Matrices

CHAPTER 8.1 ~ 8.8

Ch8.1-8.8_2

Contents

8.1 Matrix Algebra8.2 Systems of Linear Algebra Equations8.3 Rank of a Matrix8.4 Determinants8.5 Properties of Determinants8.6 Inverse of a Matrix8.7 Cramer’s Rule8.8 The Eigenvalue Problem

Ch8.1-8.8_3

8.1 Matrix Algebra

MatrixThe forms

(x1 x2 … xn) or(1)

Each array in (1) is called a matrix.

nx

x

x

2

1

Ch8.1-8.8_4

Entry or Element: amn, the members in the arraySize: m nSquare matrix: n n, n is called the order.Main diagonal entries: ann

A matrix is any rectangular array of numbers or

functions

(2)

DEFINITION 8.1Matrix

mnmm

n

n

aaa

aaa

aaa

21

22221

11211

Ch8.1-8.8_5

A n 1 matrix is called a column vector.

A 1 n matrix is called a row vector.

DEFINITION 8.2Column and Row Vectors

na

a

a

2

1

)( 21 naaa

Two matrices A, B are equal if aij = bij .

DEFINITION 8.4Equality of Matrices

Ch8.1-8.8_6

Example 1

(a) The following matrix are not equal, since they are not of the same size.

(b) The following matrix are not equal, since the corresponding entries are not all equal.

11

11

111

111

43

21

34

21

Ch8.1-8.8_7

The sum of two mn matrices A, B is

A + B = (aij + bij)m n

DEFINITION 8.4Matrix Addition

Ch8.1-8.8_8

Example 2

(a)

211

539

874

,

5106

640

312

BA

395

1179

566

25)1(1016

563490

)8(37142

BA

Ch8.1-8.8_9

Example 2 (2)

(b)

The sum is not defined.

01

01,

654

321BA

Ch8.1-8.8_10

If k is a real number, then the scalar multiple is

DEFINITION 8.5Scalar Multiple of a Matrix

nmij

mnmm

n

n

ak

kakaka

kakaka

kakaka

k

)(

21

22221

11211

A

Ch8.1-8.8_11

If A, B, C are mn matrices, k1 and k2 are scalars(i) A + B = B + A(ii) A + (B + C) = (A + B) + C (iii) (k1k2) A = k1(k2A)

(iv) 1 A = A (v) k1(A + B) = k1A + k1B

(vi) (k1 + k2) A = k1A + k2A

THEOREM 8.1Matrix Multiplication

Ch8.1-8.8_12

If A is mp, B is pn, then the product is

DEFINITION 8.6Matrix Multiplication

pnpp

n

n

mpmm

p

p

bbb

bbb

bbb

aaa

aaa

aaa

21

22221

11211

21

22221

11211

AB

nm

p

kkjikba

1

Ch8.1-8.8_13

Example 3

(a)

(b)

Note: In general, AB BA

86

29,

53

74BA

3457

4878

65)2(36593

87)2(46794

．．．．．．．．

AB

02

34,

72

01

85

BA

66

34

154

07)3(227)4(2

00)3(120)4(1

08)3(528)4(5

．．．．．．．．．．．．

AB

Ch8.1-8.8_14

We can write a system of linear equations asthe form of product. eg:

Associate Law: A(BC) = (AB)C

Distributive Law: A(B + C) = AB + BC

(3) 83

24

83

24

21

21

2

1

xx

xx

x

x

Ch8.1-8.8_15

The transpose of (2) is

DEFINITION 8.7Transpose of a Matrix

mnnn

m

m

T

aaa

aaa

aaa

21

22212

12111

A

Ch8.1-8.8_16

Note: (A + B + C)T = AT + BT + CT (ABC)T = CTBTAT

Suppose A and B are matrices and k a scalar, (i) (AT)T = A(ii) (A + B)T = AT + BT (iii) (AB)T = BTAT

(iv) (kA)T = kAT

THEOREM 8.2Properties of Transpose

Ch8.1-8.8_17

Zero Matrices

A + 0 = A(4)

and A + (–A) = 0(5)

00

00

00

,00

00,

0

0000

Ch8.1-8.8_18

Triangular Matrices

For square n n matrices

TriangularLower TriangularUpper 143215

02111

00398

00061

00002

1000

9800

7650

4321

Ch8.1-8.8_19

Diagonal Matrices

For square n n matrices, i≠j, aij = 0

100

00

007

21

Ch8.1-8.8_20

Scalar Matrices

Diagonal matrices with equal aii

10

015

50

05

Ch8.1-8.8_21

Identity Matrices

A: m n, then Im A = A In = A

1000

0010

0001

Ch8.1-8.8_22

Symmetric

An n × n matrix A is said to be symmetric if AT = A.

467

652

721

A

AA

467

652

721T

Ch8.1-8.8_23

8.2 Systems of Linear Algebraic Equations

General Form

(1)mnmnmm

nn

nn

bxaxaxa

bxaxaxa

bxaxaxa

2211

22222121

11212111

Ch8.1-8.8_24

Solution

A solution of a (1) is said to be consistent if it has at least one solution and inconsistent if it has no solutions.

If a linear system is consistent:(i) a unique solution(ii) infinitely many solutions

See Fig 8.2

Ch8.1-8.8_25

Fig 8.2

Ch8.1-8.8_26

Example 1

Verify that x1 = 14 + 7t, x2 = 9 + 6t, x = t, where t is any real number, is a solution of

2x1 – 3x2 + x3 = 1 x1 – x2 – x3 = 5

Solution2(14 + 7t) – 3(9 + 6t) + 4t = 1 14 + 7t – (9 + 6t) – t = 5

For each number of t, we obtain a different solution.

2,3,0

4,33,42

0,9,14

321

321

321

xxx

xxx

xxx

txtxtx 321 ,69,714

Ch8.1-8.8_27

Solving Systems: Elementary Operations

(i) Multiply an equation by a nonzero constant.(ii) Interchange the position of equations(iii) Add a nonzero multiple of one equation to another one.

Ch8.1-8.8_28

Example 2

Solve 2x1 + 6x2 + x3 = 7 x1 + 2x2 – x3 = –1 5x1 + 7x2 – 4x3 = 9

Solution(i) Interchange positions

x1 + 2x2 – x3 = –1 … (a)2x1 + 6x2 + x3 = 7 … (b)

5x1 + 7x2 – 4x3 = 9 … (c)(ii) −2(a) + (b)

x1 + 2x2 – x3 = –1 … (a)2x2 + 3x3 = 9 … (b)’

5x1 + 7x2 – 4x3 = 9 … (c)

Ch8.1-8.8_29

Example 2 (2)

(iii) −5(a) + (c) x1 + 2x2 – x3 = –1 … (a)2x2 + 3x3 = 9 … (b)’

–3x2 + x3 = 14 … (c)’(iv) From (b)’ and (c)’, we have

x2 = −3, x3 = 5, then x1 = 10

Ch8.1-8.8_30

Augmented Matrix

To solve (1) we can use the augmented matrix

(2)

mmnmm

n

n

baaa

baaa

baaa

21

222221

111211

Ch8.1-8.8_31

Example 3

(a) The augmented matrix represents

x1 – 3x2 + 5x3 = 24x1 + 7x2 – x3 = 8

(b) x1 – 5x3 = – 1 x1 + 0x2 – 5x3 = – 12x1 + 8x2 = 7 and 2x1 + 8x2 + 0x3 = 7

x2 + 9x3 = 1 0x1 + x2 + 9x3 = 1are the same. Thus the matrix of the system is

8174

2531

1910

7082

1501

Ch8.1-8.8_32

Elementary Row Operations

(i) Multiply a row by a nonzero constant.(ii) Interchange the position of rows(iii) Add a nonzero multiple of one row to another one.

Matrices after elementary row operations are called row equivalent. The procedure is called row reduction.

Ch8.1-8.8_33

Example 4

(a) and

are in row-echelon form.

(b) and

are in reduced row-echelon form.

0000

1010

2051

410000

226100

0000

1010

7001

410000

606100

Ch8.1-8.8_34

Example 5

Solve the equations in example 2.

Solution(a)

9475

7162

1121

9475

1121

716212R

14130

10

1121

14130

9320

1121

29

23

52

221

3121

RRRRR

Ch8.1-8.8_35

Example 5 (2)

and

x3 = 5, x2 = –3, x1 = 10

5100

10

1121

00

10

1121

29

23

255

211

29

23

3 3112

32 RRR

529

23

12

3

32

321

x

xx

xxx

Ch8.1-8.8_36

Example 5 (3)

(b)

we have the same solution.

5100

3010

10001

5100

10

10401

5100

10

1121232

3134

29

23

122

29

23

RRRR

RR

Ch8.1-8.8_37

Example 6

Use Gauss-Jordan method to solvex1 + 3x2 – 2x3 = – 74x1 + x2 + 3x3 = 52x1 – 5x2 + 7x3 = 19

Solution

0000

3110

2101

3110

3110

7231

3311110

3311110

7231

19752

5314

7231

3212

3111

2111

3121

3

24

RRRR

RR

RRRR

Ch8.1-8.8_38

Example 6 (2)

We have x2 – x3 = –3 x1 + x3 = 2

Let x3 = t, then x2 = –3 + t, x1 = 2 – t.

Ch8.1-8.8_39

Example 7

Solve x1 + x2 = 14x1 − x2 = −62x1 – 3x2 = 8

Solution

We have 0 + 0 = 16, no solutions.

1600

210

101

832

614

111

Ch8.1-8.8_40

Networks

From Fig 8.3, we have

or(3) 0

0

0

3322

2211

321

RiRi

RiRiE

iii

0

0

3322

2211

321

RiRi

ERiRi

iii

Ch8.1-8.8_41

Fig 8.3

Ch8.1-8.8_42

Example 8

Solve (3) where R1 = 10 ohms, R2 = 20 ohms, R2 = 10 ohms, E = 12 volts

SolutionFrom the data, we have i1 – i2 – i3 = 0

10i1 + 20i2 = 1220i2 – 10i3 = 0

Ch8.1-8.8_43

Example 8 (2)

Use Gauss-Jordan method

We have i1 = 18/25, i2 = 6/25, i3 = 12/25.

2512

256

2518

operationsrow

100

010

001

010200

1202010

0111

Ch8.1-8.8_44

Homogeneous Systems

The system of equations

(4)

is always consistent, since x1 = x2 = … = xn = 0 will satisfy the system.

0

0

0

2211

2222112

1212111

nmnmm

nn

nn

xaxaxa

xaxaxa

xaxaxa

Ch8.1-8.8_45

The system (4) possesses nontrivial solutions if the

number m of equations is less than the number n of

unknowns.

THEOREM 8.3Existence of Nontrivial Solutions

Ch8.1-8.8_46

Example 9

Solve 2x1 − 4x2 + 3x3 = 0 x1 + x2 − 2x3 = 0

SolutionUse Gauss-Jordan method

0760

0211

0342

0211

0211

0342

21

12

2 RR

R

Ch8.1-8.8_47

Example 9 (2)

Let x3 = t, then x1 = (5/6)t, x2 = (7/6)t, and the system has nontrivial solutions.

010

001

010

0211

67

65

67

12

261

RR

R

Ch8.1-8.8_48

Example 10: Chemical Equations

Balance C2H6 + O2 CO2 + H2O

SolutionAssuming x1C2H6 + x2O2 x3CO2 + x4H2O, we have 2x1 = x3 (for C)

6x1 = 2x4 (for H)2x2 = 2x3 + 2x4 (for O)

Then

0100

0010

0001

01220

02006

00102

32

67

31

Ch8.1-8.8_49

Thus x4 = t, x1 = t/3, x2 = 7t/6, x3 = 2t/3. We choose t = 6, then

x1 = 2, x2 = 7, x3 = 4, x4 = 62C2H6 + 7O2 4CO2 + 6H2O

Example 10 (2)

Ch8.1-8.8_50

8.3 Rank of a Matrix

Introduction

Row vectors: u1 = (a11 a12 … a1n), u2 = (a21 a22, … a2n),…, um = (am1 am2 … amn)

mnmm

n

n

aaa

aaa

aaa

21

22221

11211

A

Ch8.1-8.8_51

Column Vectors:

mn

n

n

n

mm a

a

a

a

a

a

a

a

a

2

1

2

22

12

2

1

21

11

1 ,,, vvv

Ch8.1-8.8_52

The rank of a m n matrix A, denoted by rank(A), is

the maximum linearly independent row vectors.

DEFINITION 8.8Rank of a Matrix

Ch8.1-8.8_53

Example 1

Consider

(1)

The row vectors are in turn denoted as u1, u2, u3. Since 4u1 – ½u2 + u3 = 0, they are linearly dependent. In addition, u1 and u2 are not constant multiple couples, so they are linearly independent.

rank(A) = 2

8753

8622

3111

A

Ch8.1-8.8_54

Row Space

As in Example 1, Span(u1, u2, u3) is called the Row Space of A.

If a matrix A is row equivalent to a row-echelon form B, then(i) Row space of A = Row space of B(ii) The nonzero rows form a basis for the row

space of A(iii) rank(A) = the number of nonzero rows in B

THEOREM 8.4Rank of a Matrix by Row Reduction

Ch8.1-8.8_55

Example 2

Consider the matrix in (1)

rank(A) = 2

0000

210

3111

1420

2840

3111

8753

8622

3111

21

32

241

3221

31

21R

RRRRRR

A

Ch8.1-8.8_56

Example 3

Determine whether the set ofu1 = <2, 1, 1>, u2 = <0, 3, 0>, u3 = <3, 1, 2>

is linearly independent.

SolutionForm a matrix A by using u1, u2, u3, and reduce it.

We have rank(A) = 3, so the set of vectors is linearly independent.

100

010

001

213

030

112operations

row

A

Ch8.1-8.8_57

Rank and Linear Systems

Consider

Make an augmented matrix and reduce it:

832

64

1

21

21

21

xx

xx

xx

00

10

01

32

14

11operations

row

Ch8.1-8.8_58

We have rank(A|B) = 3. Since

then rank(A) = 2

100

010

001

1600

210

101

832

614

111

00

10

01

32

14

11operations

row

Ch8.1-8.8_59

AX = B is consistent if and only if rank(A|B) = rank(A)

THEOREM 8.5Consistence of AX = B

Suppose AX = B with m equations and n unknowns is

consistent. If rank(A) = r, then the solution contains

n – r parameters.

THEOREM 8.6Number of Parameters in a Solution

Ch8.1-8.8_60

Example

x1 + 3x2 – 2x3 = –7 4x1 + x2 + 3x3 = 5 (3)2x1 – 5x2 + 7x3 = 19

We already know that (3) is consistent and has infinitely many solutions. Since

We have rank(A|B) = rank(A) = 2, then the number of parameter of the solution is 3 – 2 = 1.

0000

3110

2101

19752

5314

7231operarions

row

Ch8.1-8.8_61

Flow chart

AX = 0

Always consistent

Unique solution X = 0

rank(A) = n

Infinity of solutions

Rank(A) < n

n – r parameters

Ch8.1-8.8_62

AX = B, B≠0

Inconsistent

rank(A) < rank(A│B)

consistent

rank(A) = rank(A│B)

Unique solution

rank(A) = nInfinity of solutions

rank(A) < n

n – r parameters

Ch8.1-8.8_63

8.4 Determinants

Notation

nnnn

n

n

aaa

aaa

aaa

21

22221

11211

A

nnnn

n

n

aaa

aaa

aaa

21

22221

11211

det A

Ch8.1-8.8_64

eg:

The determinant is

(1)

DEFINITION 8.9Determinant of 2 × 2 Matrix

2221

1211

aa

aaA

211222112221

1211det aaaaaa

aaA

695)3()9(695

36)det(,

95

36

AA

Ch8.1-8.8_65

The determinant is

(1)

DEFINITION 8.10Determinant of 3 × 3 Matrix

333231

232221

131211

aaa

aaa

aaa

A

.

det

332112322311

312213322113312312332211

333231

232221

131211

aaaaaa

aaaaaaaaaaaa

aaa

aaa

aaa

A

Ch8.1-8.8_66

In view of (1), we have

(4)

where A has been expanded by cofactors along the first row, with the cofactors of a11, a12, a13:

Thus det A = a11C11 + a12C12 + a13C13 (5)

3231

222113

3331

232112

3332

232211det

aa

aaa

aa

aaa

aa

aaa

A

3231

222113

3331

232112

3332

232211 aa

aaC

aa

aaC

aa

aaC

Ch8.1-8.8_67

In general, the cofactors of aij is Cij = (–1)i+ j Mij (6)

where Mij is called a minor determinant.From (3), we have

(8)

Similarly, we can expand by cofactors along the third rows:

det A = a31C31 + a32C32 + a33C33 (9)Note: We can expand by cofactors along any rows or any columns.

323222221212

2321

131132

3331

131122

3331

232112

211323113231133311223123332112 )()()(det

CaCaCa

aa

aaa

aa

aaa

aa

aaa

aaaaaaaaaaaaaaa

A

Ch8.1-8.8_68

Example 1

Find the determinant of

SolutionAlong the first row:

351

306

742

A

131211 742

351

306

742

det CCC A

35

30)1(

351

306

742

)1( 111111

C

Ch8.1-8.8_69

Example 1 (2)

31

36)1(

351

306

742

)1( 212112

C

51

06)1(

351

306

742

)1( 313113

C

120)]1(0)5(6[7)]1(3)3(6[4)]5(3)3(0[2

51

06)1(7

31

36)1(4

35

30)1(2det 312111

A

Ch8.1-8.8_70

We also can expanded along the second row, since it has zero entry.

120)6(3)23(6

51

42)1(3

35

74)1(6

306det

3221

232221

CCCA

Ch8.1-8.8_71

Example 2

Find the determinant of

SolutionAlong the third column:

238)]2(5)4(6[7

42

56)1)(7(

042

781

056

)1)(7(

0)7(0

042

781

056

det

3232

332313

CCCA

042

781

056

A

Ch8.1-8.8_72

Let A = (aij)n × n ne an n × n matrix. For each

the cofactor expansion of det A along the ith row is

For each the cofactor expansion of det A

along the ith column is

THEOREM 8.7Consistence of AX = B

,1 ni

,1 nj ininiiii CaCaCa 2211det A

njnjjjjj CaCaCa 2211det A

Ch8.1-8.8_73

Example 3

Find the determinant of

SolutionAlong the fourth row

where

4001

1611

3201

4215

A

44434241 )4(00)1(

4001

1611

3201

4215

det CCCC

A

Ch8.1-8.8_74

Example 3 (2)

161

320

421

)1( 1441

C

611

201

215

)1( 4444 C

1861

21)1(3

11

41)1(2

16

42)1(0

161

320

421

)1(

322212

41

C

Ch8.1-8.8_75

Example 3 (3)

411

15)1(2

61

25)1(0

61

21)1)(1(

611

201

215

322212

44

C

34)4)(4()18)(1(

4001

1611

3201

4215

det

A

Ch8.1-8.8_76

8.5 Properties of Determinants

If AT is the transpose of the n × n matrix A, then

det AT = det A

THEOREM 8.8Determinant of a Transpose

If any two rows (columns) of an n × n matrix A are the

same, then det A = 0.

THEOREM 8.9Two Identical Rows

4143

75det

A 41

47

35det

TA

Ch8.1-8.8_77

Example 1

229

224

226

A

0

229

224

226

det A

Ch8.1-8.8_78

If all the entries in a row (column) of an n × n matrix A

are all zeros, then det A = 0.

THEOREM 8.10Zero Row or Column

If B is the matrix obtained by interchanging any two

rows (columns) of an n × n matrix A, then

det B = −det A

THEOREM 8.11Interchanging Rows

Ch8.1-8.8_79

If B is obtained by interchanging the first and third row of

AB det

312

706

914

914

706

312

det

Ch8.1-8.8_80

If B is obtained from an n × n matrix A by multiplying

a row (column) by a nonzero real number k, then

det B = k det A

THEOREM 8.12Constant Multiple of a Row

A

B

A

det)(

det

rowth thealong cofactorsby det ofexpansion

2211

2211

kCaCaCak

CkaCkaCka

i

ininiiii

ininiiii

Ch8.1-8.8_81

Example 2

(a)

(b)

80)21(8012

11285

24

1185

164

815

1620

85

．．

．

00)2(

227

115

114

)2(

247

125

124

．

Ch8.1-8.8_82

If A and B are both n × n matrices, then

det AB = det A det B.

THEOREM 8.13Determinant of a Matrix Product

Ch8.1-8.8_83

Example 3

Suppose

then

Now det AB = −24, det A = −8, det B = 3, Thus det AB = det A det B.

53

43,

11

62BA

96

2212AB

Ch8.1-8.8_84

Suppose B is the matrix obtained from an n × n matrix A by multiplying the entries in a row (column) by Nonzero real number k and adding the result to the corresponding entries in another row (column). Then det B = det A.

THEOREM 8.14Determinant is Unchanged

Ch8.1-8.8_85

Example 4

We have det A = 45 = det B = 45.

BA

2411

703

215

414

703

215313 RR

414

703

215

A

Ch8.1-8.8_86

Proof

Suppose A is an n × n matrix (upper or lower). Then

det A = a11 a22 … ann

where a11, a22, …, ann are the entries on the main

diagonal of A.

THEOREM 8.15Determinant of a Triangular Product

333231

2221

11

0

00

aaa

aa

a

A

332211323322113332

2211 )0(

0det aaaaaaa

aa

aa ．A

Ch8.1-8.8_87

Example 5

(a)

2427

0495

0062

0003

A

144)2()4(63

2427

0495

0062

0003

det

．．．A

Ch8.1-8.8_88

(b)

400

060

003

A

7246)3(

400

060

003

det

．．A

Ch8.1-8.8_89

Example 6

Find the determinant of

Solution

842

234

726

A

726

234

421

2

421

234

726

2

842

234

726

det

A

Ch8.1-8.8_90

Example 6 (2)

)19)(5)(1)(2(

1900

1850

421

2

17100

1850

421

2

726

1850

421

2

Ch8.1-8.8_91

Suppose A is a n n matrix. If ai1, ai2, …, ain are the

entries in the ith row and Ck1, Ck2, …, Ckn are the

cofactors in the kth row, then ai1 Ck1 + ai2 Ck2 + …+ ain Ckn = 0, for i k

If a1j, a2j, …, anj are the entries in the jth column and

C1k, C2k, …, Cnk are the cofactors in the kth column,

then a1j C1k + a2j C2k + …+ anj Cnk = 0, for j k

THEOREM 8.16A Property of Cofactors

Ch8.1-8.8_92

THEOREM 8.16

ProofLet B be the matrix obtained from A by letting the entries in the ith row of A be the same as the ones in the kth row, that is,

ai1 = ak1, ai2 = ak2, …, ain = akn then there are two identical rows in B, det B = 0,then

kninkiki

knknkkkk

CaCaCa

CaCaCa

2211

2211det0 B

Ch8.1-8.8_93

Example 7

Consider the matrix

we have

842

234

726

A

0)10(7)40(2)25(6

34

267

24

762

23

726

331332123111

CaCaCa

Ch8.1-8.8_94

8.6 Inverse of a Matrix

Note: If A has no inverse, then A is called singular.

Let A be an n n matrix. If there exists an n n matrixB such that

AB = BA = I (1)where I is the n n identity, then A is said to be nonsingular or invertible. Then B is said to be the inverse of A.

DEFINITION 8.11Inverse of a Matrix

Ch8.1-8.8_95

Proof(i) A-1A = AA-1 = I, A = (A-1)-1

(ii) (AB)(AB)-1 = I, B-1A-1(AB)(AB)-1 = B-1A-1

(AB)-1 = B-1A-1

Let A, B be nonsingular matrices. (i) (A-1)-1 = A (ii) (AB)-1 = B-1A-1 (iii) (AT)-1 = (A-1)T

THEOREM 8.17Properties of the Inverse

Ch8.1-8.8_96

Let A be an n × n matrix. The matrix that is the

Transpose of the matrix of cofactors corresponding to

the entries of A:

is called the adjoint of A and is denoted by adj A.

DEFINITION 8.12Adjoint Matrix

nnnn

n

nT

nnnn

n

n

CCC

CCC

CCC

CCC

CCC

CCC

21

22212

12111

21

22221

11211

Ch8.1-8.8_97

ProofFor n = 3,

(3)

Let A be an n × n matrix. If det A 0, then

(2)

THEOREM 8.18Finding the Inverse

AA

A adjdet

11

A

A

A

AA

det00

0det0

00det

)adj(

332313

322212

312111

333231

232221

131211

CCC

CCC

CCC

aaa

aaa

aaa

Ch8.1-8.8_98

From Theorem 8.16,

we have

Thus A-1 = adj A/det A

00

00

00

333322322131133312321131

332332223121132312221121

331332123111231322122111

CaCaCaCaCaCa

CaCaCaCaCaCa

CaCaCaCaCaCa

IAAAA )(det

100

010

001

)(det)adj(

Ch8.1-8.8_99

For 22 matrix

(4)

2221

1211

aa

aaA

1121

1222

1112

2122

2221

1211adjaa

aa

aa

aa

CC

CC TT

A

1121

12221

det1

aa

aa

AA

Ch8.1-8.8_100

For 33 matrix

(5)

3231

222113

3331

232112

3332

232211

333231

232221

131211

aa

aaC

aa

aaC

aa

aaC

aaa

aaa

aaa

A

332313

322212

3121111

det1

CCC

CCC

CCC

AA

Ch8.1-8.8_101

Example 1

Find the inverse of

Solution

Check

102

41A

21

1

1

25

12

410

21

A

10

01

541010

2245

1

25

102

41

21

1AA

10

01

5411

202045

102

41

1

25

21

1AA

Ch8.1-8.8_102

Example 2

Find the inverse of

SolutionSince

103

112

022

A

612

222

12

022

11

02

603

222

13

022

10

02

303

125

13

121

10

11

333231

232221

131211

CCC

CCC

CCC

Ch8.1-8.8_103

Example 2 (2)

We have

21

21

41

61

61

125

61

61

121

1

663

225

221

121

A

Ch8.1-8.8_104

An n n matrix A is nonsingular if and only if det A 0

THEOREM 8.19Nonsingular Matrices and det A

Ch8.1-8.8_105

Example 3

has no inverse.

A is singular, since det A = 0

33

22A

Ch8.1-8.8_106

An nn matrix A can be transformed into the n n identity matrix I by a sequence of elementary row operations, then A is nonsingular. The same operations that transforms A into the identity I will also transform I into A-1.

THEOREM 8.20Finding the Inverse

Ch8.1-8.8_107

First we construct the augmented matrix (A|I), and the process for finding A-1 is outlined.

100

010

001

)|(

21

22221

11211

nnnn

n

n

aaa

aaa

aaa

IA

Ch8.1-8.8_108

Row operations on A until I is obtained

)IA( | )A|I( 1

Simultaneously applying the same operations

Ch8.1-8.8_109

Example 4

Find the inverse of

Solution

655

432

102

A

1050

011530

0001

100655

010432

0001

100655

010432

001102

25

217

21

21

52

21

21

3121

121

RRRR

R

Ch8.1-8.8_110

Example 4 (2)

6105100

010

0001

00

010

0001

010

010

0001

31

31

35

21

21

30

51

31

61

301

31

31

35

21

21

51

21

1017

31

31

35

21

21

3

32

351

231

R

RR

RR

Ch8.1-8.8_111

Example 4 (3)

Thus

6105100

10178010

352001233

5132

1

RRRR

6105

10178

3521A

Ch8.1-8.8_112

Example 5

Find the inverse of

Solution

306

542

211

A

100306

012960

001211

100306

010542

001211

212 RR

Ch8.1-8.8_113

Example 5 (2)

There is a row of zeros, it is singular

114000

012960

001211

106960

012960

001211

32

316

RR

RR

Ch8.1-8.8_114

Using the Inverse

A system of m linear equations in n unknowns

(6)

can be written as AX = B, where

mnmnmm

nn

nn

bxaxaxa

bxaxaxa

bxaxaxa

2211

22222121

11212111

,

21

22221

11211

mnmm

n

n

aaa

aaa

aaa

A ,2

1

nx

x

x

X

mb

b

b

2

1

B

Ch8.1-8.8_115

Special Case

When m = n, if A is nonsingular, then

X = A-1B(7)

Ch8.1-8.8_116

Example 6

Use the inverse to solve

SolutionThe system can be written as

Since , it is nonsingular. From (4)

1663

1592

21

21

xx

xx

16

15

63

92

2

1

x

x

03963

93

23

96

391

63

92 1

Ch8.1-8.8_117

Example 6 (2)

Using (7)

Thus

316

13

234

391

16

15

23

96

391

2

1

x

x

3/1,6 21 xx

Ch8.1-8.8_118

Example 7

Use the inverse to solve

SolutionFrom the equations we have

4432

1655

22

321

321

31

xxx

xxx

xx

655

432

102

A

Ch8.1-8.8_119

Example 7 (2)

Thus (7) gives

Thus

36

62

19

1

4

2

6105

10178

352

1

4

2

655

432

102 1

3

2

1

x

x

x

36,62,19 321 xxx

Ch8.1-8.8_120

A homogeneous system of n linear equations in n unknowns AX = 0 has only the trivial solution if and only if A is nonsingular.

THEOREM 8.21Trivial Solution Only

A homogeneous system of n linear equations in n unknowns AX = 0 has a nontrivial solution if and only if A is singular.

THEOREM 8.22Existence of Nontrivial Solutions

Ch8.1-8.8_121

8.7 Cramer’s Rule

IntroductionFor example, the equations

(1)

possesses the solution

(2)

where a11 a22 – a12 a21 0

2222121

1212111

bxaxa

bxaxa

,21122211

2122211 aaaa

baabx

21122211

2112112 aaaa

abbax

Ch8.1-8.8_122

Rewrite (2) as determinant forms, we have

(3)2221

1211

221

111

2

2221

1211

222

121

1 ,

aa

aaba

ba

x

aa

aaab

ab

x

Ch8.1-8.8_123

Special Matrix

For the following system

(4)nnnnnn

nn

nn

bxaxaxa

bxaxaxa

bxaxaxa

2211

22222121

11212111

Ch8.1-8.8_124

We define a special matrix

nnnknnknn

nkk

nkk

k

aabaaa

aabaaa

aabaaa

1121

2122122221

1111111211

A

Ch8.1-8.8_125

Let A be the coefficient matrix of the system (1).

If det A 0, then the solution of (1) is given by

where Ak, k = 1, 2, …, n, is defined in (5)

THEOREM 8.23Cramer’s Rule

,detdet

,,detdet

,detdet 2

21

1 AA

AA

AA n

nxxx

Ch8.1-8.8_126

Proof

nnnnn

nn

nn

nnnnn

n

n

cbCbCb

cbCbCb

cbCbCb

b

b

b

CCC

CCC

CCC

2211

2222121

1212111

2

1

21

22212

12111

det1

det1

A

ABAX 1-

Ch8.1-8.8_127

Now the entry in the kth row is

(7)

AA

A

detdet

det2211

k

nknkkk

CbCbCbx

Ch8.1-8.8_128

Example 1

Solve

Solution1245

33

723

321

321

321

xxx

xxx

xxx

,78

215

331

173

det

,13

245

311

123

det

2

A

A

52

145

311

723

det

39

241

313

127

det

3

1

A

A

Ch8.1-8.8_129

Example 1 (2)

From (3), we have

,3detdet 1

1 AA

x ,6detdet 2

2 AA

x 4detdet 3

3 AA

x

Ch8.1-8.8_130

8.8 The Eigenvalue Problems

Let A be n n matrix. A number is said to be an

eigenvalue of A if there exists a nonzero solution

vector K of AK = K (1)

The solution vector K is said to be an eigenvector

corresponding to the eigenvalue .

DEFINITION 8.13Eigenvalues and Eigenvectors

Ch8.1-8.8_131

Example 1

Verify that is an eigenvector of the matrix

SolutionSince

we conclude that K is an eigenvector of A.

1

1

1

K

112

332

310

A

KAK )2(

1

1

1

)2(

2

2

2

1

1

1

112

332

310eigenvalue

Ch8.1-8.8_132

From (1), we have(A – I)K = 0

(2)However, (2) is the same as a homogeneous system of linear equations. Since we want K to be nontrivial, we must have

det (A – I) = 0(4)Inspection of (4) shows det (A – I) results in an nth-degree polynomial in , and is called the characteristic equation.

Ch8.1-8.8_133

Example 2

Find the eigenvalues and eigenvectors of

Solution

121

016

121

A

0

121

016

121

)det(

IA

Ch8.1-8.8_134

Example 2 (2)

We have –3–2 + 12 = 0 or ( + 4)( – 3) = 0then = 0, −4, 3. To find the eigenvectors,(i) 1 = 0,

0000

010

001

0000

010

0121

0000

06130

0121

0121

0016

0121

)|0(

136

131

2

136

6

122131

3121

RRR

RRRR

0IA

Ch8.1-8.8_135

Example 2 (3)

Choose k3 = −13, then

3231 136

,131

kkkk

13

6

1

1K

Ch8.1-8.8_136

Example 2 (4)

(ii) For 2 = −4,

01680

01890

0321

0125

0036

0321

0321

0036

0125

)|4(

3121

133

56

RRRR

RR

0IA

Ch8.1-8.8_137

Example 2 (5)

implies k1 = −k3 , k2 = 2k3. Choose k3 = 1, then

0000

0210

0101

0210

0210

03213212

381

291

22

RRRR

RR

1

2

1

2K

Ch8.1-8.8_138

Example 2 (6)

(iii) 2 = 3,

implies k1 = – k3, k2 = –(3/2)k3. Choose k3 = –2,

0000

010

0101

0421

0046

0122

)|3( 23

operationsrow

0IA

2

3

2

3K

Ch8.1-8.8_139

Example 3

Find the eigenvalues and eigenvectors of

Solution

We see 1 = 2 = 5 is an eigenvalue of multiplicity 2.From (A – 5I|0), we get

71

43A

0)5(71

43)det( 2

IA

02

042

21

21

kk

kk

Ch8.1-8.8_140

Example 3 (2)

Choose k2 = 1, we have k1 = 2, then

We can have only one eigenvector though A is a 22 matrix.

1

21K

Ch8.1-8.8_141

Example 4

Find the eigenvalues and eigenvectors of

Solution

We see 1 = 11, 2 = 3 = 8 is of multiplicity 2.

911

191

119

A

0)8)(11(

911

191

119

)det( 2

IA

Ch8.1-8.8_142

Example 4 (2)

(i) For 1 = 11, Gauss-Jordan method gives

Hence k1 = k3, k2 = k3. If k3 = 1, then

0000

0110

0101

0211

0121

0112

)|11( 0IA

1

1

1

1K

Ch8.1-8.8_143

Example 4 (3)

(ii) Now for 2 = 8, we have

For k1 + k2 + k3 = 0, we can select two of them arbitrarily. We choose: k2 = 1, k3 = 0, and k2 = 0, k3 = 1, then

0000

0000

0111

0111

0111

0111

)|8( 30IA

,

0

1

1

2

K

1

0

1

3K

Ch8.1-8.8_144

ProofSince AK = K,The proof is completed.

Let A be a square matrix with real entries. If = + i,

0, is a complex eigenvalue of A, then its conjugate

is also an eigenvalue of A. If K is an

eigenvector corresponding to , then is an eigenvector

corresponding to .

THEOREM 8.24Complex Eignvalues and Eigenvectors

iK

,KKA KKA

Ch8.1-8.8_145

Example 5

Find the eigenvalues and eigenvectors of

Solution

For 1 = 5 + 2i,

45

16A

0291045

16)det( 2

IA

ii 25,25 121

0)21(5

0)21(

21

21

kik

kki

Ch8.1-8.8_146

Example 5 (2)

Since k2 = (1 – 2i) k1, after choosing k1 = 1, then

From Theorem 8.24, then

i21

11K

i21

112 KK

,2512 i

Ch8.1-8.8_147

The eigenvalues of an upper triangular, lower triangular,

or diagonal matrix are the main diagonal entries.

THEOREM 8.25Triangular and Diagonal Matrices