第一章 行列式
1. 利用对角线法则计算下列三阶行列式: 201 (1)1?4?1;
?183201 解 1?4?1
?183 =2′(-4)′3+0′(-1)′(-1)+1′1′8 -0′1′3-2′(-1)′8-1′(-4)′(-1) =-24+8+16-4=-4. abc (2)bcacab
abc 解 bca
cab =acb+bac+cba-bbb-aaa-ccc
=3abc-a3-b3-c3.
111 (3)abc;
a2b2c2111 解 abc
a2b2c2 =bc+ca+ab-ac-ba-cb
222222
(a-b)(b-c)(c-a).
xyx?y (4)yx?yx.
x?yxyxyx?y 解 yx?yx
x?yxy =x(x+y)y+yx(x+y)+(x+y)yx-y-(x+y)-x =3xy(x+y)-y-3x y-x-y-x =-2(x+y).
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:
23
3
3
2
3
3
3
3
3
3
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
(
42
tt1)a11a23a3ra4s,
这种排列共有两个
即24和
t其中rs是2和4构成的排列所以含因子a11a23的项分别是 (1)a11a23a32a44 (1)a11a23a34a42
(1)a11a23a32a44(1)a11a23a34a42
2
1
a11a23a32a44 a11a23a34a42
4. 计算下列各行列式:
线性代数同济大学第四版习题答案Word版
