49
??x1??x3?x2? x3? ??x4?0 当x3?1时? 得对应的齐次方程组的基础解系??(?1? 1? 1? 0)T?
??x1?5x2?2x3?3x4?11 (2)?5x1?3x2?6x3?x4??1?
??2x1?4x2?2x3?x4??6 解 对增广矩阵进行初等行变换? 有
?1?52?311?r?109/7?1/21? B??536?1?1? ~ ?01?1/71/2?2??
?2421?6??000?00???? 与所给方程组同解的方程为
?x1??(9/7)x3?(1/2)x4?1? ?x?(1/7)x?(1/2)x?2?234 当x3?x4?0时? 得所给方程组的一个解
??(1? ?2? 0? 0)T?
与对应的齐次方程组同解的方程为
?x1??(9/7)x3?(1/2)x4? ?x?(1/7)x?(1/2)x?234 分别取(x3? x4)T?(1? 0)T? (0? 1)T? 得对应的齐次方程组的基础解系
?1?(?9? 1? 7? 0)T? ?2?(1? ?1? 0? 2)T?
29? 设四元非齐次线性方程组的系数矩阵的秩为3? 已知?1? ?2? ?3是它的三个解向量? 且
?1?(2? 3? 4? 5)T? ?2??3?(1? 2? 3? 4)T?
求该方程组的通解?
解 由于方程组中未知数的个数是4? 系数矩阵的秩为3? 所以对应的齐次线性方程组的基础解系含有一个向量? 且由于?1? ?2? ?3均为方程组的解? 由非齐次线性方程组解的结构性质得
2?1?(?2??3)?(?1??2)?(?1??3)? (3? 4? 5? 6)T
为其基础解系向量? 故此方程组的通解?
x?k(3? 4? 5? 6)T?(2? 3? 4? 5)T? (k?R)?
30? 设有向量组A? a1?(?? 2? 10)T? a2?(?2? 1? 5)T? a3?(?1? 1? 4)T? 及b?(1? ?? ?1)T? 问?? ?为何值时
(1)向量b不能由向量组A线性表示?
(2)向量b能由向量组A线性表示? 且表示式唯一?
(3)向量b能由向量组A线性表示? 且表示式不唯一? 并求一般表示式?
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第一章 行列式
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)?3?0?(?1)?(?1)?1?1?8 ?0?1?3?2?(?1)?8?1?(?4)?(?1) ??24?8?16?4??4?
(2)abbcccaab?
解 abbcccaab
?acb?bac?cba?bbb?aaa?ccc ?3abc?a3?b3?c3?
(3)1aa12bb12cc2?
解 111aa2bb2cc2
?bc2?ca2?ab2?ac2?ba2?cb2 ?(a?b)(b?c)(c?a)?
xyx?y (4)yx?yxx?yxy?
xyx?y 解 yx?yxx?yxy
?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?
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?
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??at11a23a32a44? (?1)a11a23a34a42?(?1)2a11a23a34a42?a11a23a34a42? 4? 计算下列各行列式?
4 (1)101122042? 05121074 解 101122042c2?c34?12?10??????1202?41?21?102?0512107c4?7c(?1)4?3 31003021?140103?14 2
?4?110c2?c3991010123?142??????c00?2?0?
1?12c31717142 (2)31?114211?5203 62221414 解 3c1?121??4???c?223?114202r??4?r??221?3?1205203622152036021221320 40 ?r?4???r12?31?114202?0? 0203000 (3)?bdabbf?accfcdae?deef?
解 ?bdabbf?accfcdae?deef?adf?bbceb?cc?ee
?adfbce?111?1111?11?4abcdef?
a100 (4)?01b10? 0?01?c11da100 解 ?r1?ar010?b011?c011?????2?0d011?0?baba00110 ?c11d)(?1)2?11??aba0c3?dc ?(?121?abaad01?c11d??????01?c11?0cd
?(?1)(?1)3?21?abad ?abcd?ab?cd?ad?1? ?11?cd 5 ? 证明
:
(1)2a21aaab?1bb2 21b?(a?b)3;
证明
a2abb2 21aa?1b21b?cc?2???c?12a21aabb??aa22b2b??a2 2a
3?c100 ?(?1)3?1abb??aa22b2b??a22a?(b?a)(b?a)a1b?2a?(a?b)3 ?3
ax?byay?bzaz?bxxyz (2)ay?bzaz?bxax?by?(a3?b3)yzxaz?bxax?byay?bzzxy;
证明
ax?byay?bzaz?bx ay?bzaz?bxax?byaz?bxax?byay?bz
xay?bzaz?bxyay?bzaz?bx ?ayaz?bxax?by?bzaz?bxax?byzax?byay?bzxax?byay?bz
xay?bzzyzaz?bx ?a2yaz?bxx?b2zxax?byzax?byyxyay?bz
xyzyzx ?a3yzx?b3zxyzxyxyz
xyzxyz ?a3yzx?b3yzxzxyzxy
xyz ?(a3?b3)yzxzxy?
a2(a?1)2(a?2)2(a?3)2 (3)b2(b?1)2(b?2)2(b?3)2c2(c?1)2(c?2)2(c?3)2?0; d2(d?1)2(d?2)2(d?3)2 证明
a2(a?1)2(a?2)2(a?3)2 b2(b?1)2(b?2)2(b?3)2c2(c?1)2(c?2)2(c?3)2(c4?c3? c3?c2? c2?c1得) d2(d?1)2(d?2)2(d?3)2a222 ?b22a?12adc2222bdc??1?12b?c?3?322ab??55(c?c? 1?3322dc??543c3?c2得) 2d5a22 ?b22ab??11222dc2222dc??1122?0? 222
4
1111abcd (4)a2b2c2d2a4b4c4d4 ?(a?b)(a?c)(a?d)(b?c)(b?d)(c?d)(a?b?c?d); 证明
