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?1404?110c2?c39910 ?12?2??????00?2?010314c1?12c3171714
23 (2)151?12042361122
23 解 151?12042361c4?c221?????321521?12042301?12042360r4?r222?????310221?121423402 00r4?r123 ?????1002?000
?abacae (3)bd?cddebfcf?ef
?abacae?bce 解 bd?cdde?adfb?ce
bfcf?efbc?e?111 ?adfbce1?11?4abcdef11?1
a1 (4)?001b?1001c?1001d
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