2024年1月8日发(作者：曹县一模数学试卷)

七夕，古今诗人惯咏星月与悲情。吾生虽晚，世态炎凉却已看透矣。情也成空，且作“挥手袖底风”罢。是夜，窗外风雨如晦，吾独坐陋室，听一曲《尘缘》，合成诗韵一首，觉放诸古今，亦独有风韵也。乃书于纸上。毕而卧。凄然入梦。乙酉年七月初七。

-----啸之记。

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

183

解

2011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)bcacab

解

abcbcacab acbbaccbabbbaaaccc

3abca3b3c3

111(3)abca2b2c2

解

111abca2b2c2

bc2ca2ab2ac2ba2cb2

(ab)(bc)(ca)

(4)xyxyyxyxxyxy

解

xyxyyx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

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 3 2 (1个)

n(n1)2

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

41(1)107

4cc41232120c7c10330074210411002122(1)43

2141031410 解

411

23(2)154110c2c399101220020

cc1123

11224236

1cc214224230 解

11202315112042360rr4220223121

r4r121311200020

00

abacae(3)bdcddebfcfef

解

bceabacaebdcddeadfbcebcebfcfef

111adfbce1114abcdef1111b1001c1001d

a1(4)00

0rar2011d01aba1b101c001001d 解

a1001b1001c1

1aba0c3dc21abaad21c1cd(1)(1)1c1101001d(1)(1)321abad11cd

5 证明:

abcdabcdad1

a2abb2(1)2aab2b111(ab)3;

证明

a22a1abab131b2c2c1a2aba2b2a22b2aba2b2a001c3c11

(ab)3 

(1)

aba2bab2a2aba(ba)(ba)122b2aaxbyaybzazbxxyz(2)aybzazbxaxby(a3b3)yzxazbxaxbyaybzzxy;

证明

axbyaybzazbxaybzazbxaxbyazbxaxbyaybz

xaybzazbxyaybzazbxayazbxaxbybzazbxaxbyzaxbyaybzxaxbyaybzxaybzzyzazbxa2yazbxxb2zxaxbyzaxbyyxyaybzxyzyzxa3yzxb3zxyzxyxyzxyzxyza3yzxb3yzxzxyzxyxyz(a3b3)yzxzxy

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)2 证明

a2b2c2d2(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(c3)2(d3)2(c4c3 c3c2 c2c1得)

a2b22cd2a2b22cd22a12b12c12d12a12b12c12d12a32b32c32d322222a52b52c52d5(c4c3 c3c2得)

220

22

1a(4)a2a41bb2b41cc2c41dd2d4

(ab)(ac)(ad)(bc)(bd)(cd)(abcd);

证明

1aa2a41bb2b41cc2c41dd2d4

11110bacada0b(ba)c(ca)d(da)0b2(b2a2)c2(c2a2)d2(d2a2)

111(ba)(ca)(da)bcd222b(ba)c(ca)d(da)

111(ba)(ca)(da)0cbdb0c(cb)(cba)d(db)(dba)1

(ba)(ca)(da)(cb)(db)c(c1

ba)d(dba) =(ab)(ac)(ad)(bc)(bd)(cd)(abcd)

x100x1(5)      000anan1an2          0000  x1a2xa1xna1xn1    an1xan 

证明 用数学归纳法证明

x1x2a1xa2 命题成立 当n2时

D2axa12 假设对于(n1)阶行列式命题成立 即

Dn1xn1a1 xn2    an2xan1

则Dn按第一列展开 有

10DnxDn1an(1)n1x1      

11    00    00         

   x1

xD n1anxna1xn1    an1xan 

因此 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

an1  annD1      a11  a1n

a1n  annD2      a11  an1 

ann  a1nD3      an1  a11

证明D1D2(1)n(n1)2D D3D 

证明 因为Ddet(aij) 所以

a11an1  annaD1      (1)n1n1  a11  a1na21        a1nann  a2n

a11a21(1)n1(1)n2an1  a31          a1na2nann   

  a3nn(n1)2

(1)12  (n2)(n1)D(1) 同理可证

D2(1)

D3(1)

n(n1)2D

a11  an1n(n1)n(n1)      (1)2DT(1)2Da1n  annn(n1)n(n1)

n(n1)2D2(1)2(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 

1, 其中对角线上元素都是a 未写出的元素a都是0

解

a000a000aDn      000100  01  00  00(按第        a0  0a n行展开)

0an1(1)0  000a  0000  0          000  a1a0

(1)2na 

0  a(n1)(n1)0(n1)(n1)a

(1)n1(1)n 

anaa(n2)(n2)nan2an2(a21)

x(2)Dn

a aax  a        aa;

  x 解 将第一行乘(1)分别加到其余各行 得

xaaaxxa0Dnax0xa      ax00  a  0  0    0xa

再将各列都加到第一列上 得

x(n1)aaa0xa0Dn00xa      000  a  0  0    0xa [x(n1)a](xa)n1

an(a1)nan1(a1)n1(3)Dn1    aa111  (an)n  (an)n1;

      an  1 解 根据第6题结果 有

11a1n(n1)aDn1(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

n(n1)

Dn1(1)

(1)

(1)

an 

 

2n1ij1[(ai1)(aj1)]

j)]

n(n1)2n1ij1n(n1)2[(i2n(n1)  1(1)n1ij1(ij)

n1ij1(ij)

bn (4)D2ncna1b1c1d1 

 

;

dn 解

an 

 

 

 

bn

D2ncna1b1c1d1(按第1行展开)

dn

an1 

 

ancn10a1b1c1d1 

 

bn10

dn100dn0an1 

 

a1b1c1d1 

 

bn1

(1)2n1bncn1cn

dn10 再按最后一行展开得递推公式

D2nandnD2n2bncnD2n2 即D2n(andnbncn)D2n2

于是

D2n(aidibici)D2

i2n而

D2a1b1a1d1b1c1

c1d1n

所以

D2n(aidibici)

i1 (5) Ddet(aij) 其中aij|ij|;

解 aij|ij|

Dndet(aij)23210        n1n2n3n4            n1n2n3n4  0

111111r1r21111111111r2r3           

n1n2n3n4            111

1  0  0  0  0

  0      n1

1000200c2c112201222c3c11           

n12n32n42n5 (1)n1(n1)2n2

解

1a1111a2Dn    11  1  1      1an1a11(6)Dn11a2    11  1  1      1an, 其中a1a2    an0

a10a2a2c1c20a3c2c3    00   

0000a3  00  001  001  001          an1an11  0an1an

110a1a2  an  00100a1a2  an  00011  00010  00n001  00001  0                      000  10000  00a1110a210a3    11an1111an000  1a111a2a31  1an1ni1

0  001ai1

(a1a2an)(11)

i1ai

8 用克莱姆法则解下列方程组

x1x2x3x45x12x2x34x42(1)

2x13x2x35x423xx2x11x01234

解 因为

11D231231111214142511

151D212122130214284511

5D12201231111214142511 

1D3123D1D1231522014426511D2D

1D4123D3D3123111125214220D4D1

所以

x11

x22

x3

x4

15x16x20x15x26x3x25x36x40(2)x35x46x50x45x51

解 因为

51D00065100066565

10D100151D300051D55101507650007036510021201

51D211456500039565

51D4000

所以

x11507665

x21145

x3703

x4395

x4212

665665665665 9 问

取何值时

零解？

解 系数行列式为

xx2x301齐次线性方程组x1x2x30有非x12x2x3011D11121

令D0 得

0或1

于是 当0或1时该齐次线性方程组有非零解

10 问取何值时

有非零解？

解 系数行列式为

124134D231211111101(1)x12x24x30齐次线性方程组2x1(3)x2x30x1x2(1)x30

(1)3(3)4(1)2(1)(3)

(1)32(1)23

令D0 得

0

2或3

于是 当0

2或3时 该齐次线性方程组有非零解


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