PDF《第一章部分习题解答》

第一章部分习题解答 P.

13 习题 1.2 9.证明:

11 1

1 11

12 1

21 22( )

2 21

2 2

1 1

2 1 j n n n t n j n j n n nn n nj d a t a t a t dt a t a t a t d a t a a t a t a t a t d dt dt a t a t a t d a t a t a t dt = = ∑ " " " " " " # # # # # " " " nn # 证:由(1.2.7)有 左边=

1 2

1 2

1 2 ( )

1 2 ( 1) n n n i i i i i i n i i i d a a a dt τ ? ∑ " " " =

1 2

1 2

1 2

1 2

1 2 ( )

1 2

1 2

1 2 ( 1) n n n n i i i i i i n i i i n i i i n i i i d d a a a a a a a a a dt dt dt τ ? ? ? + + + ? ? ? ? ∑ " " " " " " n d =

1 2

1 2

1 2

1 2

1 2

1 2

1 2

1 2

1 2

1 2

1 2

1 2 ( 1) ( 1) ( 1) n n n n n n n n i i i ii i i i i i i i n i i i n i i i n i i i i i i ii i d d a a a a a a a a a dt dt dt τ τ τ ∑ ∑ ∑ " " " " " " " " " n d " =

11 12

1 11

12 1

11 12

1 21 22( )

2 21 22( )

2 21 22( )

2 1

2 1

2 1

2 n n t n t n t n n nn n n nn n n d d a t a t a t a t a t a t a t a t a t dt dt dt d d a t a a t a t a a t a t a a t dt dt dt d d a t a t a t a t a t a t a t a t dt dt + + + " " " " " " " n n d d " " ) ( nn d a t dt " ) =右边.得证. 有不少同学是这样做的: 左边=

1 2

1 2

1 2 ( )

1 2 ( 1) n n n j j j j j n j j j d a a a dt τ ? ∑ " " " j =

1 2

1 2

1 2

1 2

1 2 ( )

1 2

1 2

1 2 ( 1) n n n n j j j j j nj j j nj j j nj j j j d d a a a a a a a a a dt dt dt τ ? ? ? + + + ? ? ? ? ∑ " " " " " " n d =

11 12

1 11

12 1

11 12

1 21 22( )

2 21 22( )

2 21 22( )

2 1

2 1

2 1

2 ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( n n n t n t n t n n n n n nn n n nn d d d a t a t a t a t a t a t a t a t a t dt dt dt d d d a t a a t a t a a t a t a a t dt dt dt d d a t a t a t a t a t a t a t a t dt dt + + + " " " " " " " ) ( nn d a t dt " ) =

11 12

1 1

2 1

1 2 n n i i i i n n nn a t a t a t d d d a t a t a t dt dt dt a t a t a t = ∑ " # # # " # # # " n (*) =右边 (*)=右边这一步应该是没有道理的. P.

21 习题 1.3 4. 求n阶行列式

1 1

1 1

1 1

1 1

1 1

1 1

1 1

1 1 D ? ? ? ? ? = ? " " # # # # " " 展开后正项的个数. 解:设D展开后正项和负项的项数分别为 P 和N,则1!2n P N n P N D ? + = ? = = 解这个方程组得:

1 !

2 2 n n P ? + = . 9. 题目略. 证明:原式=

5 1

2 3

4 1

8 0

5 18055

8 3

2 8

83283 10000

1000 100

10 6

1 0

4 61042

4 8

5 7

48576 5

7 7

7 57776 C C C C C + + + + 其中第五列元素都能被

23 整除,所以得证. 习题 1.4 3. (1)

1 1

1 1

1 0

0 0

0 0

0 0

0 0

0 ( 1) ( 1) ( 1)

0 0

0 0

0 0

0 0

0 0

0 0

0 n n n n n n n n n x y y y y z x z x z x z x D y xD yz xD x x z z x + + ? ? ? " " " " " # # # # " " " " 这样按第n列展开最方便,不过有不少同学按第n行展开, 虽然也能算得结果,但是过程稍微麻烦一些.

1 n + ? + P.

44 习题 1.6 3(2) 原式=

1 1

2 1

2 2

1 1

1 2

1 1

2 0

1 1

1 1

1 1 ( 1) ( 1)

2 2

0 1

1 1

1 3

1 0

1 1

1 1

1 1

1 1

1 1

1 1

1 1

1 1

1 0

0 0 ( 1) ( 1)

1 1

1 1

1 0

0 0

2 2

1 1

1 1

1 n n n n n i i n n n n R R n R R n n n n n C C R R n n n n n n n n n n ? ? ? = ? ? ? ? ? ? ? ? + + + ? ? ? ? ? ? ? + + = = ∑ " " " " " " " " " " " " " "

1 1

1 2

0 0

0 ( 1) ( 1) ( 1)( 1) ( 1)

2 2 n n n n n n n n n n n n n ? ? ? ? + + " % P.45, 5. 设11

12 13

1 1

2 0

0 .

1 0

3 0

1 0

0 n a a a a D n = " " " # # # # " 求 其中 是 的代数余子式 (

1 11

12 1 . n A A A + + + " 1i A 1i a ,2, , ) i n . = " 解:

11 12

1 1

2 1

1 1

1 1

2 0

0 1

0 3

0 1

0 0

1 1

1 1

1 2

0 2

0 ( 1/ 2) ( 1/ )

0 0

3 0

0 0

1 1 (1 ) !

2 n n A A A n n C C n C n n n + + + = ? ? ? = ? ? ? " " " " # # # # " " " " " " # # # " "

1 0

0 #