北大版-线性代数第一章部分课后标准答案详解
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北大版-线性代数第一章部分课后答案详解
————————————————————————————————作者:————————————————————————————————日期:
习题1.2:
1 .写出四阶行列式中
11121314212223243132333441
42
43
44
a a a a a a a a a a a a a a a a 含有因子1123a a 的项
解:由行列式的定义可知,第三行只能从32a 、34a 中选,第四行只能从42a 、44a 中选,所以所有的组合只有()
()
13241τ-11233244a a a a 或()
()
13421τ-11233442a a a a ,即含有因子1123a a 的项
为11233244a a a a 和11233442a a a a
2. 用行列式的定义证明111213141521
22232425
31
3241425152
000000000
a a a a a a a a a a a a a a a a =0 证明:第五行只有取51a 、52a 整个因式才能有可能不为0,同理,第四行取41a 、42a ,第三行取31a 、32a ,由于每一列只能取一个,则在第三第四第五行中,必有一行只能取0.以第五行为参考,含有51a 的因式必含有0,同理,含有52a 的因式也必含有0。
故所有因式都为0.原命题得证.。
3.求下列行列式的值:
(1)01000020;0001000
n n -L L M M M O
M L L
(2)00100200100000
n n
-L L M O M O M L L
; 解:(1)0100
0020
0001
000
n n -L
L
M M M O
M L
L
=()()23411n τ-L 123n ⨯⨯⨯⨯L =()1
1!n n --
(2)0010
02001000
00
n n
-L L
M O
M O M L L
=()()()()
12211n n n τ---L 123n ⨯⨯⨯⨯L =()()()
1221!n n n --- 4.设n 阶行列式:A=1111n
n nn
a a a a L
M O
M L
,B=111112122122
21212n n n
n n n n n nn
a a
b a b a b a a b a b a b a -----L L M
M
O
M L
,其中0b ≠,试
证明:A=B 。
证明:
B=
111112122122
21212n n n
n n n n n nn
a a
b a b a b a a b a b a b a -----L L M
M
O
M L
=
()
()
[]1212121212121n n n n s s s s n s s s s s n s s s n a b a b a b τ---∈-∑L L L !
=
()
(
)
[]1212121212121()n n n n s s s s n s s s s s n s s s n a a a b b b τ---∈-∑L L L L !
=()(
)
[]12121212(1)(2)()121n n n n s s s s s s n s s s n s s s n a a a b τ-+-+-∈-∑
L L L L !
=
()(
)
[]121212121n n n s s s s s s n s s s n a a a τ∈-∑
L L L !
=A
命题得证。
5.证明:如下2007阶行列式不等于0:
D=
22
22
33
332007
20072007
2007
1
220062007232007200834200820082007200820082008L L L
M M O M M L
; 证明:最后一行元素,除去2007
2007是奇数以外,其余都是偶数,故含2007
2008
的因式也都
是偶数。
若最后一行取2007
2007
,则倒数第二行只有取2006
2007
才有可能最后乘积为奇数,
以此类推,只有次对角线上的元素的积为奇数,其余项的积都为偶数。
故原命题得证。
习题1.3
1求下列行列式的值:
(1)
3111
131111311113
; (2)
0111
101111011110
; (3.)A=
+c 23243236310+6b 3a b c d a
a b
a b c
a b d
a a
b a b
c a b c
d a a b a b c a c d
++++++++++++++++,
解:
(1)
3111
131111311113
3423
12
λλλλλλ-+-+-+−−−→
31112200022
22
---4332
21
c c c c c c +++−−−→
6321020000200002
=48
;
(2)
0111101111011110
3423
12
λλλλλλ-+-+-+−−−→
011111
0001100
1
1
---4332
21
c c c c c c +++−−−→
332101
0000100
1
---=3-;
(3.).A=
+c 23243236310+6b 3a b c d a
a b
a b c
a b d
a a
b a b
c a b c
d a a b a b c a c d
++++++++++++++++,
+c 23243236310+6b 3a b c d a
a b
a b c
a b d a a b a b c a b c d
a a
b a b
c a c
d ++++++++++++++++==
023*********+63a c d a
a
a b c
a b c d a a a b c
a b c d
a a a
b
c a b c
d +++++++++++++++
=324326310+63a b c d a b
a b c
a b c d
a b a b c a b c d a b a b c a b c d
++++++++++++++0023243236310+63a
d a
a
a b
a b c d a a a b
a b c d
a a a
b a b
c d
+++++++++++
0000+=2432232432310+6336310+63a c d a a
a
c
a b c d a
a
a b
a b c a a c a b c d a a a b a b c a a c a b c d a a a b a b c
++++++++++++++++00000000+
=+=
23223432224323633610+633310+63a
d a a a
a
a b
d a
a
a
a b c a
a
b
a b c a a a b d a a a a b c
a a
b a b c
a a a
b d
a a a a
b
c a a
b a b c
+++++++++++++
000000000000+
=
23432322342333610+63633610366a a a a a
a
a
a b a
a
a
c a
a
a
a a
a
a
b a a a a b
a a a c a a a a
a a a
b a a a a b a a a c
a a a a
a a a b
++
=+
24
32431
+62
2
+3
5
000
000
00011
1100002
6
26234343
23405
55
361036103610a a a a
a a a a a a a a a a a a a a a a
a a a
a a a
a a
a a
a
a a a
λλλλλλ-+−−−−→−−−−→=
—
413
1065
a a a a a ⨯⨯⨯=
2.求下列n 阶行列式的值:
(1)()()2
1
21
2221
22311
12
n n n n n n n n n n n n ++++-+-+L L L M M O M L
;(2)32222322
22322223
L L L M M M O M L
;(3)
123103120
1230
n n n ------L L L
M M M O M L ;(4)1231131211231
n x n x n x +++L L L M M M O M L 解:(1)n D =()()2
121
2221
22311
12
n n n n n n n n n n n n ++++-+-+L L L M M O M L
; (1) 若n=1;则n D =1; (2) 若n=2;则n D =1234
=2-;
(3) 若3n ≥,则
n D =()()2
1
21
2221
2231112
n n n n n n n n n n n n ++++-+-+L L L M M O M L 2312
λλλλ-+-+−−−→
()()2
12
11
12
n n n n
n n n n n n n n -+-+L L L M M O M L
=0; 综上:n D =112
203
n n n =⎧⎪
-=⎨⎪≥⎩
(2)
3222232222322223
L L L M M M O M L
1i i
λλ--+−−−−−−−→
L 其中,i 先后取n,n-1,2
3
2
22110001
1
00011
---L L L
M M O O M
1
i i c c -+−−−−−−→
L i 依次取n,n-12
()()3212222
20100
00100000
01
n n +--⨯L
L L M O O M =2n+1; (3)
1
2
3103
12
1230
n n n ------L L L
M M M O M L
1n,n-1,2
i
i λλ+−−−−−→
L 依次取123223232n n n n
⨯L L
L
O
M =n!; (4)
1
2
3
1131211231
n x n x n x +++L
L L M M M O M L
123n
i
i ic c -+−−−−−→
L 依次取、、1
11121
1
x x x n ---+M O
=()()()121x x x n ---+L ;
习题1.4
1. 计算下列行列式:
(1)0
000
00
000
x
a b c
y d
c z f g h k u
l v
;(2)211212
21222
1
2
1+x 11n
n n n n
x x x x x x x x x x x x x x ++L
L
M M O M
L
;
(3)
765432978943
749700
536100
005600
006800
;(4)
00010000000000001000n n
a a a a a ⨯L L L M M M O M M L L
解:(1)
00000000x a b c
y d c z f g h k u l v 2424
c c λλ↔↔−−−→
00000000000
x
b a
c g u k h l z
c
f y
v
1212
c c λλ↔↔−−−→
000000000
00
u
g
k h
l x b a c
z c f y v
=xyzuv; (2)
D=
211212
21222
1
2
1+x 11n
n n n n
x x x x x x x x x x x x x x ++L
L
M M O M
L
=
21122
2121
2
1+x 0
101n n x x x x x x x x x +L
L
M M O M
L
+
211212
21222121+x 1n
n n n n
x x x x x x x x x x x x x x +L
L
M M O M
L
=
()
211212
21
222
11121
1+x 111n
n n
n n n n x x x x x x x x x x x x x x +---+-+L L M M
O M
K +2
n
x
21
12
1
2
2122
1
21+x 11
x x x x x x x x x +L
L M M O
M L
=211211
2
21
221
2
11
121
1+x 11n n n n n x x x x x x x x x x x x x x -----++L
L M M
O
M
L
+2n x =
2112
122
21
2
22
2
21
222
1+x 11n n n n n x x x x x x x x x x x x x x -----++L L M M
O
M
L
+2
n-1x +2n x =L =1+
2212x x +++L 2
n x ;
(Q 21
12
12
21
221
2
1+x 11
x x x x x x x x x +L
L M
M O
M L
i 12n 1
n i
i x λλ--+−−−−−−→
L 依次取、、1
2
3
1
1
1
1
x x x O L
=1)
(3)
765432978943
749700536100
005600
006800
=()()()56347632
5
6974316874005300+++-=
()
()()
3+4+1+2567432-168
5343
=
566874325343
=4;
(4)00010000000000001000n n
a a a a a ⨯L L L
M M M O
M M L L
=n
a +
()()()
231121n n n a τ---L =()2
2
1n a
a
--;
2.试用拉普拉斯定理计算:A=12342
22212
3
4
1
11001
23000
1
11100x x x x x x x x ; 解:
()()()()()()12121+2+1+32
341
342
22
2221234224134
2
2221234
11100123001
111
1
111
11
011111+-1121300x x x x x x x x x x x x x x x x x x x x +++=-⨯⨯()
()()
()()()()()12233
443324231412
23
4
011
11
102230x x x x x x x x x x x x x x ++++-⨯=------⎡⎤⎣⎦ 2. 利用范德蒙行列式计算:
(1)()
()
()
()
1
1
1
1111
1
1
n
n
n
n n n a
a a n a a a n a a a n ---------L L M
M L M L L
;(2)
11
11111111222
22
211111111
n
n n n n n n n n n n n
n n n n n n a a b a b b a a b a b b a a b a b b ------++++++L
L M
M
O
M
M
L ,
(0,i a ≠1,2,,1i n =+L )
解:(1)()
()
()
()
1
1
1
111111n
n
n n n n a a a n a a a n a a a n ---------L L M
M L M L L
i 1
n-12
i i λλ-↔−−−−−−→L 依次取n 、、
()
n-1
1-()
()
1
1
1
1
1
11n n n a a a n a
a a n ------L
L M M O M L −−−→
同理()
()()()
()
n-121
111111n n
n
n
a a a n
a a a n +-++-----L L L M M
O
M
L
=()
()()
12
11
1n n n i j j i -+≥>≥--∏
2)
11
11111111222
22
21
11
11
11
1
n n n n n n n n n n n n n n n n n n a a b a b b a a b a b b a
a b
a b
b
------++++++L L M M O M M L
=
()
n
121n a a a +L 2
1111112
2222222
3333332
1
111
111
1
11n
n
n
n
n n n n n n b b b a a a b b b a a a b b b a a a b b b a a a ++++++⎛⎫⎛⎫ ⎪ ⎪⎝⎭
⎝⎭
⎛⎫⎛⎫ ⎪ ⎪⎝⎭⎝⎭⎛⎫⎛⎫ ⎪ ⎪⎝⎭⎝⎭⎛⎫⎛⎫
⎪ ⎪⎝⎭
⎝⎭
L
L
L M
M
M
O
M
L
=()
n
121n a a a +L
n 11j i i j i j b b a a +≥>≥⎛⎫
- ⎪ ⎪⎝⎭∏=()n 11
i
j
i j i j b a
a b +≥>≥-∏
习题1.5
1. 用克莱姆法则解下列方程:
(1)1234124
2341234258
369
2254760
x x x x x x x x x x x x x x +-+=⎧⎪--=⎪⎨
-+=-⎪⎪+-+=⎩ 解:
D=
2
151130602121
4
7
6
-----24
λλ-+−−−→2
1511306021
2
7
712
-----=()(
)()
34231+++-2121
7716
---+
()
()()
34242225
17121
+++--+()
()()
34341221
171213
+++----=27;
同理:x D =91,y D =108-,z D =27-,
w D =27;∴1x =x D D =3;2x =y D D =4-;3x =z D D =1-;4w D x D
==1;
总复习题一
1.计算行列式D=
2111 4211 2011029998 1212
-
-
-
;
2.计算行列式D=
246427327 1014543443 342721621
-
;
3.计算行列式D=1111 1111 1111 1111
x
x
y
y
+
-
+
-
;
4.计算行列式D=
1111 1111 1111
1111
x
x
x
x
--
-+-
--
+--
;
5.计算行列式D=1333 3233 3333
333n
L
L
L
M M M O M
L
;
6.计算行列式A=
11121
21222
12
n
n
n n n n
a b a b a b
a b a b a b
a b a b a b
+++
+++
+++
L
L
M M O M
L
;
7.计算行列式
D=
()
100
120
123
1
2
1
n
n
--
---N
;
8证明D=123
1111111
1
1111111
1n
a a a a ++++L
L L M M M O M L
; 9.证明:
2cos 10001
2cos 1
00
01
2cos 000002cos 10
1
2cos x x x x x
L L
L M M M O M M L L
=sin(1)sin n x
x + 10.试证明
()()()()()
()()()
()
111212122212n n n n nn a t a t a t a t a t a t d dt a t a t a t L L M
M O M L
=
()()()()()()()()()
111121221
1j n n
j n j n nj nn d
a t a t a t dt d
a t a t a t dt d
a t a t a t dt
=∑L
L L L M
L
M L M L
L
11.一个n 阶行列式n D 的元素满足,则称为反对称行列式,证明:奇数阶反对称行列式为零。
12计算由杨辉三角规律给出的n 阶横列式
D=
11111123
3
136141L L L L L
L M O
L
L M M M O
L
M M M M M O
解:1、
D=
2
111421120110299981
2
1
2
---=331
124
c c i
i c c ++−−−−−→−−−→依次取、、42
1263
103399133
1
1
---
34
λλ-+−−−→42126
3
1
033991
00100
--=()4+3
422
1100630331
--=1800-
2.
D=246
427327
1014543443342721621
-=()11
1+-246
543443721621
+()12
10144431427
342621
+--+
()
13
10145431327
342721
+--=246()()()()50043600214004370021++-++⎡⎤⎣⎦
()4271014621342443-⨯+⨯+()3271014721342543⨯+⨯=246⨯17800+()()1014721427100427721100---⎡⎤⎣⎦+342()()543427100427543100---⎡⎤⎣⎦=
4378800-29811600-3967200=-29400000
3、D=1111111111111
1
1
1x
x y y
+-+-=
4123
i
i c c -+−−−−−→
依次取、、x 001001001
1x
y
y
y
y y
--=
x 001001
0011x y y y y -+
x 000000000
x y
y
y
y y
--1234
i
i λλ-+−−−−−→
依次取、、x 00100000
x x x y x y
y
y ----++2
2
x y
34
+λλ-−−−→x
00100
00y
00
x x x y y
---+2
2
x y =2
2
x y
4、
D=
1111111111111
1
1
1
x x x x ---+---+--+1i i
i c c +−−−−−→依次取1、2、3
00
10
10
100
1
x x x
x
x
x x ----1321i i
i λλ+-+−−−−−→
依次取、、x 10
001
x x x --=()
()314
42
1x -⨯-= 4
x
5、D=133332333333333n
L
L L
M M M O M L 312456n
i
i c c -+−−−−−−→
L 依次取、、、、2
313
3133
3
n ---M O
=
3
2
1
13
n ---O
=6()3n -!
6.计算行列式A=
111212122212n n n n n n
a b a b a b a b a b a b a b a b a b +++++++++L L M M O M L
Q A=
111212122212n n n n n n
a b a b a b a b a b a b a b a b a b +++++++++L L
M M O M L
;
1)若n=1,则A=11a b +; 2)若n=2,则A=
()()1112
21122122
a b a b a a b b a b a b ++=--++;
3)若n ≥3,则A=
11121122312122212232121223n n n n n n n n
n n
a b a b a b b b b b a b a b a b a b b b b b a b a b a b a b b b b b a b +++--++++--+=
+++--+L L L L M M O M M M O M L
L
=0
∴()()1121121203a b n A a a b b n n +=⎧⎪
=--=⎨⎪≥⎩
7.计算行列式
D=
()100120123
121n n
-----N
2312
λλλλ-+-+−−−→
=
()
100020003121n n
-----N
=()
1+1
1n +-()20031
21n n
-----N
=
()
1+n 1
1+-()
1+n 1
1+-2⨯⨯
()31
21n n
-----N
=
()(
)
321n +-()12
61n n
--⨯
---N =
()
()+3(+4)
2
13!!n n n -=()
(1)
2
13!!n n n --
8证明
D=123
11111111
11111111n a a a a ++++L L L M M M O M L =12111n
n i i a a a a =⎛⎫+ ⎪⎝⎭
∑L ,
12n a a a L 0≠;
D=123
111111*********
1n
a a a a ++++L L L M M M O M L
=+1
i n-1n-221
i i λλ-+−−−−−−−→
L 依次取、、、、11
22
3211
1+a 111
1
n n n n
a a a a a a a a -------L
O
O i -1
n n-12
i i λλ+−−−−−−→
L 依次取、、、n 1
n 2n n-1n
1111+a 00a a 0a a a a ---L L O
M O
M =12n-1
111100000a 0
a a ---L L O M O
M +n 1n 2n n-1n
111a 00a a 0a a a a ---L L O
M O
M =()()
1+n
1
12111n n a a a ----L +
n
a 1
2n-1
111100110a 1
a a ---L L O
M O M 111n
i i i a c c ---+−−−−−−→L 依次取2、3、、n =121n a a a -L + n
a 111
1211
2n-11
111000
0a 0
n a a a a a ----++++---L L L O
M O
M =121n a a a -L + n a ()
11n
+-(111
1211n a a a ----++++L )()
n-1
1-121n a a a -L
=12111n
n i i a a a a =⎛⎫
+ ⎪⎝⎭
∑L
9.证明:
2cos 10001
2cos 1
00
12cos 000002cos 10
1
2cos x x x x x
L L
L M M M O M M L L
=sin(1)sin n x
x +; Q n D =
2cos 100012cos 1000
12cos 000002cos 10
1
2cos n n
x x x x x ⨯L L L
M M M O M M L L
=
2cosx
()()
112cos 100012cos 1000
12cos 000002cos 10
1
2cos n n x x x x x --L L L
M M M O M
M L L
-
()()
22110002cos 100
012cos 10012cos n n x x x --L
O M M
M M O L
=2cosx 1n D --2n D - 下面用归纳假设法证明:1):当n=1时,1D =2cos x =sin 2sin x
x
; 当n=2时,2D =2cos 11
2cos x x
=2
2cos x +cos2x =
sin 3sin x
x
;(Q
sin 3sin x x =
sin 2cos cos 2sin sin x x x x
x
+=22cos x +cos2x ) 当n=3时,3D =
2cos 1012cos 10
1
2cos x
x x
=3
8cos x 4cos x -=
sin 4sin x
x
(同理可证) 又3D =2cos x 2D 1D -=3
8cos x 4cos x -=
sin 4sin x
x
;
2):假设,当n=k ()3k ≥时,有k D =sin(k 1)sin x x +;1k D -=sin sin kx
x
成立
则当n=k+1时。
有
1k D +=
()()
112cos 10001
2cos 100012cos 000002cos 10
1
2cos k k x x x x x ++L L L
M M M O M M L L
==2cos x k D 1k D --=
2cos x
()sin k+1sin x x sin k sin x x
-
=[]
2cos sin(1)sin(1)cos cos(1)sin sin x k x k x x k x x x +-+-+=sin(1)cos cos(1)sin sin k x x k x x x
+++=()sin 2sin k x
x +;满足。
则原命题得证。
10.试证明
()()
()()()
()()()
()
111212122212n n n n nn a t a t a t a t a t a t d dt a t a t a t L L M
M O M L
=
()()()()()()()()()
111121221
1j n n
j n j n nj nn d
a t a t a t dt d
a t a t a t dt d
a t a t a t dt
=∑L
L L L M
L
M L M L
L
10.证明:
()()()()()
()()()
()
111212122212n n n n nn a t a t a t a t a t a t d dt a t a t a t L L M
M O M L
=d dt
()()
()()()12
1
2
121n
n
s s s s s s n a t a t a t τ-∑L L =()
()
()()()1212121[]n n s s s s s s n d
a t a t a t dt
τ-∑L L +()
()
()()()1212121[
]n n s s s s s s n d
a t a t a t dt
τ-∑L L ++L
()
()
()()()1212121[]n n s s s s s s n d
a t a t a t dt
τ-∑L L =
()()()
()()()
()()()
11112122n1j n j n nj nn d
a t a t a t dt d
a t a t a t dt d
a t a t a t dt
L L L L M M M M M
L L +L
()()()()()()()()()11112122n1j n j n nj nn d
a t a t a t dt d
a t a t a t dt d
a t a t a t dt
L L L L M M M M M L L +
+
L ()()()()()()()()
()11112122n1j n j n nj nn d a t a t a t dt
d a t a t a t dt d a t a t a t dt
L L L L M
M
M M M L
L =()()()()()()()
()()
1111212211j n n j n j n nj nn d
a t a t a t dt d
a t a t a t dt d
a t a t a t dt
=∑
L L L L M
L M L M L
L
11.一个n 阶行列式n D 的元素满足ij ji a a =-,则称为反对称行列式,证明:奇数阶反对称行列式为零。
证明:设D=
111212122212n n n n nn
a a a a a a a a a L L M M O M L
为反对称行列式,(其中n 为奇数)。
则必有
D=
()
()
121
2
121n n
s s s s s s n a a a τ-∑L L =()
()
1212121()()()n n s s s s s ns a a a τ----∑L L =
()(
)
()
12121211n n s s s n
s s ns a a a τ--∑
L L =-()
()
1212121n n s s s s s s n a a a τ-∑L L ⇒D=D -。
即D=0。
12计算由杨辉三角规律给出的n 阶横列式
D=
11111123
3
136141L L L L L L M O
L
L M M M O
L
M M M M M O
解:令
n D =
11111123
3
136141L L L L L L M O
L
L M M M O
L
M M M M M O
=121
12
00230123
012123
1
123
2
03
20
12221
2
2
2
2n n n n n n n n n n n
n n n n n
n n n n n n c c c c c c c c c c c
c c c
c
c
c
c
c
λλλλλλ---++-+-++--+--+++-+-−−−−→
M L L L O L L M M O
L M M
M M M O L
02310
123
10
12112
3
01123
1
03
30
12
331
1
3
3
2n n n n n n n
n n n n n n
n n n n n c c c c c c c c c c c c c c c c c c c ----++----++-+-L L L O L L M M O
L M
M
M M
M
O
L =1n D -=2n D -=L =1D =1
制作者:聂斌 时间:2011-9-29。