同济大学线性代数第六版答案(全)1 利用对角线法则计算下列三阶行列式201(1)1 4*****解1 41832 ( 4)3 0 ( 1) ( 1) 1 1 8 0 1 3 2 ( 1) 8 1 ( 4) ( 1) 24 8 16 4 4 abc(2)bcacababc解bcacabacb bac cba bbb aaa ccc 3abc a3 b3 c3111(3)abca2b2c2111解abca2b2c2bc2 ca2 ab2 ac2 ba2 cb2 (a b)(b c)(c a)xyx y(4)yx yxx yxyxyx y解yx yxx yxyx(x y)y yx(x y) (x y)yx y3 (x y)3 x3 3xy(x y) y3 3x2 y x3 y3 x3 2(x3 y3)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4 41 43 42 32 (3)3 4 2 1解逆序数为5 3 2 3 1 4 2 4 1, 2 1 (4)2 4 1 3解逆序数为3 2 1 4 1 4 3 (5)1 3 (2n 1) 2 4 (2n)n(n 1)解逆序数为2 3 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)1a11a23a32a44 a11a23a32a44 ( 1)ta11a23a34a42 ( 1)2a11a23a34a42 a11a23a34a42 4 计算下列各行列式41 (1)***-*****142 0741 解***-*****c2 c***** 1***** 104 1 102 122 ( 1)43 *****c4 7c*****3 14 4 110c2 c*****123 142c00 2 01 2c*****2 (2)31 1***** 224 解31 ***** c 4 c3 223 1202r 4 r ***-*****06***-*****62***-***** 40r4 140r12***-*****300 0 0(3) bdabbf acaecfcd deef解bdabbf accfcdaedeefad bbb ccce eeadfbce 11111 11 1 4abcdefa1 (4) 001b 1001c 100 1da1 解001b 1001c 10r1 ar201 ab0 1b10 1d00a 1c 100 1daba0c3 dc2 abaad( 1)( 1)2 1 1c1 1c1 cd0 100 1d5 证明:abad abcd ab cd ad 1 ( 1)( 1)3 2 11 cda2abb2(1)2aa b2b (a b)3;111 证明a2abb2c2 c1a2ab a2b2 a22aa b2b 2ab a2b 2a00111c3 c11ab ab aab a (a b)3 (b a)(b a)1 ( 1)2b a2b 2a 3 1ax byay bzaz bxxyz(2)ay bzaz bxax by (a3 b3)yzx;az bxax byay bzzxy证明ax byay bzaz bxay bzaz bxax byaz bxax byay bzxay bzaz bxyay bzaz bxayaz bxax by bzaz bxax byzax byay bzxax byay bzxay bzzyzaz bxa2yaz bxx b2zxax byzax byyxyay bzxyzyzxa3yzx b3zxyzxyxyzxyzxyza3yzx b3yzxzxyzxyxyz(a3 b3)yzxzxya2b2(3)2cd2(a 1)2(b 1)2(c 1)2(d 1)2(a 2)2(b 2)2(c 2)2(d 2)2(a 3)2(b 3)20; (c 3)2(d 3)2证明a2b2cd2(a 1)2(b 1)2(c 1)2(d 1)2(a 2)2(b 2)2(c 2)2(d 2)2(a 3)2(b 3)2(c c c c c c得) (c 3)***-*****(d 3)2a22b c2d22a 12b 12c 12d 12a 32b 32c 32d 32a 52b 5(c c c c得) 2c *****d 5a22b c2d2 1a (4)a2a41bb2b42a 12b 12c 12d 11cc2c41d d2d4*****2 0 22(a b)(a c)(a d)(b c)(b d)(c d)(a b c d); 证明1a a2a41bb2b41cc2c41d d2d41110b ac ad a0b(b a)c(c a)d(d a)0b2(b2 a2)c2(c2 a2)d2(d2 a2)111(b a)(c a)(d a)b cd2(b a)c2(c a)d2(d a)11(b a)(c a)(d a)0 c bd b0c(c b)(c b a)d(d b)(d b a)1 (b a)(c a)(d a)(c b)(d b)c(c 1b a)d(d b a)=(a b)(a c)(a d)(b c)(b d)(c d)(a b c d) x0an1x 0an 10 1 0an 20000xn a1xn 1 an 1x an x 1a2x a1用数学归纳法证明x 1 x2 ax a 命题成立当n 2时D2 a122x a1假设对于(n 1)阶行列式命题成立即Dn 1 xn 1 a1 xn 2 an 2x an 1 则Dn按第一列展开有1Dn xDn 1 an( 1)n 1 x10 1 100 x00 1xD n 1 an xn a1xn 1 an 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an1 anna1n annann a1nD1 D2 D3a11 a1na11 an1an1 a11证明D1 D2 ( 1)n(n 1)2D D3 Da1nann a2n证明因为D det(aij) 所以a11an1 annD1 ( 1)n 1an1a21a11a21( 1)n 1( 1)n 2an1a31a1na2nann a3nn(n 1)2( 1)1 2 (n 2) (n 1)D ( 1) 同理可证D2 ( 1) D3 ( 1)n(n 1)2Da11 an1n(n 1)n(n 1)( 1)2DT ( 1)2D a1n annn(n 1)2D2 ( 1)n(n 1)2( 1)n(n 1)2D ( 1)n(n 1)D D7 计算下列各行列式(Dk为k阶行列式) (1)Dn 都是0 解a0 Dn 0010a0 0000a 00000 a0100(按第n行展开) 0aa1 1a, 其中对角线上元素都是a 未写出的元素( 1)0000a 0000 0 000 a1a0( 1)2n a 0a(n 1) (n 1)0(n 1) (n 1)an 1n( 1) ( 1)a(n 2)(n 2)an an an 2 an 2(a2 1)x(2)Dn aaax a aa; xa 0 0 0x a解将第一行乘( 1)分别加到其余各行得xaaa xx a0 Dn a x0x aa x00再将各列都加到第一列上得x (n 1)aaa0x a0Dn 00x a000an(a 1)nan 1(a 1)n 1(3)Dn 1aa 111a0n 1 0 [x (n 1)a](x a) 0x a(a n)n (a n)n 1;a n 1解根据第6题结果有11a 1n(n 1)a Dn 1 ( 1)2an 1(a 1)n 1an(a 1)n1 a nn 1(a n) (a n)n此行列式为范德蒙德行列式Dn 1 ( 1) ( 1) ( 1)ann(n 1)2n 1 i j 1n(n 1)2[(a i 1) (a j 1)]n 1 i j 1n(n 1)2[ (i j)]n (n 1) 12( 1)n 1 i j 1(i j)n 1 i j 1(i j)bn(4)D2ncna1b1c1d1; dn解anbnD2ncna1b1c1d1(按第1行展开) dnan 1ancn 10a1b1c1d1bn 10dn 100dn0an 1( 1)2n 1bncn 1na1b1c1d1bn 1dn 10再按最后一行展开得递推公式D2n andnD2n 2 bncnD2n 2 即D2n (andn bncn)D2n 2 于是D2n (aidi bici)D2i 2n而D2a1b1a1d1 b1c1 c1d1ni 1所以D2n (aidi bici) (5) D det(aij) 其中aij |i j|; 解aij |i j| 01 Dn det(aij) 23n 1***-*****1210 n 2n 3n 4n 1n 2n 3 n 4 01r1 r2 11 1r2 r3n 11 1 1 1 n 211 11 11 1 1 n 3n 4 111 1 01c2 c1 11 1c3 c1n 1 a11(6)Dn 11 a211000200 2 20 2 2 2 2n 32n 42n 5 000 0 n ( 1)n 1(n 1)2n 21 1, 其中a1a2 an 01 an解a11Dn 11 a21111 1 ana1c1 c2 a200a2 a3 0000a3 00 0 0 0 an 1 00***** an 11 an1 an1 1a1a2 an00001 1 00001 00 000 100a1 110a210a311an 1111 an100a1a2 an0010 0001 0 000 0000 1a1 1 1a2 1a3 1an 1ni 1000 001 ai 1(a1a2 an)(1 1)i 1ai8 用克莱姆法则解下列方程组x1 x2 x3 x4 5 x 2x2 x3 4x4 2 (1) 12x1 3x2 x3 5x4 2 3x x 2x 11x 0 1234n解因为1312 311 1 1214 142 51152 D1 20 D3 23所以x112 311 1 1214 142 D 2 52 11352 201 1 1214 284 51112 315 2 ***** 426 D 1 42 511312 311 1 1252 142 20DDDD1 x2 2 x3 3 x4 1DDDD1 5x1 6x20 x1 5x2 6x3(2) x2 5x3 6x4 0x3 5x4 6x5 0x4 5x5 1解因为51D 000***-********-*****0 665 65D1 00 D3 ***-*****00***-*****01***-*****5105010 1507 D2 ***-*****10 703 D4 060501***-*****0***-*****10***-*****01000 1145 65000 395 655 D5 0***-********-*****1100 212 01所以x1 1507 x2 1145 x3 703 x4 395 x4 212***-********-*****x1 x2 x3 09 问取何值时齐次线性方程组x1 x2 x3 0有非x1 2 x2 x3 0零解？解系数行列式为1 D 112 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(1 )x1 2x2 4x3 010 问取何值时齐次线性方程组2x1 (3 )x2 x3 0x1 x2 (1 )x3 0有非零解？解系数行列式为24 3 4D 23 1 21 1111 101(1 )3 ( 3) 4(1 ) 2(1 )( 3 ) (1 )3 2(1 )2 3 令D 0 得0 2或3于是当0 2或3时该齐次线性方程组有非零解矩阵及其运算1 已知线性变换x1 2y1 2y2 y3x2 3y1 y2 5y3 x3 3y1 2y2 3y3求从变量x1 x2 x3到变量y1 y2 y3的线性变换解由已知x1 221 y1x2 315 y2x 323 y2 31y1 221 x1 7 49 y1故y2 315 x2 63 7 y2y 323 x 32 43 y3 2y1 7x1 4x2 9x3y2 6x1 3x2 7x3y3 3x1 2x2 4x32 已知两个线性变换x1 2y1 y3x2 2y1 3y2 2y3x3 4y1 y2 5y3y1 3z1 z2y2 2z1 z3 y3 z2 3z3求从z1 z2 z3到x1 x2 x3的线性变换解由已知x1 201 y1 201 31x2 232 y2 232 20x 415 y 415 0 12 30 z11 z2 z3 3613 z1 12 49 z210 116 z 3x1 6z1 z2 3z3所以有x2 12z1 4z2 9z3x3 10z1 z2 16z3111 1233 设A 11 1 B 1 24 求3AB 2A及ATB1 11 051 111 123 111解3AB 2A 3 11 1 1 24 2 11 11 11 051 1 11 058 111 *****3 0 56 2 11 1 2 1720290 1 11 429 2 111 123 058AB 11 1 1 24 0 561 11 051 290T4 计算下列乘积431 7(1) 1 23 2570 1431 7 4 7 3 2 1 1 35解1 23 2 1 7 ( 2) 2 3 1 6 570 1 5 7 7 2 0 1 493(2)(123) 213解(123) 2 (1 3 2 2 3 1) (10) 1 2(3) 1 ( 12)32 ( 1)2 2 2 2解1 ( 12) 1 ( 1)1 2 13 3 ( 1)3 2 342 6102140 (4) 11 134 43 1 301 2 1 212 6 7820 5 6 1 2102140 解1 134 143 1 30a11a12a13 x1(5)(x1x2x3) a12a22a23 x2aaa ***** x3解a11a12a13 x1(x1x2x3) a12a22a23 x2aaa ***** x3x1(a11x1 a12x2 a13x3 a12x1 a22x2 a23x3 a13x1 a23x2 a33x3) x2 x 322 a33x3 2a12x1x2 2a13x1x3 2a23x2x3 a11x12 a22x2。