线性代数第五版答案(全)..doc

第一章 行列式 1( 利用对角线法则计算下列三阶行列式( (1)( 解 (2(((4)(3(0(((1)(((1)(1(1(8 (0(1(3(2(((1)(8(1(((4)(((1) ((24(8(16(4((4( (2)( 解 (acb(bac(cba(bbb(aaa(ccc (3abc(a3(b3(c3( (3)( 解 (bc2(ca2(ab2(ac2(ba2(cb2 ((a(b)(b(c)(c(a)( (4)( 解 (x(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)( 解 逆序数为( 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( (1)( 解 ( (2)( 解 ( (3)( 解 ( (4)( 解 (abcd(ab(cd(ad(1( 5( 证明: (1)((a(b)3; 证明 ((a(b)3 ( (2); 证明 ( (3); 证明 (c4(c3( c3(c2( c2(c1得) (c4(c3( c3(c2得) ( (4) ((a(b)(a(c)(a(d)(b(c)(b(d)(c(d)(a(b(c(d); 证明 =(a(b)(a(c)(a(d)(b(c)(b(d)(c(d)(a(b(c(d)( (5)(xn(a1xn(1( ( ( (