(高等教育)线性代数课后习题1-4作业答案

高等教育出版社线性代数课后习题1-4作业答案第一章行列式1利用对角线法则计算下列三阶行列式(1)381141102解3811411022(4)30(1)(1)1180132(1)81(4)(1)2481644(3)222111cbacba解222111cbacbabc2ca2ab2ac2ba2cb2(ab)(bc)(ca)4计算下列各行列式(1)71100251020214214解71100251020214214010014231020211021473234cccc34)1(14310221101414310221101401417172001099323211cccc(2)2605232112131412高等教育出版社线性代数课后习题1-4作业答案解2605232112131412260503212213041224cc041203212213041224rr0000003212213041214rr(3)efcfbfdecdbdaeacab解efcfbfdecdbdaeacabecbecbecbadfabcdefadfbce4111111111(4)dcba100110011001解dcba100110011001dcbaabarr10011001101021dcaab101101)1)(1(1201011123cdcadaabdcccdadab111)1)(1(23abcdabcdad16.证明:(1)1112222bbaababa(ab)3;证明1112222bbaababa00122222221213ababaabaabacccc高等教育出版社线性代数课后习题1-4作业答案abababaab22)1(2221321))((abaabab(ab)3(2)yxzxzyzyxbabzaybyaxbxazbyaxbxazbzaybxazbzaybyax)(33;证明bzaybyaxbxazbyaxbxazbzaybxazbzaybyaxbzaybyaxxbyaxbxazzbxazbzayybbzaybyaxzbyaxbxazybxazbzayxabzayyxbyaxxzbxazzybybyaxzxbxazyzbzayxa22zyxyxzxzybyxzxzyzyxa33yxzxzyzyxbyxzxzyzyxa33yxzxzyzyxba)(338.计算下列各行列式(Dk为k阶行列式)(1)aaDn11,其中对角线上元素都是a未写出的元素都是0解高等教育出版社线性代数课后习题1-4作业答案aaaaaDn00010000000000001000(按第n行展开))1()1(100000000000010000)1(nnnaaa)1()1(2)1(nnnaaannnnnaaa)2)(2(1)1()1(anan2an2(a21)(2)xaaaxaaaxDn;解将第一行乘(1)分别加到其余各行得axxaaxxaaxxaaaaxDn0000000再将各列都加到第一列上得axaxaxaaaanxDn0000000000)1([x(n1)a](xa)n第二章矩阵及其运算1.计算下列乘积(5)321332313232212131211321)(xxxaaaaaaaaaxxx解高等教育出版社线性代数课后习题1-4作业答案321332313232212131211321)(xxxaaaaaaaaaxxx(a11x1a12x2a13x3a12x1a22x2a23x3a13x1a23x2a33x3)321xxx322331132112233322222111222xxaxxaxxaxaxaxa2.设111111111A150421321B求3AB2A及ATB解1111111112150421321111111111323AAB2294201722213211111111120926508503092650850150421321111111111BAT3.已知两个线性变换32133212311542322yyyxyyyxyyx323312211323zzyzzyzzy求从z1z2z3到x1x2x3的线性变换解由已知221321514232102yyyxxx321310102013514232102zzz321161109412316zzz高等教育出版社线性代数课后习题1-4作业答案所以有3213321232111610941236zzzxzzzxzzzx4.设3121A2101B问(1)ABBA吗?解ABBA因为6443AB8321BA所以ABBA(3)(AB)(AB)A2B2吗?解(AB)(AB)A2B2因为5222BA1020BA906010205222))((BABA而718243011148322BA故(AB)(AB)A2B25.举反列说明下列命题是错误的(1)若A20则A0解取0010A则A20但A0(2)若A2A则A0或AE解取0011A则A2A但A0且AE(3)若AXAY且A0则XY解取0001A1111X1011Y高等教育出版社线性代数课后习题1-4作业答案则AXAY且A0但XY7.设001001A求Ak解首先观察0010010010012A2220020123232323003033AAA43423434004064AAA545345450050105AAAkAkkkkkkkkkk0002)1(121用数学归纳法证明当k2时显然成立假设k时成立，则k1时，0010010002)1(1211kkkkkkkkkkkkAAA11111100)1(02)1()1(kkkkkkkkkk高等教育出版社线性代数课后习题1-4作业答案由数学归纳法原理知kkkkkkkkkkkA0002)1(1218.设AB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ATA所以(BTAB)TBT(BTA)TBTATBBTAB从而BTAB是对称矩阵11求下列矩阵的逆矩阵(1)5221解5221A|A|1故A1存在因为1225*22122111AAAAA故*||11AAA1225(3)145243121解145243121A|A|20故A1存在因为214321613024*332313322212312111AAAAAAAAAA高等教育出版社线性代数课后习题1-4作业答案所以*||11AAA1716213213012(4)naaa0021(a1a2an0)解naaaA0021由对角矩阵的性质知naaaA1001121112.利用逆矩阵解下列线性方程组(1)3532522132321321321xxxxxxxxx解方程组可表示为321153522321321xxx故0013211535223211321xxx从而有001321xxx19.设P1AP其中1141P2001求A11解由P1AP得APP1所以A11A=P11P1.高等教育出版社线性代数课后习题1-4作业答案|P|31141*P1141311P而11111120012001故31313431200111411111A6846832732273120.设APP其中111201111P511求(A)A8(5E6AA2)解()8(5E62)diag(1158)[diag(555)diag(6630)diag(1125)]diag(1158)diag(1200)12diag(100)(A)P()P1*)(||1PPP1213032220000000011112011112111111111421.设AkO(k为正整数)证明(EA)1EAA2Ak1证明因为AkO所以EAkE又因为EAk(EA)(EAA2Ak1)所以(EA)(EAA2Ak1)E由定理2推论知(EA)可逆且(EA)1EAA2Ak1证明一方面有E(EA)1(EA)高等教育出版社线性代数课后习题1-4作业答案另一方面由AkO