1 資訊科學數學 13 : Solutions of Linear Systems 陳光琦助理教授 (Kuang-Chi Chen) [email protected]

11

資訊科學數學資訊科學數學 13 13 ::

Solutions of Linear SystemsSolutions of Linear Systems

陳光琦助理教授陳光琦助理教授 (Kuang-Chi Chen)(Kuang-Chi Chen)[email protected]@mail.tcu.edu.tw

22

Linear Equations and MatricesLinear Equations and Matrices

Solutions of Linear Systems of EquationsSolutions of Linear Systems of Equations

33

Solutions of Linear Systems of Solutions of Linear Systems of EquationsEquations

1.6 Solutions of Linear Systems of Equations1.6 Solutions of Linear Systems of Equations

63

5

42

43

42

41

xx

xx

xx

63100

51010

42001

44

Row Echelon FormRow Echelon Form

•Definition – Row echelon form (r.e.f.)Definition – Row echelon form (r.e.f.)

An An mmnn matrix matrix AA is said to be in is said to be in row echelon formrow echelon form if if

(a) All zero rows, if there are any, appear at the bottom (a) All zero rows, if there are any, appear at the bottom of the matrixof the matrix

(b) The first nonzero entry from the left of a nonzero ro(b) The first nonzero entry from the left of a nonzero row is a 1; a leading one of the roww is a 1; a leading one of the row

(c) For each nonzero row, the leading one appears to the (c) For each nonzero row, the leading one appears to the right and below any leading one’s in preceding rowsright and below any leading one’s in preceding rows

55

Reduced Row Echelon FormReduced Row Echelon Form

• Definition – Reduced row echelon formDefinition – Reduced row echelon form

An An mmnn matrix matrix AA is said to be in is said to be in reduced row echeloreduced row echelon formn form if if

(a) (a) AA is in row echelon form is in row echelon form

(b) If a column contains a leading one, then all other entr(b) If a column contains a leading one, then all other entries in that column are zeroies in that column are zero

(( 列梯形式列梯形式 ; ; 簡化之列梯形式簡化之列梯形式 ))

66

Example 1 - in row echelon formExample 1 - in row echelon form• E.g. 1E.g. 1

1 5 0 2 2 4

0 1 0 3 4 8

0 0 0 1 7 2

0 0 0 0 0 0

0 0 0 0 0 0

C

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

A

0 0 1 3 5 7 9

0 0 0 0 1 2 3

0 0 0 0 0 1 2

0 0 0 0 0 0 1

0 0 0 0 0 0 0

B

77

Example 2 – Example 2 – reducedreduced row echelon row echelon formform

• E.g. 2E.g. 21 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

D

1 0 0 0 2 4

0 1 0 0 4 8

0 0 0 1 7 2

0 0 0 0 0 0

0 0 0 0 0 0

E

1 2 0 0 1

0 0 1 2 3

0 0 0 0 0

F

88

E.g. 2 – not E.g. 2 – not reducedreduced row echelon row echelon formform• E.g. 2E.g. 2 1 2 0 4

0 0 0 0

0 0 1 3

G

1 0 3 4

0 2 2 5

0 0 1 2

H

1 0 3 4

0 1 2 5

0 1 2 2

0 0 0 0

I

1 2 3 4

0 1 2 5

0 0 1 2

0 0 0 0

J

,

,

,

Nonzero element above leading 1 in row 2

99

Three Basic Types of Elementary Three Basic Types of Elementary Row OperationsRow Operations

• Type 1 – InterchangeType 1 – Interchange

row row ii and row and row jj are interchanged are interchanged

• Type 2 – MultiplyType 2 – Multiply

row row ii = row = row ii times times cc

• Type 3 – AddType 3 – Add

Add Add dd times row times row rr of of AA to row to row ss of of AA

row row ss = row = row ss + + dd row row rr

1010

Example 3Example 3• E.g. 3E.g. 3

9633

2032

2100

A

2100

2032

9633

B

3211

2032

2100

C

5631

2032

2100

D

E2

E1⇒

E3

1111

Row EquivalenceRow Equivalence

• Definition – Row EquivalenceDefinition – Row Equivalence

An An mmnn matrix matrix AA is said to be is said to be row equivalencerow equivalence to an to an mmnn matrix matrix BB if if BB can be obtained by ap can be obtained by applying a finite sequence of elementary row opeplying a finite sequence of elementary row operations to the matrix rations to the matrix AA . .

1212

Example 4Example 4• E.g. 4E.g. 4

3221

2312

3421

A

3221

8734

3421

B

8734

3221

3421

C

8734

3221

6842

D

E1

E3⇒

E2⇒

1313

Theorem 1.5Theorem 1.5• Theorem 1.5Theorem 1.5

Every Every mmnn matrix is matrix is row equivalentrow equivalent to a matr to a matrix in ix in row echelon formrow echelon form . .

1414

E.g. 5 - Procedure of Row Echelon E.g. 5 - Procedure of Row Echelon FormForm

• E.g. 5E.g. 5

Step 1 – Find the pivotal columnStep 1 – Find the pivotal column

Step 2 – Identify the pivot in the pivotal columnStep 2 – Identify the pivot in the pivotal column

79602

42522

43200

14320

A

Pivot column

Pivot

1515

(cont’d)(cont’d)

• E.g. 5E.g. 5

Step 3 – Interchange if necessary so that the pivot is in Step 3 – Interchange if necessary so that the pivot is in the 1the 1stst row row

Step 4 – Multiply so that the pivot equals to 1Step 4 – Multiply so that the pivot equals to 1

79602

14320

43200

42522

1A

pivot

79602

14320

43200

2111 25

2A

1616


• E.g. 5E.g. 5

Step 5 – Make all entries in the pivot column, except Step 5 – Make all entries in the pivot column, except the entry where the pivot was located, equal to zerothe entry where the pivot was located, equal to zero

37120

14320

43200

2111 25

3A

1717


• E.g. 5E.g. 5

Step 6 – Ignore the first row and repeatStep 6 – Ignore the first row and repeat

⇒⇒

⇒ … … … ⇒ ⇒ … … … ⇒

37120

14320

43200 2 1 1 1 2

5

B

37120

43200

14320 21 1 1 2

5

1B

00000

2100

210

2111

23

21

23

25

H

1818

Example 6Example 6• Example 6Example 6

13

32A

21

321A

32

212A

70

213A

1919

RemarkRemark

• RemarkRemark

- There may be - There may be more than onemore than one matrix in matrix in row echrow echelon formelon form that is that is row equivalentrow equivalent to a given mat to a given matrix.rix.

- A matrix in row echelon form (r.e.f.) that is ro- A matrix in row echelon form (r.e.f.) that is row equivalent to w equivalent to AA is called is called

“ “a row echelon form of a row echelon form of AA”.”.

2020


- Every - Every mmnn matrix is matrix is row equivalentrow equivalent to a to a uniuniqueque matrix in matrix in reduced row echelon formreduced row echelon form..

2121

Example 7 – r.e.f. to Example 7 – r.e.f. to reduced r.e.f.reduced r.e.f.• E.g. 7E.g. 7

00000

2100

010

2111

23

25

417

25

1J

00000

2100

010

7011

23

25

417

419

2J

00000

2100

010

9001

23

25

417

219

K

00000

2100

210

2111

23

21

23

25

H

2222


Let Let AxAx = = bb and and CxCx = = dd be two linear systems e be two linear systems each of ach of mm equations in equations in nn unknowns. If the aug unknowns. If the augmented matrices [mented matrices [AA||bb] and [] and [CC||dd] of these syste] of these systems are ms are row equivalentrow equivalent, then both linear syste, then both linear systems have the ms have the same solutionssame solutions..

2323

Corollary 1.1Corollary 1.1• Corollary 1.1Corollary 1.1

If If AA and and CC are row equivalent are row equivalent mmnn matrices, t matrices, then the linear system hen the linear system AxAx = 0 and = 0 and CxCx = 0 have = 0 have exactly the same solutions.exactly the same solutions.

2424

Gauss-Jordan Reduction Gauss-Jordan Reduction ProcedureProcedure

• The Gauss-Jordan reduction procedureThe Gauss-Jordan reduction procedure

Step 1. Form the Step 1. Form the augmented matrix [augmented matrix [AA||bb]]

Step 2. Obtain the Step 2. Obtain the reduced row echelon form reduced row echelon form [ [CC||dd]] of the augmented matrix [ of the augmented matrix [AA||bb] by u] by using elementary row operationssing elementary row operations

Step 3. For each nonzero row of [Step 3. For each nonzero row of [CC||dd], solve the c], solve the corresponding equation.orresponding equation.

(augmented matrix (augmented matrix 擴增矩陣擴增矩陣 ))

2525

Gauss Elimination ProcedureGauss Elimination Procedure

• The Gauss elimination procedureThe Gauss elimination procedure

Step 1. Step 1. Form the augmented matrix Form the augmented matrix [[AA||bb]]

Step 2. Obtain a Step 2. Obtain a row echelon form [row echelon form [CC||dd]] of of the augmented matrix [ the augmented matrix [AA||bb] by using elem] by using elementary row operationsentary row operations

Step 3. Solving Step 3. Solving the linear system corresponding to the linear system corresponding to [[CC||dd], ], by back substitution (by back substitution ( 後代入法後代入法 ).).

2626

Example 8Example 8• E.g. 8E.g. 8 Solve the linear system by Gauss-Jordan reductionSolve the linear system by Gauss-Jordan reduction

- Step 1- Step 1

3 3

8 2

932

zx

zyx

zyx

3103

8112

9321

2727


• E.g. 8 - Solve the linear system by Gauss-Jordan reductionE.g. 8 - Solve the linear system by Gauss-Jordan reduction

- Step 2- Step 2

3103

8112

9321

241060

10550

9321

241060

2110

9321

12400

2110

9321

3100

2110

9321

3100

1010

9321

3100

1010

0021

3100

1010

2001

2828


• E.g. 8 - Solve the linear system by Gauss-Jordan reductionE.g. 8 - Solve the linear system by Gauss-Jordan reduction

- Step 3 - Step 3 xx = 2 = 2 yy = -1 = -1 zz = 3 = 3

2929

Example 9Example 9

• Example 9Example 9

- Solve the linear system by Gauss-Jordan reduction- Solve the linear system by Gauss-Jordan reduction

xx + + yy + 2 + 2zz – 5 – 5w w = 3= 3

22xx + 5 + 5yy – – zz – 9 – 9w w = -3= -3

22xx + + yy – – zz + 3 + 3w w = -11= -11

xx – 3 – 3yy + 2 + 2zz + 7 + 7w w = -5= -5

3030



- Step 1- Step 1

- Step 2- Step 2

57231

113112

39152

35211

00000

32100

23010

52001

3131


• E.g. 9 - Step 3E.g. 9 - Step 3

leading variablesleading variables

a free variablea free variable

3 2

2 3

52

wz

wy

wx

wz

wy

wx

23

32

25

rw

rz

ry

rx

23

32

25

3232


- Solve the linear system by Gauss-Jordan reduction- Solve the linear system by Gauss-Jordan reduction

xx11 + 2 + 2xx22 – 3 – 3xx44 + + xx55 = 2 = 2

xx11 + 2 + 2xx22 + + xx33 – 3 – 3xx44 + + xx55 + 2 + 2xx66 = 3 = 3

xx11 + 2 + 2xx22 – 3 – 3xx44 + 2 + 2xx55 + + xx66 = 4 = 4

33xx11 + 6 + 6xx22 + + xx33 – 9 – 9xx44 + 4 + 4xx55 + 3 + 3xx66 = 9 = 9

3333



- Step 1- Step 1

- Step 2- Step 2

9349163

4123021

3213121

2013021

0000000

2110000

1200100

0103021

3434


• Example 10 - Step 3Example 10 - Step 3

leading variablesleading variables

free variablesfree variables

2

12

0 3 2

65

63

6421

xx

xx

xxxx

65

63

2461

2

21

23

xx

xx

xxxx

rx

rx

sx

rx

tx

tsrx

6

5

4

3

2

1

2

21

23

3535


- Solve the linear system by Gauss elimination- Solve the linear system by Gauss elimination

xx + 2 + 2yy + 3 + 3zz = 9 = 9

22xx – – yy + + zz = 8 = 8

33xx – – zz = 3 = 3

3636



- Step 1- Step 1

- Step 2- Step 2

1 2 3 9

2 1 1 8

3 0 1 3

1 2 3 9

0 1 1 2

0 0 1 3

3737


• Example 11 - Step 3Example 11 - Step 3

- By back substitution- By back substitution

2 3 9

2

3

x y z

y z

z

3

1

2

z

y

x

3838


- Solve the linear system by Gauss elimination- Solve the linear system by Gauss elimination

xx + 2 + 2yy + 3 + 3zz + 4 + 4w w = 5= 5

xx + 3 + 3yy + 5 + 5zz + 7 + 7w w = 11= 11

xx – – zz – – ww = -6 = -6

3939


• Example 12Example 12 - Step 1- Step 1

- Step 2- Step 2

- Step 3 - Step 3 ⇒⇒ 0 0xx + 0 + 0yy + 0 + 0zz + 0 + 0w w = 1 = 1 No solutions !!⇒ No solutions !!⇒

1 2 3 4 5

1 3 5 7 11

1 0 1 2 6

1 0 1 2 0

0 1 2 3 0

0 0 0 0 1

4040

Consistent and InconsistentConsistent and Inconsistent

• Consistent and inconsistentConsistent and inconsistent

- Consistent: Linear systems with - Consistent: Linear systems with at least one at least one solutionsolution

- Inconsistent: Linear systems with - Inconsistent: Linear systems with no solutionsno solutions

4141

Homogeneous SystemsHomogeneous Systems• A system of linear equations is said to be A system of linear equations is said to be homogeneohomogeneo

usus if all the constant terms are zeros. if all the constant terms are zeros.

aa1111xx11 + + aa1212xx22 + … + + … + aa11nnxxnn = 0 = 0

aa2121xx11 + + aa2222xx22 + … + + … + aa22nnxxnn = 0 = 0

……

aamm11xx11 + + aamm22xx22 + … + + … + aamnmnxxnn = 0 = 0

⇒ ⇒ AxAx = 0 = 0

Thus, a homogeneous system always has the solution Thus, a homogeneous system always has the solution xx11 = = xx22 = … = = … = xxnn = 0 = 0 → the trivial solution→ the trivial solution

4242


2 3 0

3 2 0

2 2 0

x y z

x y z

x y z

1 2 3 0

1 3 2 0

2 1 2 0

1 0 0 0

0 1 0 0

0 0 1 0

0x y z

4343


0

0

2 0

x y z w

x w

x y z

1 1 1 1 0

1 0 0 1 0

1 2 1 0 0

1 0 0 1 0

0 1 0 1 0

0 0 1 1 0

x r

y r

z r

w r

4444


A homogeneous system of A homogeneous system of mm equations in equations in nn u unknowns has nknowns has a non-trivial solutiona non-trivial solution if if mm < < nn, th, that is, if the number of unknowns exceeds the nat is, if the number of unknowns exceeds the number of equations.umber of equations.

namely, a homogeneous system has more varinamely, a homogeneous system has more variables than equations has many solutions.ables than equations has many solutions.

(a homogeneous system(a homogeneous system 齊次系統齊次系統 ; non-trivial solution ; non-trivial solution 非零非零解解 ))

4545

Example 15 - A Homogeneous Example 15 - A Homogeneous SystemSystem

• E.g. 15E.g. 15

5723

113 2

39 52

352

wzyx

wzyx

wzyx

wzyx

3 2

2 3

52

wz

wy

wx

rw

rz

ry

rx

23

32

25

4646


x

yx

z

w

5 2 5 2

2 3 2 3

3 2 3 2

0

r r

r rx

r r

r r

5 2

2 3 and

3 2

0

p h

r

rx x

r

r

• If letIf let

4747

A Homogeneous System A Homogeneous System ExampleExample

xx = = xxpp + + xxhh

xxpp is a particular solution to the given system is a particular solution to the given system

AxAxpp = = bb , where , where bb = [3 -3 -11 -5] = [3 -3 -11 -5]TT

xxhh is a solution to the associated is a solution to the associated

homogeneous system homogeneous system AxAxhh = 0 = 0 . .

4848

Polynomial InterpolationPolynomial Interpolation• Polynomial InterpolationPolynomial Interpolation

- The general form- The general form

yy = = aann – 1 – 1xxnn – 1 – 1 + + aann – 2 – 2xxnn – 2 – 2 + … + + … + aa11xx + + aa00

E.g. E.g. nn = 3, = 3, yy = = aa22xx22 + + aa11xx + + aa00

Given three distinct points (Given three distinct points (xx11 , , yy11), (), (xx22 , , yy22), (), (xx33 , , yy33),),

we havewe have

yy11 = = aa22xx1122 + + aa11xx11 + + aa00

yy22 = = aa22xx2222 + + aa11xx22 + + aa00

yy33 = = aa22xx3322 + + aa11xx33 + + aa00

4949

Example 16Example 16• Example 16 - Find the quadratic polynomial that Example 16 - Find the quadratic polynomial that

interpolates the points (1, 3), (2, 4), (3, 7)interpolates the points (1, 3), (2, 4), (3, 7)

2 1 0

2 1 0

2 1 0

3

4 2 4

9 3 7

a a a

a a a

a a a

2 1 01 , 2 , 4a a a

2 2 4y x x

5050

Example 17 – Temperature Example 17 – Temperature DistributionDistribution

TT11 = (260 – 100 + = (260 – 100 + TT22 + + TT33 )/4 or 4 )/4 or 4TT11 – – TT22 – – TT33 = 160 = 160

TT22 = ( = (TT11 + 100 + 40 + + 100 + 40 + TT44 )/4 or - )/4 or -TT11 + 4 + 4TT22 – – TT44 = 140 = 140

TT33 = (60 + = (60 + TT11 + + TT44 + 0)/4 or - + 0)/4 or -TT11 + 4 + 4TT33 – – TT44 = 60 = 60

TT44 = ( = (TT22 + + TT33 + 40 + 0)/4 or - + 40 + 0)/4 or -TT22 – – TT33 + 4 + 4TT44 = 40 = 40

⇒ ⇒

⇒ ⇒ TT11 = 65, = 65, TT22 = 60, = 60, TT33 = 40, = 40, TT44 = 35 . = 35 .

4 1 1 0 160

1 4 0 1140

1 0 4 1 60

0 1 1 4 40

A b

5151

Linear Equations and MatricesLinear Equations and Matrices

The Inverse of A MatrixThe Inverse of A Matrix

5252

The Inverse of A MatrixThe Inverse of A Matrix• 1.7 The inverse of a matrix1.7 The inverse of a matrix

DefinitionDefinition

- An - An nnnn matrix matrix AA is called is called nonsingularnonsingular (or (or invinvertible ertible 可逆的可逆的 ) if there exists an ) if there exists an nnnn matrix matrix BB such that such that ABAB = = BABA = = IInn . .

- The matrix - The matrix BB is called the is called the inverseinverse of of AA

- - If there exists If there exists no such matrix no such matrix BB, then , then AA is calle is called d singularsingular (or (or noninvertiblenoninvertible))

- - AA is also an is also an inverseinverse of of BB

5353

Example 1Example 1


⇒ AB = BA = I2

- - BB is an inverse of is an inverse of AA and and AA is nonsingular. is nonsingular.

22

32A

1 3/ 2

1 1B

5454


An inverse of a matrix, if exists, is An inverse of a matrix, if exists, is uniqueunique..(proof)(proof)

Let Let BB and and CC be inverses of be inverses of AA..

Then Then ABAB = = BABA = = IInn, and , and ACAC = = CACA = = IInn..

Thus, Thus, CC((ABAB) = ) = CICInn

((CACA))BB = = CC

IInnBB = = CC , i.e., , i.e., BB = = CC . .

5555

Example 2 - Find the InverseExample 2 - Find the Inverse

For the matrix For the matrix AA, find the inverse, find the inverse

If exists, let the inverse If exists, let the inverse AA-1-1 be be

such thatsuch that

43

21A

dc

baA 1

10

01

43

212

1 Idc

baAA

5656


10

01

4343

22

dbca

dbca

043

12

ca

ca

143

02

db

db and and

12 1

3/ 2 1/ 2

a bA

c d

2 1 1 2 1 0

3 / 2 1/ 2 3 4 0 1

5757


⇒ ⇒ No solution; singularNo solution; singular

42

21A

dc

baA 1

10

01

42

212

1 Idc

baAA

10

01

4242

22

dbca

dbca

042

12

ca

ca142

02

db

db and and

5858

Theorem 1.10Theorem 1.10

• Thm. 1.10 - Properties of an inverseThm. 1.10 - Properties of an inverse

- If - If AA is is nonsingularnonsingular, then , then AA-1-1 is nonsingular is nonsingular

and and ((AA-1-1))-1-1 = = AA ; ;

- If - If AA and and BB are nonsingular matrices, then are nonsingular matrices, then ABAB is nonsingular and is nonsingular and ((ABAB))-1-1 = = BB-1 -1 AA-1-1 ; ;

- If - If AA is a nonsingular matrix, then is a nonsingular matrix, then ((AATT))-1-1 = ( = (AA-1-1))TT . .

5959


43

21A

21

23

1 12A

21

23

1

1

2)( TA

42

31TA

21

23

1

1

2)( TA ⇒ ⇒ and and

6060

Corollary1.2Corollary1.2• Corollary 1.2Corollary 1.2

- If - If AA11 , , AA22 , … , , … , AArr are are nnnn nonsingular matric nonsingular matric

es, then es, then ((AA11 AA22 … … AArr) is nonsingular) is nonsingular and and ((AA11 AA

22 … … AArr))-1-1 = = AArr-1-1 … … AA22

-1-1 AA11-1-1 . .

6161


Suppose that Suppose that AA, , BB are are nnnn matrices, matrices,

- If - If ABAB = = IInn , then , then BABA = = IInn ; ;

- If - If BABA = = IInn , then , then ABAB = = IInn ..

6262

The Way to Find The Way to Find AA-1-1

• A practical method for finding A practical method for finding AA-1-1

Step 1. Form the 2Step 1. Form the 222nn matrix matrix [[AA | | IInn]] obtained by obtained by adjoining the identity matrix adjoining the identity matrix IInn to the given matrix to the given matrix AA

Step 2. Compute the Step 2. Compute the reduced row echelon formreduced row echelon form of the of the matrix obtained in Step 1 by using elementary row matrix obtained in Step 1 by using elementary row operationsoperations

Step 3. Suppose that Step 2 has produced the matrix Step 3. Suppose that Step 2 has produced the matrix [[CC | | DD] in reduced row echelon form:] in reduced row echelon form:

• If If CC = = IInn , then , then DD = = AA-1-1 ; ;

• If If CC ≠≠ IInn , then , then CC has a row of zeros and the matrix has a row of zeros and the matrix A A is singular .is singular .

6363

Example 5 – Find the InverseExample 5 – Find the Inverse• E.g. 5 – Find the inverseE.g. 5 – Find the inverse

155

320

111

A

100

010

001

155

320

111

3IA

100

010

001

155

320

111

105

010

001

400

320

111

105

00

001

400

10

111

21

23

41

45

21

23

0

00

001

100

10

111

41

45

83

21

815

41

41

0

0

100

010

011 13 1 18 2 8

15 318 2 8

5 14 4

1 0 0

0 1 0

0 0 1 0

41

45

83

21

815

81

21

813

1

0

A

6464

Example 6 – Find the InverseExample 6 – Find the Inverse• E.g. 6 - Find the inverseE.g. 6 - Find the inverse

325

121

321

A

100

010

001

325

121

321

3IA

132

011

001

000

440

321

100

010

001

325

121

321

100

011

001

325

440

321

105

011

001

12120

440

321

The left-half matrix cannot have a one in the (3, 3) location, the reduced echelon form cannot be I3. Thus A-1 does not exist.

6565

Theorem 1.12 & 1.13Theorem 1.12 & 1.13• Theorem 1.12Theorem 1.12

An An nnnn matrix is matrix is nonsingularnonsingular iff it is iff it is row equirow equivalence to valence to IInn . .

• Theorem 1.13Theorem 1.13

If If AA is an is an nnnn matrix, the matrix, the homogeneous systehomogeneous system m AxAx = 0 = 0 has a has a nontrivial solutionnontrivial solution iff iff

AA is singular is singular..

6666

Proof of Theorem 1.13Proof of Theorem 1.13• Proof of Theorem 1.13Proof of Theorem 1.13

Suppose that Suppose that AA is nonsingular, then is nonsingular, then AA-1-1 exists and exists and

AA-1-1((AAxx) = ) = AA-1-1 0 0

((AA-1-1AA))xx = 0 = 0

IInn xx = 0 = 0

xx = 0 = 0 ⇒ ⇒ AxAx = 0 has a trivial solution = 0 has a trivial solution

(contradiction to a non-trivial solution, hence (contradiction to a non-trivial solution, hence AA must must be singular)be singular)

6767


Consider the homogeneous system Consider the homogeneous system AxAx = 0, where = 0, where AA is is the matrix (the matrix (AA is nonsingular) is nonsingular)

155

320

111

A

0155

0320

0111

0100

0010

0001Gauss-Jordan reductionGauss-Jordan reduction

The trivial solutionThe trivial solution x x = 0 = 0

6868


- Consider the homogeneous system - Consider the homogeneous system AxAx = 0, where = 0, where AA is the matrix (is the matrix (AA is singular) is singular)

325

121

321

A

0325

0121

0321

0000

0110

0101

rz

ry

rx

6969


If If A A is an is an nnnn matrix, then matrix, then A A is nonsingularis nonsingular if iff the linear system f the linear system AxAx = = bb has has a unique solutioa unique solutionn for every for every nn1 matrix 1 matrix bb . .

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SummarySummary

The Symmetry, Singularity, The Symmetry, Singularity, Inverse of A MatrixInverse of A Matrix

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Some Special MatrixSome Special Matrix

4. 4. A square matrix A square matrix AA is said to be is said to be antisymmetricantisymmetric if - if -AATT = = AA. (i) If . (i) If AA is square, prove that is square, prove that AA + + AATT is symmetri is symmetri

c and c and AA – – AATT is antisymmetric is antisymmetric;;

(ii) (ii) any square matrix any square matrix AA can be decomposed into the can be decomposed into the sum of a symmetric matrix sum of a symmetric matrix BB and an antisymmetric and an antisymmetric matrix matrix CC: : AA = = BB + + CC . .

5.5. G Given two symmetric matrices of the same size, iven two symmetric matrices of the same size, AA a and nd BB, then a necessary and sufficient condition for th, then a necessary and sufficient condition for the product e product ABAB to be symmetric is that to be symmetric is that ABAB = = BABA..

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Some Special MatrixSome Special Matrix

1. 1. A square matrix A square matrix AA is said to be is said to be normalnormal if if AAAATT = = AATTAA. .

All symmetric matrices are normal All symmetric matrices are normal ;;

2.2. A square matrix A square matrix AA is said to be is said to be idempotentidempotent if if AA22 = = AA..

If If AA is idempotent then is idempotent then AATT is also idempotent is also idempotent ;;

3.3. A square matrix A square matrix AA is said to is said to nilpotentnilpotent if there is a po if there is a positive integer sitive integer pp such that such that AApp = = OO. The least integer su. The least integer such that ch that AApp = = OO is called the is called the degree of nilpotencydegree of nilpotency of th of the matrix. If e matrix. If AA is nilpotent, then is nilpotent, then AATT is also nilpotent w is also nilpotent w

ith the same degree of nilpotencyith the same degree of nilpotency..

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List of Nonsingular List of Nonsingular EquivalencesEquivalences

• Nonsingular equivalencesNonsingular equivalences

1. 1. A A is nonsingular ;is nonsingular ;

2.2. xx = 0 is the only solution to = 0 is the only solution to AxAx = 0 ; = 0 ;

3.3. AA is row equivalence to is row equivalence to IInn ; ;

4.4. The linear system The linear system AxAx = = bb has a unique has a unique solution for every solution for every nn1matrix 1matrix bb . .

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Properties of Matrix Properties of Matrix InverseInverse

• Properties of Matrix InverseProperties of Matrix Inverse

1. (1. (AA-1-1))-1-1 = = AA ; ;

2.2. ( (ccAA))-1-1 = (1/ = (1/cc))AA-1-1 , where , where cc is a nonzero is a nonzero scalar;scalar;

3.3. ( (ABAB))-1-1 = = BB-1-1AA-1-1 ; ;

4.4. ( (AAnn))-1-1 = ( = (AA-1-1))nn ; ;

5.5. ( (AATT))-1-1 = ( = (AA-1-1))TT , where , where TT : transpose. : transpose.

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Conditions of Matrix Conditions of Matrix InverseInverse

• A matrix has no inverse, ifA matrix has no inverse, if

(i) two rows are equal;(i) two rows are equal;

(ii) two columns are equal; (Use the transpose)(ii) two columns are equal; (Use the transpose)

(iii) it has a column of zeros. (iii) it has a column of zeros.

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The Inverse of 2The Inverse of 22 Matrix2 Matrix

• If If AA = , show that = , show that AA-1-1 = . = .

Note: The cancellation law doesn’t hold.Note: The cancellation law doesn’t hold.That is, That is, ABAB = = ACAC doesn’t imply that doesn’t imply that BB = = CC . .

Also, Also, ABAB = = OO doesn’t imply that doesn’t imply that AA = = OO or or BB = = OO..

However, if However, if AA is an invertible matrix is an invertible matrix, then, then

if if ABAB = = ACAC , then , then BB = = CC ; ;

if if ABAB = 0, then = 0, then BB = 0 . = 0 .

a b

c d

1

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d b

c aad bc

Documents

1 資訊科學數學 13 : Solutions of Linear Systems 陳光琦助理教授 (Kuang-Chi Chen) [email protected]