同济第五版线性代数课后习题答案.pdf

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3811411022 ( 4) 3 0 ( 1) ( 1) 1 1 8 0 1 3 2 ( 1) 8 1 ( 4) ( 1) 24 8 16 44(2)bacacbcba解bacacbcbaacb bac cba bbb aaa ccc 3abc a3b3c3(3)222111cbacba解222111cbacbabc2ca2ab2ac2ba2cb2(a b)(b c)(c a)(4)yxyxxyxyyxyx解yxyxxyxyyxyxx(x y)y yx(x y) (x y)yx y3(x y)3x33xy(x y) y33x2y x3y3x32(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)1 2 3 4解逆序数为 0 (2)4 1 3 2解逆序数为 441 43 42 32(3)3 4 2 1解逆序数为 53 2 3 1 4 2 4 1, 2 1(4)2 4 1 3解逆序数为 32 1 4 1 4 3(5)1 3 (2n 1) 2 4 (2n)解逆序数为2) 1(nn3 2 (1 个) 5 2 5 4(2 个) 7 2 7 4 7 6(3 个) (2n 1)2 (2n 1)4 (2n 1)6 (2n 1)(2n 2)(n 1 个) (6)1 3 (2n 1) (2n) (2n 2) 2解逆序数为 n(n 1) 3 2(1 个) 5 2 5 4 (2 个) (2n 1)2 (2n 1)4 (2n 1)6 (2n 1)(2n 2)(n 1 个) 4 2(1 个) 6 2 6 4(2 个) (2n)2 (2n)4 (2n)6 (2n)(2n 2)(n 1 个) 3 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 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)ta11a23a32a44( 1)1a11a23a32a44a11a23a32a44( 1)ta11a23a34a42( 1)2a11a23a34a42a11a23a34a424 计算下列各行列式(1)71100251020214214解71100251020214214010014231020211021473234cccc34) 1(14310221101414310221101401417172001099323211cccc(2)2605232112131412解2605232112131412260503212213041224cc041203212213041224rr0000003212213041214rr(3)efcfbfdecdbdaeacab解efcfbfdecdbdaeacabecbecbecbadfabcdefadfbce4111111111(4)dcba100110011001解dcba100110011001dcbaabarr10011001101021dcaab101101) 1)(1(1201011123cdcadaabdcccdadab111) 1)(1(23abcd ab cd ad 15 证明: (1)1112222bbaababa(a b)3; 证明1112222bbaababa00122222221213ababaabaabaccccabababaab22) 1(2221321)(abaabab(a b)3(2)yxzxzyzyxbabzaybyaxbxazbyaxbxazbzaybxazbzaybyax)(33; 证明bzaybyaxbxazbyaxbxazbzaybxazbzaybyaxbzaybyaxxbyaxbxazzbxazbzayybbzaybyaxzbyaxbxazybxazbzayxabzayyxbyaxxzbxazzybybyaxzxbxazyzbzayxa22zyxyxzxzybyxzxzyzyxa33yxzxzyzyxbyxzxzyzyxa33yxzxzyzyxba)(33(3)0) 3()2() 1()3()2() 1()3()2() 1()3()2() 1(2222222222222222ddddccccbbbbaaaa; 证明2222222222222222) 3()2() 1() 3()2() 1()3()2() 1() 3()2() 1(ddddccccbbbbaaaa(c4c3c3c2c2c1得) 5232125232125232125232122222ddddccccbbbbaaaa(c4c3c3c2得) 022122212221222122222ddccbbaa(4)444422221111dcbadcbadcba(a b)(a c)(a d)(b c)(b d)(c d)(a b c d); 证明444422221111dcbadcbadcba)()()(0)()()(001111222222222addaccabbaddaccabbadacab)()()(111)()(222addaccabbdcbadacab)()(00111)()(abdbddabcbccbdbcadacab)()(11)()()()(abddabccbdbcadacab=(a b)(a c)(a d)(b c)(b d)(c d)(a b c d)(5)1221100000100001axaaaaxxxnnnxna1xn 1an 1x an证明用数学归纳法证明当 n 2 时2121221axaxaxaxD命题成立假设对于 (n 1)阶行列式命题成立即Dn 1xn 1a1xn 2an 2x an 1则 Dn按第一列展开有1110010001) 1(11xxaxDDnnnnxDn 1anxna1xn 1an 1x an因此 对于 n 阶行列式命题成立6 设 n 阶行列式 D det(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nnnnaaaaD1111111112nnnnaaaaD11113aaaaDnnnn证明DDDnn2)1(21) 1(D3D证明因为 D det(aij) 所以nnnnnnnnnnaaaaaaaaaaD2211111111111) 1() 1() 1(331122111121nnnnnnnnaaaaaaaaDDnnnn2) 1() 1()2(21) 1() 1(同理可证nnnnnnaaaaD) 1(11112)1(2DDnnTnn2) 1(2) 1() 1() 1(DDDDDnnnnnnnn)1(2) 1(2)1(22)1(3) 1() 1() 1() 1(7 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aaaaaDn00010000000000001000(按第 n 行展开 ) ) 1() 1(100000000000010000) 1(nnnaaa)1() 1(2) 1(nnnaaannnnnaaa)2)(2(1) 1() 1(anan 2an 2(a21)(2)xaaaxaaaxDn; 解将第一行乘 ( 1)分别加到其余各行得axxaaxxaaxxaaaaxDn0000000再将各列都加到第一列上得axaxaxaaaanxDn0000000000) 1(x (n 1)a(x a)n 1(3)1111)() 1()() 1(1111naaanaaanaaaDnnnnnnn; 解根据第 6 题结果 有nnnnnnnnnnaaanaaanaaaD)() 1()() 1(1111) 1(1112) 1(1此行列式为范德蒙德行列式112)1(1)1() 1() 1(jinnnnjaiaD112)1()() 1(jinnnji1121) 1(2) 1()() 1() 1(jinnnnnji11)(jinji(4)nnnnndcdcbabaD11112; 解nnnnndcdcbabaD11112(按第 1 行展开 ) nnnnnnddcdcbabaa00001111111100) 1(1111111112cdcdcbababnnnnnnn再按最后一行展开得递推公式D2nandnD2n 2bncnD2n 2即 D2n(andnbncn)D2n 2于是niiiiinDcbdaD222)(而111111112cbdadcbaD所以niiiiincbdaD12)(5) D det(aij) 其中 aij|i j|; 解 aij|i j|0432140123310122210113210)det(nnnnnnnnaDijn04321111111111111111111112132nnnnrrrr15242321022210022100021000011213nnnnncccc( 1)n 1(n 1)2n 2(6)nnaaaD11111111121, 其中 a1a2an0解nnaaaD11111111121nnnnaaaaaaaaacccc1000010001000100010000113322121321111312112111000011000001100001100001nnnaaaaaaaaniinnaaaaaaaa11111312112110000010000001000001000001)11)(121niinaaaa8 用克莱姆法则解下列方程组(1)01123253224254321432143214321xxxxxxxxxxxxxxxx解因为14211213513241211111D142112105132412211151D284112035122412111512D426110135232422115113D14202132132212151114D所以111DDx222DDx333DDx144DDx(2)150650650651655454343232121xxxxxxxxxxxxx解因为6655100065100065100065100065D150751001651000651000650000611D114551010651000650000601000152D70351100650000601000051001653D39551000601000051000651010654D21211000051000651000651100655D所以66515071x66511452x6657033x6653954x6652124x9 问取何值时 齐次线性方程组0200321321321xxxxxxxxx有非零解？解系数行列式为1211111D令 D 0 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 齐次线性方程组0)1(0)3(2042)1 (321321321xxxxxxxxx有非零解？解系数行列式为101112431111132421D(1)3(3) 4(1) 2(1)( 3) (1)32(1)23令 D 0 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1 已知线性变换3213321232113235322yyyxyyyxyyyx求从变量 x1x2x3到变量 y1y2y3的线性变换解由已知221321323513122yyyxxx故3211221323513122xxxyyy321423736947yyy321332123211423736947xxxyxxxyxxxy2 已知两个线性变换32133212311542322yyyxyyyxyyx323312211323zzyzzyzzy求从 z1z2z3到 x1x2x3的线性变换解由已知221321514232102yyyxxx321310102013514232102zzz321161109412316zzz所以有32133212321116109412。