线性代数习题册行列式-习题详解

发布时间 : 星期五 文章线性代数习题册行列式-习题详解更新完毕开始阅读

a1100a143.

0a22a2300a23a32a41.( ╳

32a330?a11a22a33a44?a14aa4100a44)

二、选择题

1.若D?x2x?111?3x?11x?1,D2?1x?2,则D1与D2的大

小关系是( ).

(A)D1?D2; (B)D1?D2;(C)D1?D2;(D)随x值变化而变化.

答案:C 2.行列式

abcd(a,b,c,d???1,1,2?)的所有可能值中,

最大的是( ).

(A) 0; (B)2; (C)4; (D)6.

答案:D

三、填空题

1.

cos20?sin40?sin20?cos40?= .

解析:

cos20?sin40?sin20?cos40??cos20?cos40??sin20?sin40? ?cos60??12. 2.若x2y2xx?11?y?y,则x+y= . 解析:由x2y2xx?11?y?y,得x2?y2??2xy 即(x?y)2?0,从而x+y=0.

3.已知

x211?0,xy11?1,则y= . 解析:由x211?0,xy11?1,得x=2,x-y=1,从而y=1

1354. 若a2b2c2?a2A2?b2B2?c2C2,则C2化简后的结果246等于 . 解析:C132??24?2.

2xx125.设f(x)?1x1?132x1,则x4的系数为 ;x3的

111x系数为 .

解析:当f(x)的主对角线的4个元素相乘才能得出x4,系数为2;含x3的项只能是a12,a21,a33,a44的乘积,系数为-1. 答案:2,-1.

12345111226.设D?32146,

2221143210则(1)A31?A32?A33= ; (2)A34?A35 ;(3)A51?A52?A53?A54?A55 . 解析:A31?A32?A33?2(A34?A35)?0 2(A31?A32?A33)?(A34?A35)?0

于是A31?A32?A33?0,A34?A35?0.1234511122A51?A52?A53?A54?A55?32146

2221111111

1234511122?32146?0. 3333311111即A51?A52?A53?A54?A55?0.

1.计算下列行列式.

x1?y1x1?y2x1?y3x1?y4(1)

x2?y1x2?y2x2?y3x2?y4x3?y1x3?y2x3?y3xy;

3?4x4?y1x4?y2x4?y3x4?y4x1?y1y2?y1y3?y1y4?y1解:原式=

x2?y1y2?y1y3?y1y4?y1x3?y1y2?y1y3?y1y4?y1x4?y1y2?y1y3?y1y4?y1x1?y1y2?y1y3?y1y4?y1 =

x2?x1000x3?x1000?0.

x4?x10001?x1111(2)11?x211111?x;

311111?x41?x1?x1?x1?x1解:原式=

1x20010x

30100x41?x11?xx?x1x1x??x1?x1?x1203x4=x200 00x30000x4 =x1x2x3x4?x2x3x4?x1x3x4?x1x2x4?x1x2x3.

四、解答题

00?01000?200(3)

?????02005?000. 20060?00000?0020072006?2005解:原式=2007?(?1)22006!=?2007!

12345222112.已知D?31245?27, 1112243150求(1)A41?A42?A43;(2)A44?A45. 解:1?A41?1?A42?1?A43?2(A44?A45)?272(A41?A42?A43)?(A44?A45)?0

得A41?A42?A43??9,A44?A45?18. 3.计算下列n阶行列式.

11?1222?2n (1)Dn?332?3n; ???nn2?nn解:(利用范德蒙行列式计算)

11?112?nDTn?Dn?n!332?3n ???12n?1?nn?1?n!(2?1)(3?1)?(n?1)(3?2)(4?2)?(n?2)??n?(n?1)??n!(n?1)!(n?2)!?2!.

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