計科線性代數(shù)課件第一章

a11

x1

+

a12

x2

+

+

a1n

xn

=

b1

a

x

+

a

x

+

+

a

x

=

b21

1

22

2

2n

n

2設(shè)線性方程組

an1

x1

+

an2

x2

+

+

ann

xn

=

bn若常數(shù)項b1

,b2

,,bn不全為零,

則稱此方程組為非齊次線性方程組;

若常數(shù)項

b1

,

b2

,,bn

全為零,

此時稱方程組為齊次線性方程組.非齊次與齊次線性方程組的概念一、克拉默法則如果線性方程組

a11

x1

+

a12

x2

+

+

a1n

xn

=

b1(1)an1

x1

+

an2

x2

+

+

annxn

=

bn

a

x

+

a

x

+

+

a

x

=

b21

1

22

2

2n

n

2a1nan1

an

2

anna11

a12a22的系數(shù)行列式不等于零，即D

=a21a2

n

?

0DD

D

D

DD

D,

x

=

,

x

=

,

,

x

=

Dn

.x

=222

3

n11其中Dj

是把系數(shù)行列式D

中第j

列的元素用方程組右端的常數(shù)項代替后所得到的n

階行列式，即bnb1an1

an

,

j

-1

an

,

j

+1

anna11

a1

,

j

-1

a1

,

j

+1

a1nDj

=

那么線性方程組1)有解，并且解是唯一的，解可以表為證明用D中第j列元素的代數(shù)余子式A1

j

,A2

j

,,Anj依次乘方程組(1)的n個方程,得=

bnAnj(a

x

+

a

x

+

+

a

x

)A

=

b

A(an1

x1

+

an

2

x2

+

+

ann

xn

)Anj2 2

j21

1

22

2 2

n

n

2

j=

b1

A1

j

a11

x1

+

a12

x2

+

+

a1n

xn

)A1

j在把n

個方程依次相加，得n=

bk

Akj

,k=1由代數(shù)余子式的性質(zhì)可知,上式中x

j的系數(shù)等于D,nn

n

k=1

k=1

k=1

ak1

Akj

x1

++akj

Akj

xj

++akn

Akj

xn于是

Dxj

=

Dj

j

=

1,2,,

n).DDDDDD

D,

x

=

,

x

=x

=n,

,

x

=

Dn

.232211而其余xi

i

?

j)的系數(shù)均為0;又等式右端為Dj

.2)當D

?0

時,方程組2)有唯一的一個解由于方程組2)與方程組1)等價,故D=

Dn

.DDD

DD

D,

x

=

,

x

=x

=n,

,

x232211也是方程組的1)解.二、重要定理定理4

如果線性方程組

1)的系數(shù)行列式則1)一定有解,且解是唯一的.D

?

0,定理4’解，則它的系數(shù)行列式必為零.如果線性方程組1)無解或有兩個不同的齊次線性方程組的相關(guān)定理(

)2an1

x1

+

an

2

x2

+

+

ann

xn

=

0

a11

x1

+

a12

x2

+

+

a1n

xn

=

0a

x

+

a

x

+

+

a

x

=

021

1

22

2 2

n

nD

?0

則齊次線性方程組定理5

如果齊次線性方程組

2)的系數(shù)行列式2)沒有非零解.定理5’如果齊次線性方程組2)有非零解,則它的系數(shù)行列式必為零.系數(shù)行列式D

=0

a11

x1

+

a12

x2

+

+

a1n

xn

=

0a

x

+

a

x

+

+

a

x

=

021

1

22

2

2n

n

an1

x1

+

an2

x2

+

+

ann

xn

=

0有非零解.例1

用克拉默則解方程組2解D

=

2

x1

x

12

x

x1

+

4+

x2

-

5

x3

+

x4

=

8,-

3

x2

-

6

x4

=

9,-

x3

+

2

x4

=

-5,x2

-

7

x3

+

6

x4

=

0.21-

5107-

5131-

30-

6r1

-

2r21-

30-

602-

12r4

-

r202-

1214-

7607-

712

1

2

c

+

2c3

2c

+

2c-

3-

53-0-

10-

7-

7-

27

-

5

13=

-

2

-

1

27

-

7

12-

3

3-

7

-

2==

27,81-

519-

30-

6-

52-

1204-

76D1

==

81,22

8

-

5

11

9

0

-

6D

=0

-

5

-

1

21

0

-

7

6=

-108,2

1

8

13D

==

-27,2

1

-

5

8410-

329-

5-

62D

=10-

320-

19-

5140614-

70=

27,1=

=

3,D

27D

811\

x

=2=

-4,2=D

27D

-108x

==

-1,33=D

27D

-27x

=4D

27x

=

D4

=

27

=

1.例2

用克拉默法則解方程組

3

x

+

4

x

=

4,2

4

3

x1

+

5

x2

+

2

x3

+

x4

=

3,1

2

3

4

x1

-

x2

-

3

x3

+

2

x4

=

5

6.

x

+

x

+

x

+

x

=

11

6,解3

5

2

10

3

0D

=1

1

1

11

-

1

-

3

24

=

67

?

0,D1

=

11

6673

5

2

14

3

0

41

1 1

=

3

,5

6

-1

-3

23

3

2

10

4

0

411

6

1

11 5

6

-

3

2D2

=

1=

0,33

5

3

10

3

4D

=1

1 11

6

11

-1 5

6

224

67=

,43

5

2

30

3

0

4D

=1

1

1 11

61

-1

-3 5

6=

67,D673

1

1D

67

3= =

,1\

x

==

0,D

67D

0x=

2

=2,D67

3

1D

67

22

=3x

===

1.4

=D

67D

67x4

=1

2

3+

x

=

0,x1

+

x

2

+

(1

-

l

)x

3

=

0,

2

x

+

(3

-

l

)x例3

問

l

取何值時，齊次方程組

(1

-

l

)x1

-

2

x

2

+

4

x

3

=

0,有非零解？解1

-

l-

24D

=23

-

l1111

-

l1

-

l

-

3

+

l

4=

2 1

-

l

11

0 1

-

l=

(1

-

l

)3

+

(l

-

3

)-

4(1

-

l

)-

2(1

-

l

)(-

3

+

l=

(1

-

l

)3

+

2(1

-

l

)2

+

l