2005/7Inverse matrices-1 Inverse and Elementary Matrices.
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Transcript of 2005/7Inverse matrices-1 Inverse and Elementary Matrices.
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2005/7 Inverse matrices-1
Inverse and Elementary Matrices
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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
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2005/7 Inverse Matrices-3
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
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2005/7 Inverse Matrices-4
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
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2005/7 Inverse Matrices-5
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 ,
喬登消去法高斯
rr
1 ,3 2111 xx
1 ,4 2212 xx
1
2
11
431AX
Hence
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2005/7 Inverse Matrices-6
Note:
1110
4301
1031
0141
1
, )4(21
)1(12
AIIA
rr
喬登消去法高斯
If A can’t using row operations to be translated into identity matrix I, then A is a singular matrix.
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2005/7 Inverse Matrices-7
Ex : Find the inverse matrix of
326
101
011
A
Sol:
106326
011110
001011
)1(
12
r
100
010
001
326
101
011
IA
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2005/7 Inverse Matrices-8
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
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2005/7 Inverse Matrices-9
Thus
142
133
1321A
141100
133010
132001
)1(
21
r
1 AI
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2005/7 Inverse Matrices-10
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(
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2005/7 Inverse Matrices-11
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
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2005/7 Inverse Matrices-12
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.
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2005/7 Inverse Matrices-13
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.
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2005/7 Inverse Matrices-14
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
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2005/7 Inverse Matrices-15
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
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2005/7 Inverse Matrices-16
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
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2005/7 Inverse Matrices-17
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
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2005/7 Inverse Matrices-18
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
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2005/7 Inverse Matrices-19
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)
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2005/7 Inverse Matrices-20
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
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2005/7 Inverse Matrices-21
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
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2005/7 Inverse Matrices-22
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.
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2005/7 Inverse Matrices-23
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
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2005/7 Inverse Matrices-24
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.
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2005/7 Inverse Matrices-25
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
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2005/7 Inverse Matrices-26
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
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2005/7 Inverse Matrices-27
(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
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2005/7 Inverse Matrices-28
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
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2005/7 Inverse Matrices-29
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
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2005/7 Inverse Matrices-30
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.