阵X.111解由A*XA12X,得AXE2AX.又11022经计算可得(2EA)1011,所以X2210122
A4,有X1(2EA)1.2110440114.41014411002,且矩阵X知足AX2X,试求矩阵X.7.设矩阵A210,011012解由AX2X,得(A2E)X.(注意A2E0)又110M0101M12r2A2E210M0012M1,112M1000M02c112得方程组的解为x1x32,令x3c,得X2c1,c为随意常数.x22x31.c1aLa8.设n(n3)阶矩阵Aa1La,试求A的秩R(A).MMMaaL1解A[1(n1)a](1a)n1.当a1且an1时,A为非奇怪矩阵,所以R(A)n;111L111L1当a1时,A11L1r00L0,则R(A)1;MMMMMM11L100L0当a1时,A的n1阶子式n11aLaDn1a1La[1(n2)a](1a)n20MMMaaL1n1而A0,所以R(A)n1.x1x22x33x40,9.试求p,q取何值时,齐次线性方程组2x1x26x34x40,3x12x2px3qx4有非零解,并求通解.0,2x1x2x40解方程组的系数矩阵112311232164r0122A2pq002.312101000p2q6当p2q6时,R(A)34,方程组有非零解,且10010103r1A01,020000得方程组的解为x1x4,x23x4,x31x4.2令x42c,得方程组的通解为Xc(2,6,1,2)T,此中c为随意常数.2x1ax2x31,10.试求a取何值时,非齐次线性方程组ax1x2x32,无解、有独一解或无量多解,4x15x25x31并在有无量多解时求方程组的通解.解方程组的系数队列式2a1Aa11(a1)(5a4).455当a1且a4时,方程组有独一解.5当a1时,211M2100M1BA111M1r011M1.455M0000M0由于R(A)R(B)23,所以方程组有无量多解,且通解为Xc(0,1,1)T(1,1,0)T,此中c为随意常数.当a4时,方程组无解.512211.设矩阵A4t3,B为三阶非零矩阵.试求常数t,使得ABO.311解ABO,BOAX0有非零解A0.又A7(t3),所以t3.12.证明:(1)设A,B为矩阵,则ABBA存心义的充分必需条件是A,B为同阶矩阵.(2)对随意n阶矩阵A,B,都有ABBAE,此中E为单位矩阵.证(1)设A为mn矩阵,B为st矩阵,则ns,tm,ABBA存心义mnst,ms,t
n.即A,B
为同阶矩阵.(2)设
A
(aij)nn,B
(bij)nn,则AB
BA的主对角线上元素之和为n
n
n
n
nn
n
naikbki
bstats
aikbki
atsbst
0,i1k1
s1t1
i1
k1
t1
s1而E的主对角线上元素之和为
n,所以
AB
BA
E.13.证明:随意n阶矩阵都可表示为一个对称矩阵与一个反对称矩阵的和.证设A为随意n阶矩阵,则AATAATA,22此中为AAT对称矩阵,AAT为反对称矩阵.(你能否能联系到函数能够表示为奇函数22与偶函数之和)14n阶矩阵A,B知足ABAB,试证AE可逆,并求(AE)..已知1证由ABAB,得(AE)(BE)E,所以AE可逆,且(AE)1BE.15.设A为元素全为1的n(n1)阶方阵,证明:EA1E1A.n1证EA(En1A)EnA1A2.又A2nA,故1n1n1EA(E1A)E,11n所以EA1EA.n116.设n阶矩阵A与B等价,且A0,证明B0.证A与B等价,则存在n阶可逆矩阵P与Q,使得BPAQ,有BPAQPAQ0.注:此结论告诉我们初等变换不改变矩阵的可逆性.17.设A为n阶方阵,且A2A,证明RARAEn.证由于A(AE)A2AO,所以RARAEn.又RARAERARAER(E)n,所以RARAEn.18.设A是nm矩阵,B是mn矩阵,此中nm.若ABE,此中E为n阶单位矩阵.证明方程组BXO只有零解.证由ABE,得R(AB)n.又nR(B)R(AB)n,得R(B)n,所以方程组BXO只有零解.习题四A)1.设v1(1,1,0)T,v2(0,1,1)T,v3(3,4,0)T,求v1v2和3v12v2v3.解v1v2(1,0,1)T,3v12v2v3(0,1,2)T.2.求解以下向量方程:(1)3X,此中(1,0,1)T,(1,1,1)T.解X1()1(0,1,2)T.33(2)2X33X,此中(2,0,1)T,(3,1,1)T.解X3(3,1,4)T.3.试问向量能否由向量组1,2,3,4线性表示若能,求出由1,2,3,4线性表示的表达式.1112;111(1),2,3111111解设x11x22x33x44.由
111,41.11111000M51111M140100M11111M2(1,2,3,4|)r4,1111M110010M1111M140001M14得R(1,2,3,4)R(1,2,3,4|)4,所以可由向量组1,2,3,4线性表示,且x151,x3111(54).,x24,x44,得表达式12344401111(2)2;11,21,31,40.0110011000解设x11x22x33x44.由1111M01000M1(1,2,3,4|)1110M2r0100M11100M00010M,21000M10001M2得R(1,2,3,4)R(1,2,3,4|)4,所以可由向量组1,2,3,4线性表示,且x11,x21,x32,x42,得表达式122324.4.议论以下向量组的线性有关性:32132(1)12,21,33,41,53.45571解向量组所含向量个数大于向量的维数,所以该向量组线性有关.axayaz(2)1bx,2by,3bz,此中a,b,c,x,y,z全不为零.cxcycz解1,2对应的重量成比率,则1,2线性有关,所以该向量组线性有关.11(3)011,22,3124112解1,2,3013r121240
23.101213.001000由于R1,2,33,所以该向量组线性没关.11122,32(4)1,2,3010301110解(1,2,32220,4)01130301
004.111110r0301001.00000由于R(1,2,3,4)34,所以该向量组线性有关.5.(1)设Rn,证明:线性有关当且仅当0.(2)设1,2Rn,证明:1,2线性有关当且仅当它们对应的重量成比率.证(1)线性有关k0,k00.(2)1,2线性有关k11k220,此中k,k2不全为零.不
温馨提示
- 1. 本站所有资源如无特殊说明,都需要本地电脑安装OFFICE2007和PDF阅读器。图纸软件为CAD,CAXA,PROE,UG,SolidWorks等.压缩文件请下载最新的WinRAR软件解压。
- 2. 本站的文档不包含任何第三方提供的附件图纸等,如果需要附件,请联系上传者。文件的所有权益归上传用户所有。
- 3. 本站RAR压缩包中若带图纸,网页内容里面会有图纸预览,若没有图纸预览就没有图纸。
- 4. 未经权益所有人同意不得将文件中的内容挪作商业或盈利用途。
- 5. 人人文库网仅提供信息存储空间,仅对用户上传内容的表现方式做保护处理,对用户上传分享的文档内容本身不做任何修改或编辑,并不能对任何下载内容负责。
- 6. 下载文件中如有侵权或不适当内容,请与我们联系,我们立即纠正。
- 7. 本站不保证下载资源的准确性、安全性和完整性, 同时也不承担用户因使用这些下载资源对自己和他人造成任何形式的伤害或损失。
最新文档
- 《凡客发展现状》课件
- 2024年江沙区域产品购销协议版B版
- 房屋装修工程承包合同(2篇)
- 2024年版铁路客运服务协议范本一
- 2024实木家俱定制与家具维护保养服务合同2篇
- 2025年泸州道路运输从业资格证考哪些项目
- 2025年凉山州驾驶员货运从业资格证模拟考试题
- 2025年河北道路运输从业人员从业资格考试
- 《改善提案》课件
- 2024年建筑土建主体工程分包劳务协议样本版B版
- 工程建设监理收费标准(发改价格【2007】670号)
- 摩托车品牌文化营销与品牌故事的构建
- 2024江苏南京大数据集团有限公司招聘笔试参考题库附带答案详解
- FZT 73032-2017 针织牛仔服装
- 企业并购与资产重组智慧树知到期末考试答案2024年
- 货物包装承诺函
- 治疗用碘131I化钠胶囊-临床用药解读
- 2024人教版五年级上册数学期末口算题训练
- 2024外研版初中英语单词表汇总(七-九年级)中考复习必背
- 安徽省合肥市包河区2023-2024学年三年级上学期期末英语试卷
- 劳动争议调解仲裁法
评论
0/150
提交评论