《线性代数》同济大学版课后习题答案详解

《线性代数》同济大学版 课后习题答案详解

解

111abc222

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)?11?84?31?

2 解 ?11?084?1

31 ?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)bccaab?

a 解

bbcccaab ?acb?bac?cba?bbb?aaa?ccc ?3abc?a3?b3?c3?

111 (3)aa2bb2cc2?

abc ?bc2?ca2?ab2?ac2?ba2?cb2 ?(a?b)(b?c)(c?a)?

xyx?yyx?xy (4)?

x?yxyxyx? 解

?yyx?xyxy xy ?x(x?y)y?yx(x?y)?(x?y)yx?y3?(x?y)3?x3 ?3xy(x?y)?y3?3x2 y?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)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)? 解 逆序数为

n(n?1)2? 请浏览后下载，资料供参考，期待您的好评与关注！

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

(?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?

解

412412021052001174?1?10?122?(?1)4?3 103?14c2?c341??????10c4?7c30?12302021?102?1404?110c2?c39910 ?12?2??????00?2?0?

10314c1?12c317171423 (2)15 解

1?120423611? 221c4?c221?????321521?12042301?12042360r4?r222?????3102223151?12042361?121423402 00

r4?r123?????1002?0? 0041 (1)1001251202142? 07?abacae (3)bd?cddebfcf?ef请浏览后下载，资料供参考，期待您的好评与关注！

?

解

?abacae?bcebd?cdde?adfb?ce bfcf?efbc?e

a2abb2c2?c1a2ab?a2b2?a22aa?b2b?????2ab?a2b?2a

00111c3?c113?1?111 ?adfbce1?11?4abcdef?

11?1

222ab?ab?aab?a?(a?b)3 ?

?(b?a)(b?a)1?(?1)2b?a2b?2aa?1 (4)00 解

1b?1001c?100? 1dax?byay?bzaz?bxxyz33 (2)ay?bzaz?bxax?by?(a?b)yzx;

az?bxax?byay?bzzxya1c?100 1d

证明

a?1001b?1001c?10r1?ar201?ab0??????1b10?1d00

1?aba0c3?dc21?abaad2?1?1c1??????1c1?cd ?(?1)(?1)0?100?1d

5? 证明:

ax?byay?bzaz?bxay?bzaz?bxax?by az?bxax?byay?bzabad?(?1)(?1)3?21??11?cdxay?bzaz?bxyay?bzaz?bx ?ayaz?bxax?by?bzaz?bxax?by

zax?byay?bzxax?byay?bzxay?bzzyzaz?bx22 ?ayaz?bxx?bzxax?by

zax?byyxyay?bzxyzyzx33 ?ayzx?bzxy

zxyxyz?abcd?ab?cd?ad?1?

a2abb2 (1)2aa?b2b?(a?b);

1113

证明

请浏览后下载，资料供参考，期待您的好评与关注！