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)bca?
cababc解bca cab?acb?bac?cba?bbb?aaa?ccc ?3abc?a3?b3?c3?
111(3)abca2b2c2111解abca2b2c2?
?bc2?ca2?ab2?ac2?ba2?cb2 ?(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?y3?(x?y)3?x3 ?3xy(x?y)?y3?3x2y?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)1234? 解逆序数为0 (2)4132?
解逆序数为4?41?43?42?32? (3)3421?
解逆序数为5?32?31?42?41,21? (4)2413?
解逆序数为3?21?41?43? (5)13???(2n?1)24???(2n)? 解逆序数为32(1个) 52?54(2个) 72?74?76(3个) ??????
(2n?1)2?(2n?1)4?(2n?1)6?????(2n?1)(2n?2)(n?1个) (6)13???(2n?1)(2n)(2n?2)???2? 解逆序数为n(n?1)? 32(1个) 52?54(2个) ??????
(2n?1)2?(2n?1)4?(2n?1)6?????(2n?1)(2n?2)(n?1个) 42(1个) 62?64(2个) ??????
n(n?1)? 2(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?
41(1)10041解100125112512021202142? 074c2?c342??????10c?7c103074?12302021?104?1?102?122?(?1)4?3 ?14103?1404?110c2?c39910?12?2??????00?2?0? 10314c1?12c317171423(2)1523解151?1201?1204236423611? 221c4?c221?????321521?120423002?0? 00?
1?12042360r4?r222?????310221?121423402 00r4?r123?????10?abacae(3)bd?cddebfcf?ef?abacae?bce解bd?cdde?adfb?ce
bfcf?efbc?e?111?adfbce1?11?4abcdef?
11?1a?1(4)00a?1解001b?101b?1001c?101c?100? 1d0r1?ar201?ab0??????1b10?1d00a1c?100 1dabad?(?1)(?1)3?21??11?cd5?证明:
?abcd?ab?cd?ad?1?
a2abb2(1)2aa?b2b?(a?b)3;
111证明
222ab?ab?aab?a?(a?b)3?
?(b?a)(b?a)1?(?1)2b?a2b?2a3?1ax?byay?bzaz?bxxyz33(2)ay?bzaz?bxax?by?(a?b)yzx;
az?bxax?byay?bzzxy证明
xyz?(a3?b3)yzx?
zxya2b2(3)2cd2证明
(a?1)2(b?1)2(c?1)2(d?1)2(a?2)2(b?2)2(c?2)2(d?2)2(a?3)2(b?3)2?0;
(c?3)2(d?3)2a2b2c2d2(a?1)2(b?1)2(c?1)2(d?1)2(a?2)2(b?2)2(c?2)2(d?2)2(a?3)2(b?3)2(c4?c3? c3?c2? c2?c1得)
(c?3)2(d?3)2a22b?c2d2a22b?c2d21a(4)2aa4证明
2a?12b?12c?12d?12a?12b?12c?12d?111bcb2c2b4c42a?32a?52b?32b?5(c4?c3? c3?c2得) 2c?32c?52d?32d?52222?0? 22221d d2d4?(a?b)(a?c)(a?d)(b?c)(b?d)(c?d)(a?b?c?d); =(a?b)(a?c)(a?d)(b?c)(b?d)(c?d)(a?b?c?d)?
x0(5)? ? ?0an?1x? ? ?0an?10?1? ? ?0an?2? ? ?? ? ?? ? ?? ? ?? ? ?0000?? ? ??xn?a1xn?1?????an?1x?an? x?1a2x?a1证明用数学归纳法证明? 当n?2时?D2?x?1?x2?ax?a?命题成立?
12a2x?a1假设对于(n?1)阶行列式命题成立?即 Dn?1?xn?1?a1xn?2?????an?2x?an?1? 则Dn按第一列展开?有
同济大学工程数学线性代数第六版答案全
