文档介绍：第一章行列式 1?利用对角线法则计算下列三阶行列式?(1)381 141 102????解381 141 102????2?(? 4)?3?0?(? 1)?(? 1)?1?1?8 ?0?1?3?2?(? 1)?8?1?(? 4)?(? 1) ?? 24?8? 16?4??4?(2)bac acb cba ?解bac acb cba ? acb ? bac ? cba ? bbb ? aaa ? ccc ?3 abc ?a 3?b 3?c 3?(3) 222111cba cba ?解 222111cba cba ? bc 2? ca 2? ab 2? ac 2? ba 2? cb 2?(a?b )(b?c )(c?a)?(4)yxyx xyxy yxyx????解yxyx xyxy yxyx????x(x?y)y? yx(x?y)?(x?y) yx?y 3?(x?y) 3?x 3 ?3 xy(x?y)?y 3?3x 2y?x 3?y 3?x 3?? 2(x 3?y 3)?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)1 234?解逆序数为 0 (2)4 132?解逆序数为 4? 41? 43? 42? 32?(3)3 421?解逆序数为 5?32?31?42?4 1,21?(4)2 413?解逆序数为 3?21?41?43?(5)1 3???(2n? 1)24???(2n)?解逆序数为 2 )1(?nn ?32 (1个)52?5 4(2 个)72?74?7 6(3 个) ??????(2n? 1)2 ?(2n? 1)4 ?(2n? 1)6 ?????(2n? 1)(2 n? 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)52?54 (2个)??????(2n? 1)2 ?(2n? 1)4 ?(2n? 1)6 ?????(2n? 1)(2 n? 2)(n?1个) 4 2(1 个)62?6 4(2 个)??????(2n )2?(2n )4?(2n )6?????(2n )(2 n? 2)(n?1个) 3?写出四阶行列式中含有因子 a 11a 23的项?解含因子 a 11a 23的项的一般形式为(?1) ta 11a 23a 3ra 4s?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 a 11a 23的项分别是(?1) ta 11a 23a 32a 44?(?1) 1a 11a 23a 32a 44??a 11a 23a 32a 44?(?1) ta 11a 23a 34a 42?(?1) 2a 11a 23a 34a 42?a 11a 23a 34a 42? 4?计算下列各行列式?(1)7110 02510 2021 4214 ?解7110 02510 2021 42140100 14 2310 2021 10 2147 3234??????????? 34)1( 14 3 10 221 10 14 ???????14 310 221 10 14???0 14 17 17 200 10 99 3232 11?????????? ?(2)2605 2321 1213 1412??解2605 2321 1213 1412?2605 0321 2213 0412 24??????? cc0412 0321 2213 0412 24??????? rr00000 0321 2213 0412 14???????? rr ?(3) ef cf bf de cd bd ae ac ab????解 ef cf bf de cd bd ae ac ab???ecb ecb ecb adf ???? abcdef adfbce 4111 111 111??????(4)d c b a100 110 011 001????解d c b a100 110 011 001???d c b a ab arr100 110 011 010 21??????????d c aab10 11 01)1 )(1( 12???????010 11 1 23?????????? cd c ad a ab dcc cd ad ab???????11 1)1 )(1( 23? abcd ? ab? cd? ad?1? 5?证明: (1)111 22 22bbaa b ab a??(a?b) 3;证明 111 22 22bbaa bab a?001 222 22221213ababa abaab cc ???????????abab abaab22 )1( 22213???????21 ) )(( abaabab ?????(a?b) 3?(2)