第一學(xué)期線性代數(shù)

1.3

逆矩陣一、逆矩陣的概念與性質(zhì)二、用行初等變換求逆矩陣返回一、逆矩陣的概念與性質(zhì)數(shù)a

≠0：a

a-1

=a-1

a

=1?:?矩陣A:A

(?)=I定義

設(shè)A為n階矩陣，若存在n階矩陣B，使得AB

=

BA

=

I,則稱A為可逆矩陣，B為A的逆矩陣，記為A-1

=B.若A可逆，則A-1存在，且A

A-1

=A-1

A=I.單位陣I

:對角陣:1nn

d

d1D

=

（,

d

,

...,

d

?

0）；dn

1

1D-1

=

d1I

-1

=

I定理1

設(shè)A可逆，則它的逆是唯一的.證設(shè)有B和C

滿足AB

=

BA

=

I,AC

=

CA

=

I.注意若A,B均為方陣，且AB=I

(或BA=I),則A可逆且B=A-1.例

設(shè)2 1

,

-

1 0

A

=

求

A

的逆陣

.

c d

b

B

=

aA則1

a

0AB

=

2

b

=

1

-

1

0

c d

0

1-

b

0-

a2b

+

d

1

0

=

1

2a

+

c解

設(shè)由定義利

用待定系是數(shù)法的逆矩陣,12b

+

d

=

0,-

a

=

0,-

b

=

1,

2a

+

c

=

1,

d

=

2.

c

=

1,b

=

-1,

a

=

0,

-

1

0

1

2

2

1

0

-

1

2-

1=

0所以

1

2

1

1

0

=

,

1

2

-

1

0

0

10

-

1.-1A

=

又因為ABBA性質(zhì)設(shè)A,B

均為n階可逆矩陣，數(shù)λ≠0，則證3:(

AB)(

B-1

A-1

)

=A(

BB-1

)

A-1

=

AA-1

=

I所以AB可逆，且(AB)-1

=B-1

A-14：A-1可逆，且(A-1)-1

=A;λA可逆，且(λA)-1

=1/λ

A-1AB可逆，且(AB)-1

=B-1

A-1;AT可逆，且(AT)-1

=(A-1)T.(2)

AB

=

AC

？B

=

C問：(A+B)-1

？=A-1

+B-1若A可逆,

(1)

AB

=

0

？

B

=

0

成立消去律成立不一定成立例2

設(shè)方陣A滿足A2

-

A

-

2I

=O,

證明：A和I

-A都可逆，并求其逆矩陣；A+I

和A-2I不同時可逆.證

(1)A(

A

-

I

)

=

2I2A(

1

(

A

-

I

))

=

I2所以A可逆，且A-1

=1

(A

-I

)2(-

1

A)(

I

-

A)

=

I所以I

-A可逆，且(I

-A)-1

=-1

A(2)2(

A

+

I

)(

A

-

2I

)

=

A2

-

A

-

2I

=

O所以，A+I

和A-2I不同時可逆.為什么？例3

設(shè)

B2

=

B

,

A

=

I

+

B2證明：A可逆且A-1

=1

(3I

-A).2

22

=

3

(I

+

B)-

1

(I

+

B)22

2=

3

I

+

3

B

-

1

(I

2

+

2B

+

B2

)2

2

2=

3

I

+

3

B

-

1

I

-

B

-

1

B22

2

2

2=

I2\

A

可逆且A-1

=1

(3I

-A).證

A

1

(3I

-

A)

=

3

A

-

1

A2例4矩陣

A滿足:

Ak

=

0

,證明：I

-A

可逆；求：(I

-A)-1

.分析：(I

-

A)?

?

=

I

.(I

-

A)I

+

A

+

A2

++

Ak

-1=

I

+

A

+

A2

++

Ak

-1

)+-

A

-

A2

-

-

Ak

-1

-

Ak=

I

-

Ak

=

I\

I

-A

可逆且(I

-A)-1

=I

+A

+

+Ak

-1

.解問題：初等矩陣可逆嗎？其逆陣呢？E

-1

=

E

；

E

-1

(c)

=

E

(1

),

c

?

0;cij

ij

i

iE

-1

(c)

=

E

(-c).ij

ij為什么？定理設(shè)A為n階矩陣，則下列各命題等價：A是可逆的；AX=0只有零解；A與I

行等價；A可表為有限個初等矩陣的乘積.證

1→2:2→3:

A顯然（為什么？）行初等變換fi

B

(行階梯形矩陣)BX

=

0只有零解.

B的對角元均非零。（即非零行數(shù)等于矩陣的階數(shù)）否則B的最后一行的元全為零，則BX

=0有非零解.矛盾則，B

行初等變換fi

B的行簡化階梯形=

I1234由條件，A可經(jīng)行初等變換得I.顯然（為什么？）推論設(shè)A為n階矩陣，則AX=b有唯一解的充要條件是A可逆.證

充分性：必要性：設(shè)AX=b有唯一解X,但A不可逆.A不可逆

AX=0有非零解Z.令Y=X+Z,則Y為AX=b的解，矛盾.3→4:4→1:二、用行初等變換求逆矩陣(

A,

I

)行初等變換fi

(

I

,

A-1

)方法由例1可知用定義利用待定系數(shù)法求矩陣的逆特別是高階矩陣的逆將十分麻煩。因此，我們將推導(dǎo)求逆矩陣的簡便方法.設(shè)A可逆，所以存在初等矩陣E1,…,Ek,使得例6

求A的逆矩陣：32.

1

1

2

3A

=

2

13解

1(

A,

I

)

=

2

1

0010fi

0

1

2

3

1

0

0

1

2

3

1

0

01

2

0

1

-

3

-

4

-

2

13

3

0

0

1

0

-1

0fi0

-

4

-

5

1

0

-4

345

-

140

13

0

123100

123100

010-101

fi

010-101

4

4-

1

-

314

04

41

0

-1

0

1

，0

1

54

1

0

0

-

3

3

-

3

54

-

3

3

1

所以，A-1

=

1

-

4

0

4

-14

4-

1

-

394

01

fi

0

1

2

0

-

11

3

4

4fi

0

1

0

-1

00

1

54問：怎樣利用以上結(jié)果解方程組AX

=b

？其中A為上面矩陣，X

=（x1,

x2,

x3

)T

，

b=(1,

0,

-1)TX

=

A-1b

=‥‥‥=

(-1,

-2,

2)如果A

不可逆呢？，不是方陣呢？例7

求A的逆矩陣：21

.

0-1

1

-

2

1

04A

=解

0011

0fi

0

1

-

2

1

1

0

0(

A,

I

)

=

2

0

1

0

14

-1

0

0

1

-

2

1

1

0

04

-1

-

2

14

-1

0

00

0

00fi

0

1

-

2

1

1

0

04

-1

-

2

1

02

-1

1為什么？A不可逆例70

,

1-1-1

0

0

A

=

1

-11(2)(A

+

2I

)-1

(A

-

2I

).計算：(1)(A

+2I

)-1

(A2

-4I

),(A

+

2I

)-1

A2

-

4I

)=

(A

+

2I

)-1

(A

+

2I

)(A

-

2I

)

22-1

1

1-

3

-1

0

0

2=

A

-

2I

=

1

-1

0

-

1

-

3

0

0

=

1

-

3

0

.1解例8

設(shè)矩陣11

.

01

0

,

B

=

0

2

2

3

1

1

1A

=

1

-1

1

-1

2

0(2)XA

=

B

.例9

解矩陣方程

：(1)AX

=

B

,1

-100

2(A I

)=

11

-10fi

202310-10012100-100123102100

1解

(1)A-1

AX

=

X

=

A-1

B

.(2)

(A

+

2I

)-1

(A

-

2I

)=

?111

0

0

1

0

01

0

0

1

0

fi1

1

0

0(A

+

2I

I

)=

1

01

1

-1

011

00

1

0

-1

100100

100100

010-110

fi

010-110-

301

1

0

1

0

0

-

3

0

0

1

0

1

-

3-1

1-1A

-

2I

)=

-1(A

+

2I

)

(

0-

3

-

3

0

0

=

4

-

3

0

.4例10求A的逆矩陣A-1

.0

,ni

=1nai

?

0

.aan-1

0

a1

0

0

0

0

a2

0A

=

0

0

0

0

0

1

n

a

0

0

0

an-1

0

0

0

00

0

0

0

0

0

0a100100

0

00a20010

0解fi0

00

00

0

1n

-10

0

a0a2001

0

an0000001

0a1001000fi

010

01

11n-1n

aaa

0

0

0

0

0

0

0

0

1

0

a2

0

0

0

1

0

0

1

0

00

0

0

0

0

01A-1

=

1

7

1

41

2A-1

BA

=6

A

+BA,且A

=oo

求B.A-1

BA

-

BA

=

6

A

A-1

-

I

)BA

=

6A

A-1

-