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上海高二第一学期周练-行列式(二)、算法初步(第十五周)-006

上海高二第一学期周练-行列式(二)、算法初步(第十五周)-006
上海高二第一学期周练-行列式(二)、算法初步(第十五周)-006

《每周一练》(高二年级)

行列式(二)、算法初步

上海市金山中学 龚伟杰 【作者编号006】

一、填空题(每题5分,本大题共10题,共50分) 1.若d c b a ,,,四数成等比数列,则行列式

d

c b a 的值为 .

答案 0.

解 ∵d c b a ,,,四数成等比数列,∴0=-bc ad ,

0=-=bc ad d

c b a .

点评 本题考查双基.

2,则方程组的解的情况为 .

答案 解 无解或无数多解.

点评 本题考查三元一次方程组解的条件.

3.行列式0

134203

21的第二行第一列元素0的余子式是 .

答案

132.

解 余子式为0

132.

点评 本题考查余子式的概念,注意余子式与代数余子式之间的区别.

4.行列式0

134203

21的第二行第一列元素0的代数余子式是 .

答案 0

132-

解 代数余子式为0

1320

132)1(1

2-

=-+.

点评 本题考查代数余子式的概念,通过余子式来展开三阶行列式体现了一种化归的思想.

5.计算行列式:=1

010104

02 .

答案 2-.

解 按对角线法则或按行按列展开=

1

010104

022421

14

2-=-=. 点评 本题考查双基.

6.把

33222b a b a 33113b a b a -2

2

114b a b a +表示成一个三阶行列式为 .

答案 33

22

11

432

b a b a b a . 解 33

22

1

1

432

b a b a b a (答案不惟一). 点评 本题考查三阶行列式按行或列展开. 7.若0234150

2

=--x

,则=x .

答案 8-.

解 行列式展开,∵)8(224

12

234150

2

+=-=--x x

x

,∴8-=x . 点评 本题考查行列式的展开式.

8.方程组??

?

??-=+-=++=+-1

87552023z y x z y x z y x 的解为 .

答案 ??

?

??===121z y x .

解 ∵508752111

23=--=D ,50871215

1

20

=---=x D ,1008

152511

3

=-=y D ,501

755110

23=---=z D ,∴????

?

????

======121D D z D D y D D x z y x .

点评 本题考查用行列式解三元一次方程组.

9.若关于z y x ,,的方程组??

?

??=++=++=++4234z by x z by x z y ax 有惟一解,则b a ,满足的条件是 .

答案 1≠a 解 由01

21111

1

≠-==ab b b b a

D ,得1≠a 且0≠b .

点评 本题考查双基.

10

答案 解 按行展开0=+-=ke

kd e

d c kf kd f d b kf k

e

f e a

kf ke kd f e d c

b

a

,是充分的,但01

7551

1

00=--且没有两行成比例,故填“充分非必要”.

点评 本题考查三阶行列式的性质.

11.(2009年山东高考题)执行下面的程序框图,输出的=T .

答案 30.

解 按照程序框图依次执行为2,2,5===T n s ;642,4,10=+===T n s ;

1266,6,15=+===T n s ;20812,8,20=+===T n s ;s T n s >=+===301020,10,25,

故输出30=T .

点评 本题考查算法.

12.根据框图,得到一个数列}{n a ,则程序最后输出的a 值为 .

答案 1024

解 由题意知,???≥==-2,21

,11

n a n a n n .

当2≥n 时,12-=n n a a ,∴}{n a 是等比数列,12-=n n a ,102421011===a a . 点评 本题考查数列与算法.

13.(2010年上海高考题) 2010年上海世博会园区每天9:00开园,20:00停止入园.在下面的框图中,S 表示上海世博会官方网站在每个整点报道的入园总人数,a 表示整点报道前1个小时内入园人数,则空白的执行框内应填入 .

答案 a s s +←解 考查算法. 点评 本题考查算法.

14.如下图 “杨辉三角形”,从左上角开始的4个元素构成的二阶行列式

2

111的

值等于1;从左上角开始的9个元素构成的三阶行列式6

3

1

321

1

11

的值也等于1;猜想

从左上角开始的16个元素构成的四阶行列式

20

104110

6

3

143211111的值等于 .

答案 1.解 猜想

120

104110

6

3

14321111

1=.也可用代数余子式的概念展开行列式:

120

104110

6

3

143211111=.

点评 本题考查归纳猜想论证的思想方法和对新知识的理解能力. 二、选择题(每题5分,本大题共3题,共15分)

1.一元二次方程02=++c bx ax 的判别式ac b 42-=?用行列式表示的结果是…………………………………………………………………………………( ).

A .

b

c

a b 4- B .

1

42c

a b - C .

b

c

a b 4 D .

b

c

a b -4

答案 C . 解 ∵

b

c

a b 4ac b 42-=,故选C .

点评 本题考查双基.

1 1 1 1 1 1 1… 1

2

3

4

5 6… 1 3

6 10 15… 1 4 10 20… 1 5 15… 1 6 … 1…

2. 行列式i

h

g

f e d

c

b a 的计算方法可以如下图所示, 将三条实线相连的元素的乘积之和减去三条虚线相连的元素的乘积之和.对于这种计算方法你认为是……( ).

A .不正确D . 以上都不是

答案 B .

解 对比它的展开式知,它是正确的,故选B . 点评 提供了三阶行列式展开式的一种记忆方法.

3.1=a 是行列式03

1

113231=a

a

的…………………………………………( ). A .充分非必要条件 B .必要非充分条件

C .充要条件

答案 D .

解 ∵25923

1

1132312-+-=a a a

a

,025922=+-a a ,0

点评 本题考查双基.

4.如下图给出了一个程序框图,其功能 ……………………………… ( ). A .求第几项开始为负数 B .求前多少项的和开始为负值 C .求第几项取得最大值 D .求第几项取得最小值

答案B.

解略.

点评本题考查双基.

5.(2009年浙江高考题)某程序框图如图所示,该程序运行后输出的k的值是………………………………………………………………………………().A.4 B.5 C.6 D.7

答案 A .

解 按照程序框图依次执行为1,1==k S ;2,3==k S ;3,11==k S ;

4,1002311=>+=k S .故输出4=k ,选A .

点评 本题考查双基.

三、解答题 (14题15分,15题20分)

1.用行列式解方程组??

?

??=-++=++=++01223532732z y x z y x z y x .

答案 ??

?

??=-==213z y x .

解 ∵182131323

21-==D ,542

1121353

27

-==x D ,1821231523

7

1

==y D ,

3612

13532721-==z D ,∴????

?

?

???

==-==

==

213D

D z D D y D D x z

y

x

. 点评 本题考查用行列式解三元一次方程组.

2.用程序框图表示“计算

1

111++++ 的值”的循环结构.

答案

解 (略)

点评 本题考查双基.

3. 若关于x 、y 、z 的方程组???

??=+=++=++m z x m z m y x z y x 212有唯一解,求m 所满足的条件,并

求出唯一解.

答案 ????

?

????+=-=+-+=

112111222m z m y m m m x .

解 )1)(1(1101111

1

122-+=-==m m m m D , 当0≠D 即1±≠m 时,原方程组有唯一解.

)

122)(1(1

211

112

2

-+-==m m m m m m

D x ,

)21)(1)(1(1

2111

112m m m m m m D y --+==,

120111111-==m m m D z ,∴原方程组的唯一解是????

?

????+=-=+-+=112111222m z m y m m m x .

点评 本题考查用三阶行列式解三元一次方程组,并注意分类讨论.

4.根据框图,得到一个数列}{n a .(1)求2a 、3a 、4a 、5a ;(2)写出该数列的递推公式和通项公式;(3)求程序最后输出的a 值.

答案 (1)212=a 、313=a 、414=a 、515=a ;(2)???????≥+==-2,1

11

1,11

n a n a n n ,

n a n 1=;

(3)

100

1

. 解 (1)212=

a 、313=a 、414=a 、5

1

5=a ;

(2) 由题意知,???????≥+==-2,111

1,11

n a n a n n .

由1

111-+

=

n n a a ,得

1111-+=n n a a 1111=-?-n n a a ,∴}1{n

a 是等差数列,n n a n =?-+=1)1(11即n

a n 1

=. (3)100

1

100=

=a a . 点评 本题综合考查数列与算法.

/20101114/《每周一练》(高二年级)

上海市金山中学 龚伟杰

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