2005/7Inverse matrices-1 Inverse and Elementary Matrices.

2005/7 Inverse matrices-1

Inverse and Elementary Matrices

2005/7 Inverse Matrices-2

反矩陣 (inverse matrix)

nnMA

If there is a matrix such that nIBAAB nnMB

Note:If a matrix having no inverse matrix is called a noninvertible or

singular matrix.

Let

then (1) A is called an invertible or nonsingular matrix

(2) B is the inverse matrix of A


Theorem ： If B and C are both the inverse of A, then B = C

Pf:

CB

CIB

CBCA

CIABC

IAB

)(

)(

Since B = C, the inverse matrix of a matrix is unique.

Note:(1) The inverse matrix of A is denoted by

(2)

1A

IAAAA 11


Use the Gaussian-Jordon elimination to find the inverse matrix

1neliminatio AIIA

Ex ： Find the inverse of

31

41A

Sol: IAX

10

01

31

41

2221

1211

xx

xx

13

04

03

14

2212

2212

2111

2111

xx

xx

xx

xx

1 2

10

01

33

44

22122111

22122111

xxxx

xxxx


110

301

031

141)4(

21(1)

12 ,r

喬登消去法高斯

r

110

401

131

041)4(

21)1(

12 ,


rr

1 ,3 2111 xx

1 ,4 2212 xx

1

2

11

431AX

Hence


Note：

1110

4301

1031

0141

1

, )4(21

)1(12

AIIA

rr


If A can’t using row operations to be translated into identity matrix I, then A is a singular matrix.


Ex ： Find the inverse matrix of

326

101

011

A

Sol:

106326

011110

001011

)1(

12

r

100

010

001

326

101

011

IA


142100

011110

001011

)1(

3

r

142100

011110

001011

)4(

23

r

142100

133010

001011

)1(

32

r

106

011

001

340

110

011

)6(

13

r


Thus

142

133

1321A

141100

133010

132001

)1(

21

r

1 AI


The power of a square matrix

IA 0 )1(

0)(k )2(個

k

k AAAA

integers are , )3( srAAA srsr rssr AA )(

k

k

k

k

n d

d

d

D

d

d

d

D

3

2

1

2

1

00

00

00

00

00

00

)4(


Theorem ：

If A is invertible, then the following properties hold:

AAA 111 )(and invertible is )1(

kk

k

kk AAAAAAA )()(and invrtible is )2( 1

items

1111

0 ,1

)(and invertible is )3( 11 cAc

cAcA

TTT AAA )()(and invertible is )4( 11


111 ABAB

Theorem ：If A and B are both invertible with the size nn, then

AB is invertible and

Pf:

IBBIBBBIBBAABABAB

IAAAAIAIAABBAABAB1111111

1111111

Note:

11

12

13

11321

AAAAAAAA nn

Thus AB is invertible and the inverse matrix of AB is (BA)1.


Theorem ： Cancellation laws

If C is invertible, then the following properties hold:

(1) If AC=BC, then A=B

(2) If CA=CB, then A=B

Pf:

BA

BIAI

CCBCCA

CBCCAC

BCAC

11

11

Note:If C is noninvertible, then the cancellation laws do not hold.

Since C is invertible, C1 exists.


bx 1A

Theorem ： If A is invertible, then Ax = b has a unique solution .

Pf:

( A is nonsingular)

bx

bx

bx

bx

1

1

11

A

AI

AAA

A

If x1 and x2 are two solutions of Ax = b, then Ax1 = b = Ax2.

By the cancellation law, x1 = x2, the solution is unique. Note: bx A

bbb 1111

AIAAAA A

nA

n AAAIA bbbbbb 12

11

121

1


Three different row elementary matrices

)( )1( IrR ijij

)0( )( )2( )()( kIrR ki

ki

)( )3( )()( IrR kij

kij

row elementary matrix( 列基本矩陣 )

An nn matrix is called a row elementary matrix if it can be attained

from identity I by doing only one row elementary operation

elementary matrix

Elementary operation

identity matrix


Ex ： (a)

100

030

001

(b)

010

001

(c)

000

010

001

(d)

010

100

001

(e)

12

01

(f)

100

020

001

))((r yes 3(3)2 I )(No 非方陣

)

(No

一個非零常數必須乘上 ))((rNo 323 I

))((rYes 2(2)

12 I )(No 必須只做一次列運算


Ex ：求一序列的基本矩陣以將下列矩陣化簡成列梯形形式

0262

2031

5310

A

Sol:

100

001

010

)( 3121 IrE

102

010

001

)( 3)2(

132 IrE

21

3

)2

1(

33

00

010

001

)(IrE


0262

5310

2031

0262

2031

5310

100

001

010

)( 1121 AEArA

4200

5310

2031

0262

5310

2031

102

010

001

)( 121)2(

132 AEArA

2100

5310

2031

4200

5310

2031

21

00

010

001

)( 232

)21

(

33 AEArA

=

B

AEEEB 123 ArrrB 12)2(

13

)21

(

3 或( )列梯形矩陣


If there are finite row elementary matrices, E1, E2, …, Ek such t

hat ,AEEEEB kk 121 then B is row-equivalent to A.

列等價 (row-equivalent)


Theorem ： Elementary matrix is invertible.

If E is an elementary matrix, then E1 exists and it is also an e

lementary matrix.

Note:

ijij R1)(R )1(

)1

(1)( )( )2( ki

ki RR

)(1)( )(R )3( kij

kij R


Ex: elementary matrix inverse matrix

121

100

001

010

RE

100

001

010

)( 11

112 ER

)2(132

102

010

001

RE

102

010

001

)( 12

1)2(13 ER

)2

1(

3

21

3

00

010

001

RE

200

010

001

)( 13

1)21

(

3 ER

12R

)2(13R

)2(3R


Theorem:

A square matrix A is invertible if and only if it can be represented

as a product of a sequence of elementary matrices.

Pf: (1) Assume that A can be written as a product of a sequence

of elementary matrices. Every elementary matrix is

invertible. The product of invertible matrices is invertible.

Thus A is invertible.

(2) If A is invertible then Ax = 0 has only trivial solution.

00 IA IAEEEEk 123 11

31

21

1 kEEEEA

Thus A can be written as a product of elementary matrices.


Ex： Find a sequence of elementary matrices such that their

product is the given matrix

83

21A

Sol:

I

A

rr

rr

10

01

10

21

20

21

83

21

83

21

)2(21

)2

1(

2

)3(12

)1(1

IARRRR )1(1

)3(12

)2

1(

2)2(

21

1)2(21

1)

2

1(

21)3(

121)1(

1 )()()()( Thus RRRRA)2(

21)2(

2)3(

12)1(

1 RRRR

10

21

20

01

13

01

10

01


Theorem:

If A is an nn matrix, then the following statements are equivalent:

(1) A is invertible.

(2) For any n1 matrix b, Ax = b has only one solution.

(3) Ax = 0 has only trivial solution.

(4) A is (row) equivalent to In.

(5) A can be represented as a product of elementary matrices.


LUA L is a lower triangular matrix

U is an upper triangular matrix

Represent an nn matrix A as a product of a lower triangular

matrix L and an upper triangular matrix U .

LU-factorization (LU- 分解 )

Note:

If A is LU-factorizatiable than we can use only one elementary

operation, rij(k), to translate A into LU.

LUA

UEEEA

UAEEE

k

k

11

21

1

12


Ex: Do the LU-factorization of A.

UA

20

21

01

21 (-1)12r

01

21A

(b)

2102

310

031

A(a)

Sol:(a)

UAR )1(12

LUURA 1)1(12 )(

11

01)( )1(

121)1(

12 RRL


(b)

UA rr

1400

310

031

240

310

031

2102

310

031)4(

23)2(

13

UARR )2(13

)4(23

LUURRA 1)4(23

1)2(13 )()(

142

010

001

140

010

001

102

010

001

)()( )4(23

)2(13

1)4(23

1)2(13 RRRRL


Use LU-factorization to solve Ax=b.

. then , If bLUxLUA

Two steps:

.Then .Let bLyUxy

(1) Let y=Ux. Solve Ly=b, find y.

(2) Solve Ux=y, then we can get x.

bAx


Ex ： Use LU-factorization to solve the given linear system.

202102

13

53

321

32

21

xxx

xx

xx

Sol:

LUA

1400

310

031

142

010

001

2102

310

031

(1) Let y=Ux. Solve Ly=b, find y.

20

1

5

142

010

001

3

2

1

y

y

y

15 21 yy14)1(4)5(2204220 213 yyy


14

1

5

1400

310

031

3

2

1

x

x

x

1)2(3535

2)1)(3(131

1

21

32

3

xx

xx

x

Thus the solution of the given system is

1

2

1

x

(2) Solve Ux=y, then we can get x.

2005/7Inverse matrices-1 Inverse and Elementary Matrices.

Documents

Transcript of 2005/7Inverse matrices-1 Inverse and Elementary Matrices.