线性代数(复旦大学出版社周勇)课后习题问题详解(一)

1、实用文档第一章课后答案2 11. 2 2 1(1) 5;1 2x 1122322. 22(x 1)( x X 1) x x x 1 ；x x x 13.ab2 a2b1 1 14.3 1 4 5 32 27 8 15 36 58 9 50a05. b0c0,（第一行与第三行对应成比例）0d01 2 36. 3 1 21 8 27 6 6 6 18。2 3 1二.求逆序数1. 34215即5221002. 4312即532003. n (n 1)(n 1) (n 2)2 1 即 (n 1) (n 2)101 3(2n 1) (2n) (2n 2)4 201n 1n 1n 2104.1 2 (n

2、1) (n 1) (n 2)2 1 2* n(n 1)2二.四阶行列式中含有a11a23的项为a11a23a32a44a11a23a34a42四.计算行列式值4 12 407 240 0 94512 0 2r1 4r21202r17 r412 0210 5 2 0 10r2015 220B 15r40 0 17 850 11701170 11701.2.01101111113.11004abcdef标准5.b2b2ca2b2cab2aa2c2aa2c2a2b(ab1)(cd1)ad2a2b2aa2c2a2b2aa2c2a2b4.按第一行展开其中a2a2a2bbac2b2c2ccaba2a2a2

3、bb2b2c2ccaba02a2ba2b2c0caba 02a2b c2b2c 0 c a ba 2a 2a2b b 2b2c 2c c1 2 2abc2 1 22 2 1a 2a 02bb 02c 2c a(ab)2b2ab25abc 3ab(a b) (a c)(aa 2a 02b b 02c 2c b(ac)2ca 2a(a c)2c c a bab 3ac)2acab其余同法可求。方法2:a b c 2a 2a2b b a c 2b2c 2ccabb a c 2b(a b c)2c cab2b 2b2a2c c a b2b b a c2a2c 2c2 2416.31204 0 r2 2

4、rl2 23530 323r3 2r1 0 45114 %0 25 27031222227. 22322222212。2)0202 (n 2)!8.a 00 a0 01 0a 00 00 a0 a0 00 0(1)n10 0a 00 00 00 an 11 0n 20 10 0按第一行展开ann 1 n n 2 na ( 1) ( 1) a a a五.证明下列等式1.2 a2a12 aab2(b1)2(b2)2c2(c1)2(c2)2d2(d1)2(d2)2a(ba1(a 3)(b 3)(c 3)(d 3)a(ba)b22 aba2(ba)112)2(a3)22)2(b3)22)2(c3)22

5、)2(d3)2(a b)32一，,一一a 2a 14a 4 6a 92a2 2a 1 2 3c3 Gb2 2b 14b 4 6b 9c32c2b2 2b 1 2 3c4 c12-，一一c 2c 14c 4 6c 9c4 2c2c22c 1 2 3d2 2d 14d 4 6d 9d2 2d 1 2 303.x00Dn000an an 1 an 200x1000000x10000按第一行展开x00x00x1000x1a2x a1a n 1an 2an 3a2x0100 x 100x000anan 2an 3x 1a2 x a1同理 Dk1 xDk a-,返回代入得DnxDn 1an六、计算下列各式

6、xDn 1an ( 1)n11 (1)n2xDn 1x(xDn 2 an 1) annn 1x a1xan 1x an31.设x1，x2,x3为万程x px q 0的二个根,则由三次方程根的性质3.2ax bx cx d 0x1 x2 x3 b/a,x x2 x3x1x2x2x3x1x3c / a得 x3px0的三个根满足：x1 x2x30, xx2x3所以x1X2x3x3xix23 x1x3 3x1x2 x3p(xix2x3)3q3x1x2x30。X2x3xi2. f(x)3.求DA14A243中x3的系数为-1.A34的A)4A34Au的值,4.求D的An1An2Ann的值，A1A.2ri

7、 arn(i七.计算行列式值1.a0000x a0000x a01111xn)(x a)n 1xnn0(xim)( m)n 1i 12.123110Dn0 220 00n 1n0000各行提出公因子11(n 1)!0021103010n 1001nnk kk 1 k 2015Cj 1(j n 1,1) (n 1)!nkk 3012n 1 n0000000013.n(n 1)!( 1)n 1 k k 1LT(n1)1(1)Tnx1mx2xn为 mx2xni 1 nx1x2mxnnDnc1c jj 2x1x2xnmX m x2 mxni 1 nX mx2xn mi 1xim x2i 1riri(i2, n) 0 mD2n按第一行展开b( 1)2n2n2n 1adD2n2 bcD2n 2,同理D2n(adbc)D2nD2ad所以D2n(adbc)n4.a1a1Dnrir1(i2,n)a1a2a211a20a2ana1八.克莱姆法则解方程1.2x2y yyz2z解:1210211011202118 D11114 D22114 D3211112312132113D121/2, y1/2,z3/2；Xi2x23x34x42.x1乂23