第一章 n阶行列式

第一章 行列式

1.利用对角线法则计算下列三阶行列式：

201abc111xyx?yx?yx. （1）1?4?1； （2）bca; （3）abc； （4）y?183caba2b2c2x?yxy201解 （1）1?4?1?2?(?4)?3?0?(?1)?(?1)?1?1?8?0?1?3?2?(?1)?8?1?(?4)?(?1)

?183=?24?8?16?4=?4

abc333（2）bca?acb?bac?cba?bbb?aaa?ccc?3abc?a?b?c

cab

111222222（3）abc?bc?ca?ab?ac?ba?cb?(a?b)(b?c)(c?a)

a2b2c2

xyx?yx?yx?x(x?y)y?yx(x?y)?(x?y)yx?y3?(x?y)3?x3 （4）yx?yxy?3xy(x?y)?y3?3x2y?3y2x?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； （2）4 1 3 2； （3）3 4 2 1； （4）2 4 1 3； （5）1 3 … (2n?1) 2 4 … (2n)； （6）1 3 … (2n?1) (2n) (2n?2) … 2. 解（1）逆序数为0

（2）逆序数为4：4 1，4 3，4 2，3 2 （3）逆序数为5：3 2，3 1，4 2，4 1,2 1 （4）逆序数为3：2 1，4 1，4 3 （5）逆序数为

n(n?1)： 23 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）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

1

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 由定义知，四阶行列式的一般项为(?1)ta1p1a2p2a3p3a4p4，其中t为p1p2p3p4的逆序数．

由于p1?1,p2?3已固定，p1p2p3p4只能形如13□□，即1324或1342.对应的t分别为

0?0?1?0?1或0?0?0?2?2

??a11a23a32a44和a11a23a34a42为所求.

4.计算下列各行列式：

?4?1（1）??10??0解

12511251202120214??21?3?12??； （2）?0??127????50442c2?c310c4?7c3107042361??a?abacae????11????； （3）bd?cdde； （4）?2????0bfcf?ef??2????01b?1001c?10?0?? 1?d??41(1)

100?12302?104?1?104?110022?(?1)4?3=12?2 =122?14103?141031410c2?c3c1?12c3

991000?2=0 171714213?11250423602r4?r202213?11221423402r4?r1 00213?11200423002=0 00213?1(2)

1250

423611c4?c222?abacae?bce?111(3)bd?cdde=adfb?ce=adfbce1?11=4abcdef

bfcf?efbc?e11?1

a?1(4)

001b?1001c?1001?ab0r1?ar2?1b10?1d00a1c?101?aba002?1?1c1 =(?1)(?1)10?1ddc3?dc2

1?abaad1?abad?1c1?cd=(?1)(?1)3?2=abcd?ab?cd?ad?1

?11?cd0?10ax?byay?bzaz?bxxyza2abb23335.证明: (1)2aa?b2b=(a?b); (2)ay?bzaz?bxax?by=(a?b)yzx;

az?bxax?byay?bzzxy1112

a2b2(3)2cd2(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)21111abcd?(a?b)(a?c)(a?d)(b?c)(b?d)?(c?d)(a?b?c?d); (4)2ab2c2d2a4b4c4d4x?100x?1(5)???000anan?1an?2 证明

?????0000???xn?a1xn?1???an?1x?an. x?1a2x?a1a2ab?a2b2?a22c2?c1ab?ab2?a23?1ab?a(1)左边? ?(b?a)(b?a)2ab?a2b?2a?(?1)12b?a2b?2ac3?c1100?(a?b)3?右边

xay?bzaz?bxyay?bzaz?bx按第一列ayaz?bxax?by ?bzaz?bxax?by (2)左边分开zax?byay?bzxax?byay?bz分别再分xay?bzzyzaz?bxxyzyzx分别再分a2yaz?bxx?0?0?bzxax?bya3yzx?b3zxy zax?byyxyay?bzzxyxyzxyzxyz?a3yzx?b3yzx(?1)2?右边

zxyzxya2b2(3) 左边?2cd2a2b2c2d2?(2a?1)?(2b?1)?(2c?1)?(2d?1)abcd4a?44b?44c?44d?4abcd4444(a?2)2(b?2)2(c?2)2(d?2)2(a?3)2(b?3)2c2?c1(c?3)2c3?c1(d?3)2c4?c111114a?44b?44c?44d?46a6b?0 6c6da2b2c2d22a?12b?12c?12d?14a?44b?44c?44d?46a?96b?9

6c?96d?9a2按第二列b222分成二项cd26a?9a26b?9b2?26c?9c6d?9d29a29b2?9c29d211114a4b4c4d6a?96b?9

6c?96d?9c3?4c2a2第一项c4?6c2b2c3?4c2c2第二项c4?9c2d23

1000b?ac?ad?aab?ac?ad?ab2?a2c2?a2d2?a2 (4) 左边?2222222=

ab?ac?ad?ab2(b2?a2)c2(c2?a2)d2(d2?a2)4444444ab?ac?ad?a111c?ad?a =(b?a)(c?a)(d?a)b?ab2(b?a)c2(c?a)d2(d?a)100=(b?a)(c?a)(d?a)?b?a c?bd?bb2(b?a)c2(c?a)?b2(b?a)d2(d?a)?b2(b?a)11

(c2?bc?b2)?a(c?b)(d2?bd?b2)?a(d?b)=(a?b)(a?c)(a?d)(b?c)(b?d)(c?d)(a?b?c?d)

=(b?a)(c?a)(d?a)(c?b)(d?b)?(5) 用数学归纳法证明

当n?2时,D2?x?1?x2?a1x?a2,命题成立.

a2x?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1列展开:

?10?x?1?Dn?xDn?1?an(?1)n?1???11?所以，对于n阶行列式命题成立.

6.设n阶行列式D?det(aij),把D上下翻转、或逆时针旋转90、或依副对角线翻转，依次得

?00?x00?xDn?1?an?右边 ??1an1?anna1n?annann?a1nD1???， D2??? ，D3???，

a11?a1na11?an1an1?a11证明D1?D2?(?1)n(n?1)2D,D3?D.

证明 ?D?det(aij)

an1?D1??a11a11?a1na11?a1n?anna21?a2nn?1n?2n?1an1?annann?? ?(?1)(?1)an1??(?1)???a1n??a21?a2na31?a3na11?a1nn(n?1)n?1n?21?2???(n?2)?(n?1)?(?1)(?1)?(?1)???(?1)D?(?1)2D

an1?ann同理可证D2?(?1)n(n?1)2a11?an1n(n?1)n(n?1)T???(?1)2D?(?1)2D a1n?ann4

D3?(?1)

n(n?1)2D2?(?1)n(n?1)2(?1)n(n?1)2D?(?1)n(n?1)D?D

7.计算下列各行列式（Dk为k阶行列式）：

a(1)Dn?1?,其中对角线上元素都是a，未写出的元素都是0；

1axa(2)Dn??aax?a????aa; ?x?(a?n)n?(a?n)n?1??; 提示：利用范德蒙德行列式的结果． ?a?n?1(3) Dn?1an(a?1)nan?1(a?1)n?1???aa?111an?(4) D2n?0bn0a1b1c1d1?0??0; dncn(5)Dn?det(aij),其中aij?i?j;

1?a1111?a2(6)Dn???11解

?1?1,其中a1a2?an?0.

???1?ana00(1) Dn??010a0?0000a?00??????000?a0100 ?0a00a?0000?0?????000?a1a0 0?(?1)2n?a??a(n?1)(n?1)0(n?1)?(n?1)0a按最后一行展开(?1)n?10?0(再按第一行展开)

5

a

?(?1)n?1?(?1)n?a(n?2)(n?2)?an?an?an?2?an?2(a2?1)

(2)将第一行乘(?1)分别加到其余各行，得

xaaa?xx?a0Dn?a?x0x?a???a?x00再将各列都加到第一列上，得

?a?0?0 ??0x?ax?(n?1)aaa0x?a0Dn?00x?a???000?a?0?0?[x?(n?1)a](x?a)n?1 ??0x?a(3) 从第n?1行开始，第n?1行经过n次相邻对换，换到第1行，第n行经(n?1)次对换换到第2行…，

经n?(n?1)???1?n(n?1)次行交换，得 2Dn?111a?1n(n?1)a?(?1)2??an?1(a?1)n?1an(a?1)n?1?a?n?? ?(a?n)n?1?(a?n)n此行列式为范德蒙德行列式

Dn?1?(?1)?(?1)n(n?1)2n?1?i?j?1n(n?1)2?[(a?i?1)?(a?j?1)]

n(n?1)2n?1?i?j?1?[?(i?j)]?(?1)?(?1)n?(n?1)???12?n?1?i?j?1?[(i?j)]

?

n?1?i?j?1?(i?j)

an?(4) D2n?00?a1b1?c1d10?bn

cndn6

an?1?按第一行展开0?a1b1?c1d10??bn?100?dn?100dn2n?10an?1??(?1)bn0?cn?1cn0?a1b1c1d1?0bn?10dn?10an0cn?10

都按最后一行展开 andnD2n?2?bncnD2n?2

由此得递推公式：

D2n?(andn?bncn)D2n?2 即 D2n?而 D2??(adii?2ni?bici)D2

a1b1?a1d1?b1c1

c1d1n得 (5)aij?i?j

D2n??(aidi?bici)

i?1012310122101Dn?det(aij)?3210????n?1n?2n?3n?4??????n?1n?2n?3r1?r2n?4r2?r3,??0?1111?1?111?1?1?11?1?1?1?1????n?1n?2n?3n?4??????111 1?0?1000?1?200c2?c1,c3?c1?1?2?20?1?2?2?2c4?c1,?????n?12n?32n?42n?5?0?0?0n?1n?2=(?1)(n?1)2

?0???n?1a100?a2a200?a3a300?a4???000000?001?001?001?001 ??????an?1an?11?0?an1?an1?a1111?a2(6)Dn???11?1c1?c2,c2?c3?1??c3?c4,??1?an7

a100?a2a200?a3a3按最后一列(1?an)(a1a2?an?1)?00?a4展开（由下往上）???000000a100?a2a200?a3a3????000000n?000?000?000?000 ??????an?2an?20?00?an?00?00?00

?????an?1an?1?0?an?00?00?00?????an?1an?1?0?an????a2a200?a3a300?a4???000000?(1?an)(a1a2?an?1)?a1a2?an?3an?2an???a2a3?an ?(a1a2?an)(1??i?11) ai

8.用克莱姆法则解下列方程组：

?1,?5x1?6x2?x1?x2?x3?x4?5,?x?5x?6x?0,123?x?2x?x?4x??2,??1?234(1)?x2?5x3?6x4?0, (2)??2x1?3x2?x3?5x4??2,?x3?5x4?6x5?0,???3x1?x2?2x3?11x4?0;?x4?5x5?1.?1112解 (1)D?2?33151?22D1??2?301111111111?1401?2301?2???1?50?5?3?700?132110?2?1800?51151?1405??1?5?2?32110110?1211?5905??50?13110111111301?23???142 800?1?5414000142?1?91?50901??3?23052110?13?1?9211

09?3?231?5?1?91?5?1?90121101211??142 ??00?10?4600?138002312000014215111511151?2?140?7?230?1D2???2?2?1?50?12?3?700302110?15?180013233912?1131150?100001132??284

?1?190?2848

11511112?2412D3???426 ; D4?2?3?2?52?33101131?x1?15?1?2?142

?1?220D1DDD?1,x2?2?2,x3?3?3,x4?4??1 DDDD51(2) D?00065100065100065100按最后一行5D??0展开6551006510065100?5D??6D?? 06?5(5D???6D???)?6D???19D???30D????65D????114D?????65?19?114?5?665

?的余子式(D?为行列式D中a11的余子式,D??为D?中a11,D???,D????类推)

10D1?00151D2?00065100100010651006510006510065100按第一列D??0展开6565100651

0651006500?D??64?19D????30?????64?1507 060065

0

5600

?156?5?63 0

0156

016

0按第二列05

?001展开600

5050016?

605501

??65?1080??1145

51D3?00065100100010065100065按第三列展开100051000651056015?

60150000650

1605600

?056?6150 0

0150166

?19?6?114?703

51D4?00051D5?0006510065100065100651010001006510006510001按第四列展开10?00100051005100651065100501?60506510065105600??5?6156??395 00156按最后一列展开06?D??1?211?212 519

?x1?1507;665x2??1145;665x3?703;665x4??395;665x4?212． 665??x1?x2?x3?0?9.问?,?取何值时,齐次线性方程组?x1??x2?x3?0有非零解？

?x?2?x?x?023?111解 D3?1?1?????， 齐次线性方程组有非零解，则D3?0

12?1即 ?????0 得 ??0或??1

不难验证，当??0或??1时,该齐次线性方程组确有非零解.

??(1??)x1?2x2?4x3?0?10.问?取何值时,齐次线性方程组?2x1?(3??)x2?x3?0 有非零解？

?x?x?(1??)x?023?1解

1???241???3??4D?23??1?21??1

111??101???(1??)3?(??3)?4(1??)?2(1??)(?3??)?(1??)3?2(1??)2???3

齐次线性方程组有非零解，则D?0 得 ??0,??2或??3

不难验证，当??0,??2或??3时，该齐次线性方程组确有非零解.

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