1 利用对角线法则计算下列三阶行列式

(1)2?11?084?131

解 201?11?84?31

2(4)30(1)(1)1 0

132(1)8

1(

4)

248164

4

(2)abbcccaab

解 abbcccaab

acbbaccbabbbaaaccc

3abca3b3c3

(3)1aa112bb2cc2

解 1aa12bb12cc2

18 (1)

bc2ca2ab2ac2ba2cb2

(ab)(bc)(ca)

xyx?y (4)yx?yxx?yxyxyx?y 解 yx?yx

x?yxy

x(xy)yyx(xy)(xy)yxy3(xy)3x3

3xy(x2(x3

y)y33x2 yx3y3x3 y3)

求下列各排列的逆

2 按自然数从小到大为标准次序序数

(1)1 2 3 4

解 逆序数为0 (2)4 1 3 2

42 32

解 逆序数为4 41 43 (3)3 4 2 1

3 1

解 逆序数为5 3 2 4 2 4 1, 2 1

(4)2 4 1 3

4 1

4 3

(2n)

解 逆序数为3 2 1 (5)1 3 (2n1) 2 4

解 逆序数为

n(n?1)2

3 2 (1个) 5 2 5 4(2个) 7 2 7 4 7 6(3个)

(2n1)2 (2n1)4(2n1)(2n2) (n1个)

(6)1 3

(2n1) (2 解 逆序数为n(n1)

3 2(1个) 5 2 5 4 (2个)

(2n1)2 (2n1)4(2n1)(2n2) (n1个)

(2n1)6

n) (2n2) (2n1)6

2

4 2(1个) 6 2 6 4(2个)

(2n)(2n2)

(2n)2 (2n)4 (2n)6 (n1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)a11a23a3ra4s

t

其中rs是2和4构成的排列42

这种排列共有两个 即24和

所以含因子a11a23的项分别是 (1)a11a23a32a44 (1)a11a23a34a42

tt(1)a11a23a32a44(1)a11a23a34a42

2

1

a11a23a32a44 a11a23a34a42

4 计算下列各行列式

41 (1)100125120214207

41 解 100125120214c2?c342??????10c?7c103074?12302021?104?1?102?122?(?1)4?3 ?14103?140