1

Matrices and Determinants

Text references: Lecture Notes Series: Engineering Mathematics Volume 1, 2nd

Edition, Pearson 2006.

Matrices and Some Special Matrices

A linear system of equations

3333232131

2323222121

1313212111

bxaxaxa

bxaxaxa

bxaxaxa

=++

=++

=++

3333231

2232221

1131211

baaa

baaa

baaa

A matrix is a rectangular array of numbers (or function) enclosed in brackets

(or parentheses).

Let m, n to be positive integers,

=

mnmm

n

n

aaa

aaa

aaa

A

L

MOMM

L

L

21

22221

11211

where ija (i = 1, …, n; j = 1, …, m)

are the real numbers which are the

entries or element of A.

A also can be written as [ ]ijaA = .

The entries nnaaaa , , , , 332211 L are the main diagonal (or principal

diagonal).

A matrix containing one row is called a row matrix.

A matrix containing one column is called a column matrix.

If nm = , we call A as a nn × (or mm × ) square matrix.

If we omit some rows or columns (or both) from A, we call it as sub-matrix.

can be written as

2

Example :

1.

963

852

741

is a 33× matrix (which is a square matrix).

2.

−

−−

−

1068

3195

7332

6821

is a square matrix (or 44 × matrix).

The entries 1, 3, 1, 1 are along the main diagonal.

3.

c

b

a

is a 13× matrix (or column matrix).

4. [ ]321 is a 31× matrix (or row matrix).

5.

fed

cba contains how many 22 × sub-matrices? 12 × sub-matrices?

21× sub-matrices? y

A square matrix in which 0=ija for all ji ≠ is called diagonal matrix.

=

nna

a

a

A

L

MOMM

L

L

00

00

00

22

11

Raii ∈ , ni , ,3 ,2 ,1 L= .

A square matrix in which 1=iia and 0=ija for all ji ≠ is called identity

matrix. Usually, we write the nn × identity matrix as nI .

3

=

100

010

001

L

MOMM

L

L

nI .

A square matrix in which all the entries above the main diagonal are zero is

called lower triangular matrix.

A square matrix in which all the entries below the main diagonal are zero is

called upper triangular matrix.

A matrix, either upper triangular or lower triangular matrix is called

triangular matrix.

Example :

1.

−−

−

7914

0326

0052

0001

,

− 669

003

001

,

01

01 are the lower triangular

matrix.

2.

− 7000

8300

1400

5051

,

−

600

580

291

,

10

31 are the upper triangular

matrix.

3.

− 2000

0900

0050

0004

,

800

060

004

,

90

07 are the diagonal matrix.

4

Example :

What kinds of matrices are both upper triangular and lower triangular?

If A is both upper and lower triangular, then every entry off the main

diagonal must be zero. Hence A is a diagonal matrix.

Matrix Operations

1. Addition and subtraction: If [ ]ijaA = , [ ]ijbB = and both are the nm ×

matrices, then

[ ]ijij baBA +=+ and [ ]ijij baBA −=−

for all mi , 2, ,1 L= nj , 2, ,1 L= .

Example :(Addition and Subtraction)

Given that

−

−=

515

341A ,

−=

615

370B and

=

26

91C .

Find BA + , BA − and CA + .

−=+

1120

6111BA ,

−−

−=−

1010

031BA ,

The sum of CA + is not defined, since the matrices have different

sizes.

5

2. Scalar multiplication: If [ ]ijaA = is a nm × matrix and C is any real

number, then [ ]ijaCCA = for all mi , 2, ,1 L= nj , 2, ,1 L= .

Example : (scalar multiplication)

Compute

− 59

24

41

3 and

02

414 .

−

=

− 1527

612

123

59

24

41

3 and

=

08

164

02

41 4 .

3. Matrix multiplication: Let [ ]ijaA = be a nm × matrix and [ ]ijbB = be a

pn × matrix. Then

=

⋅

=

mpmm

ij

p

npnn

ipii

p

mnmm

inii

n

ccc

c

ccc

bbb

bbb

bbb

aaa

aaa

aaa

AB

L

MM

L

L

MMM

L

MMM

L

L

MMM

L

MMM

L

21

11211

21

21

11211

21

21

11211

where ∑=

=+++=n

k

kjiknjinjijiij babababac1

2211 L .

6

Example : (matrix multiplication)

Find the product AB for

−=

12

31A and

−

−=

623

402B .

Since A is 22× and B is 32 × , the product AB is defined as a 32 ×

matrix. To obtain the entries in the first row of AB, multiply the first

row [ ]31 of A by the column

3

2,

− 2

0 and

−

6

4 of B, respectively.

( )( ) ( )( ) ( )( ) ( )( ) ( )( ) ( )( )

+−−++=

−

−⋅

− ???

634123013321

623

402

12

31

−=

+−−+=

???

14611

???

1846092

Similarly, to obtain the entries in the second row of AB.

Thus,

−

−=

1421

14611AB .

4. Some properties of Matrix Operations

Let A, B and C be the nm × matrices, D and E be the pn × matrices, F

be a qp × matrix and 1e , 2e are any real number.

(a) ABBA +=+ ,

(b) ( ) ( ) CBACBA ++=++ ,

(c) ( ) BeAeBAe 111 +=+ ,

(d) ( ) AeAeAee 2121 +=+ ,

(e) ( ) ( )AeeAee 2121 = ,

(f) ( ) AEADEDA +=+ ,

(g) ( ) CEBEECB +=+ ,

(h) ( ) ( )FADDFA = ,

(i) ( ) ( ) ( )DeADAeADe 111 == .

7

Example :

Find x, y, z and w, if

−

+++

−=

11

21 2

wzyx

wywy

zx.

First write each side as a single matrix:

−−+

++++=

11

21

22

22

wy

wzyx

wy

zx

Set corresponding entries equal to each other to obtain the system of

four equations,

12

22

12

12

−−=

++=

+=

++=

ww

wzz

yy

yxx

or

13

2

1

1

−=

+=

=

+=

w

wz

y

yx

The solution is: 3

1

3

512 −==== wzyx ,,, .

8

Example :

Find two matrices A and B such that AB and BA are defined and have

the same size, but BAAB ≠ .

Let

−=

02

11A and

=

11

12B , then

=

⋅

−=

24

01

11

12

02

11AB

−=

−⋅

=

13

24

02

11

11

12BA

Matrix multiplication does not obey the commutative law.

Example :

Can we have 0=AB , with 0≠A and 0≠B ?

Let

=

22

11A ,

−−=

23

23B

=

−−⋅

=

00

00

23

23

22

11AB .

5. Transpose of matrix (Transposition)

If [ ]ijaA = is a nm × matrix, then the transposition of A is [ ]jiT

aA =

( TA becomes a mn × matrix).

Some properties for the Transposition

Let A and B be the nm × matrices, C be the pn × matrices, and 1e is

any real number.

(a) ( ) TTTBABA +=+ ,

(b) ( ) TTTBABA −=− ,

(c) ( ) TTAeAe 11 = ,

(d) ( ) AATT = ,

(e) ( ) TTTACAC = .

9

Example :

Find the transposes of the following vectors:

−

=

9

6

1

u , [ ]6942 −=v ,

=

9

4w , [ ]16 −=z

[ ]961 −=Tu ,

−=

6

9

4

2

Tv , [ ]94=Tw ,

−=

1

6Tz

Example :

Given

=

654

321A , find T

A and ( )TTA .

=

63

52

41T

A , ( )

=

654

321TTA .

Observe that ( ) AATT = .

Example :

Given

−

=

41

33

21

A , find TAA and AA

T .

=

1797

9189

795T

AA ,

=

297

711AAT

Observe that AAAATT ≠ .

10

Example :

Given

−=

32

11A and

=

53

12B , find ( )T

AB , TTBA and TT

AB .

( )

−

−=

174

131TAB ,

=

121

134TT BA ,

−

−=

174

131TT AB .

Observe that ( ) TTTBAAB ≠ , TTTT

ABBA ≠ and ( ) TTTABAB = .

Is it true in general that ( ) TTTABAB = ? (OPTIONAL)

If [ ]ijaA = and [ ]kjbB = , the ij-entry of AB is

mjimjiji bababa +++ L2211 (1)

Thus (1) is the ji-entry [reverse order] of ( )TAB .

On the other hand, column j of B becomes row j of TB , and row I of A

becomes column I of TA . Consequently, the ji-entry of TT

AB is

[ ] immjijiij

im

i

i

mjjij ababab

a

a

a

bbb +++=

LM

L 221

2

1

2

Thus, ( ) TTTABAB = , since corresponding entries are equal.

11

6. Let [ ]ijaA = be a nm × matrix. Then

(a) A is said to be a symmetric matrix if AAT = .

(b) A is said to be a skew-symmetric matrix if AAT −= .

Example :

Given

=

24

41A ,

−

−=

41

16B ,

−=

01

10C and

−=

03

30D .

Which of the above matrices are (i) symmetric, (ii) skew-symmetric?

A is symmetric because

=

24

41TA which is equal to A.

B is also symmetric because

−

−=

41

16TB which is equal to B.

C is skew-symmetric because

−=

01

10TC which is equal to C− .

D is also skew-symmetric because

−=

03

30TD which is equal to

D− .

Theorem

(a) If A, B are symmetric matrices and k is any constant, then kA ,

BA + , BA − , are symmetric matrices.

(b) If A, B are skew-symmetric matrices and k is any constant, then

kA , BA + , BA − , are skew-symmetric matrices.

(c) If A is any nn × matrix then

- TAA + and T

AA are symmetric matrices,

- TAA − is a skew-symmetric matrix.

12

Example :

For the matrices of example 14, find kA, BA + and BA − . Are they

symmetric?

=

kk

kkkA

24

4,

=+

63

37BA and

−

−=−

25

55BA . All of them

are symmetric matrices.

Example :

For the matrices of example 14, find kC, DC + and DC − . Are they

skew-symmetric?

−=

0

0

k

kkC ,

−=+

02

20DC ,

−=−

04

40DC . All of them

are skew-symmetric matrices.

Example :

Given

−−=

75

61A , find T

AA + , TAA and T

AA − .

−

−=

76

51TA .

−=+

141

12TAA ,

−

−=

7447

4737TAA and

−=−

011

110TAA .

Observe that TAA + and T

AA are symmetric matrices, TAA − is a

skew-symmetric matrix.

Example :

Show that the diagonal elements of a skew-symmetric matrix must be

zero.

If [ ]ijaA = is skew-symmetric then iiii aa −= . Hence each 0=iia .

13

Determinant Let A be a nn × matrix, the determinant of A is denoted by ( )Adet or A .

To evaluate determinant of A,

(1) If 2=n , then ( ) 21122211

2221

1211aaaa

aa

aaAdet −== .

(2) If 3≥n , then

( )

nnnn

n

n

aaa

aaa

aaa

Adet

L

MMM

L

L

21

22221

11211

=

( ) ( ) ( ) inin

ni

ii

i

ii

iMaMaMa

+++ −++−+−= 111 22

2

11

1L

( )∑=

+−=n

j

ijij

jiMa

1

1

where ijM is a determinant of ( ) ( )11 −×− nn matrix obtained from A

by omitting row i and column j.

ijM is introduced as minor for matrix A.

( ) ij

jiM

+−1 is introduced as cofactor for matrix A.

Example :

Find ( )5det , ( )91−det and ( )8+sdet .

The determinant is the scalar itself; hence, ( ) 55 =det , ( ) 9191 −=−det

and ( ) 88 +=+ ssdet .

Example :

Given

=

21

85A ,

=

db

caB . Find ( )Adet and ( )Bdet .

( )( ) ( )( ) 2810182521

85=−=−= ,

bcaddb

ca−= .

14

Example :

Determine the values of s for which 03

2=

s

ss.

0323

2 2 =−= sss

ss, or ( ) 032 =−ss . Hence, 0=s or

2

3=s .

Example :

Given

=

333

222

111

cba

cba

cba

A , find ( )Adet .

( ) ( ) ( ) ( )323213232132321

333

222

111

abbacaccabbccba

cba

cba

cba

Adet −+−−−==

321321321321321321 abccabbcabacacbcba −−−++=

Example :

Given

−=

431

250

112

A . Find ( )Adet .

( ) ( )( )( ) ( )( )( ) ( )( )( ) ( )( )( )

( )( )( ) ( )( )( )014223

151031121452

431

250

112

−−−

−+−+=−=Adet

4501250240 =−+−+−=

15

Some properties of determinants

(1) If matrix B is obtained from matrix A by interchanging two rows (or

two columns), then ( ) ( )AdetBdet −= .

(2) If two rows (or two columns) in matrix A are identical, then

( ) 0=Adet .

(3) If matrix B is obtained from matrix A by multiplying one complete

row (or column) by a constant λ then ( ) ( )AdetBdet λ= . Hence,

( ) ( )AdetAdetnλ=λ .

(4) If matrix A has a zeros row (or column) then ( ) 0=Adet .

(5) If matrix B is obtained from matrix A by adding to one row(column) a

multiple of another row(column) then ( ) ( )AdetBdet = .

(6) If A and B are square matrix of same size, then

( ) ( ) ( )BdetAdetABdet = .

(7) If A is a square matrix, then ( ) ( )AdetAdetT = .

Example :

Given

=

42

31A and

=

31

42B . Find ( )Adet and ( )Bdet .

( ) 264 −=−=Adet and ( ) 246 =−=Bdet .

Observe that ( ) ( )BdetAdet −= , where matrix B is obtained from matrix A by

interchanging two rows.

Example :

Given

−

−

=

111

432

111

A . Find ( )Adet .

( ) ( )( )( ) ( )( )( ) ( )( )( ) ( )( )( ) ( )( )( ) ( )( )( )211141131211141131 −−−−−−++−=Adet

0243243 =+−+−+−= Observe that row 1 and row 3 in matrix A are identical.

16

Example :

Given

=

103

21A ,

=

103

2010B and

=

5015

105C . Find ( )Adet , ( )Bdet

and ( )Cdet .

( ) 4610 =−=Adet , ( ) 4060100 =−=Bdet and ( ) 100150250 =−=Cdet .

Observe that ( ) ( )AdetBdet 10= where matrix B is obtained from matrix A

by multiplying the first row by 10.

Observe that ( ) ( )AdetCdet25= .

Example :

Given

=

103

21A and

=

103

124B . Find ( )Adet and ( )Bdet .

( ) 4610 =−=Adet and ( ) 43640 =−=Bdet .

Observe that ( ) ( )BdetAdet = where matrix B is obtained from matrix A : the

1st row of B is summation of row 1 and row 2 of matrix A.

Example :

Given

−=

12

61A and

−=

13

52B . Find ( )Adet , ( )Bdet and ( )ABdet .

−=

−⋅

−=

111

120

13

52

12

61AB ,

( ) ( )( ) ( )( ) 132611 −=−−=Adet , ( ) ( )( ) ( )( ) 171523512 −=−−=−−=Bdet and

( ) ( )( ) ( )( ) 221111120 =−−=ABdet .

Observe that ( ) ( ) ( )BdetAdetABdet = .

17

Example :

Given

−=

26

14A . Find ( )Adet and ( )T

Adet .

−=

21

64TA ,

( ) ( )( ) ( )( ) 14686124 −=−−=−−=Adet and

( ) ( )( ) ( )( ) 142624 −=−−−=TAdet .

Observe that ( ) ( )TAdetAdet = .

18

Inverse Matrix

Invertible Matrices:

AB = BA = I

A square matrix, A Identity matrix, I

If there exists a matrix B, IBAAB == ⇒ A is invertible.

B is inverse of A: BA =−1 .

1−= AB ⇔ AB =−1

Invertible matrix: nonsingular matrix ( ( ) 0det ≠A )

Not invertible: singular matrix ( ( ) 0det =A )

19

Properties of inverse matrix:

(1) AB = BA = I

AC = CA = I

CB =⇒ (CANNOT have TWO different inverses for a square matrix A.) (2) A and B are invertible ⇒ AB is invertible and

( ) 111 −−− = ABAB .

(3) If A is an invertible matirx, then

(i) A-1

is invertible and (A-1

)-1

= A.

(ii) AT is invertible and (A

T)

-1 = (A

-1)

T

(iii) An is invertible and (A

n)

-1 = (A

-1)

n for n = 0, 1,2, …

Example:

Show that

−=

01

11B is the inverse of

=

11

10A .

Solution:

=

−

=

10

01

01

11

11

10AB

=

−=

10

01

11

10

01

11BA

Hence AB = BA = I, so B is indeed the inverse of A.

Example:

Suppose A is an invertible matrix and IA =3. Find

1−A .

Solution:

AAA = I

A(AA)=I

AAA =−1

20

(iv) For any nonzero scalar k, the matrix kA is invertible and

( ) .Ak

kA11 1 −− =

Example:

Can you find the inverse of

642

531?

NO! why??

Example :

Which of the following matrices are singular?

=

11

00A ,

=

73

41B ,

−

=

292

151

131

C ,

−=

431

250

112

D .

Matrices A and C are singular.

The inverse of a nonsingular nn × matrix [ ]ijaA = is given by

(1) If

=

dc

baA then

−

−

−=−

ac

bd

bcadA

11.

(2) If 3≥n , ( )

( )TijAAdet

A11 =−

where ijA is the cofactor of A

( )

−= +

ij

ji

ij MA 1 .

21

Example :

Find the inverse for the matrices B and D in example 31.

−

−=

−

−

−=−

5

1

5

35

4

5

7

13

47

127

11B ,

( )[ ]TijA

DdetD

11 =− where ijA is the cofactor of A.

( ) 45=Ddet (from example 24)

( ) 2643

251

11

11 =−

−= +A , ( ) 2

41

201

21

12 −=−

−= +A , ( ) 5

31

501

31

13 −=−= +A

( ) 143

111

12

21 −=−= +A , ( ) 7

41

121

22

22 =−= +A , ( ) 5

31

121

32

23 −=−= +A

( ) 725

111

13

31 −=−

−= +A , ( ) 4

20

121

23

32 =−

−= +A , ( ) 10

50

121

33

33 =−= +A

−−

−

−−

=∴ −

1055

472

7126

45

11D .

Theorem

(1) A square matrix A is invertible if and only if ( ) 0≠Adet .

(2) If A and B are invertible matrices of same size, then

(a) AB is invertible.

(b) ( ) 111 −−− = ABAB .

(3) If A is an invertible matrix, then

(a) 1−A is invertible and ( ) AA =

−− 11 .

(b) nA is invertible and ( ) ( )nn

AA11 −−

= for any L 2, 1, ,0=n .

22

(c) TA is invertible and ( ) ( )TT

AA11 −−

= .

(d) For any { }0\R∈α , the matrix Aα is invertible and

( ) 111 −−− α=α AA .

Remark

If A is invertible, then ( ) ( ) 11 −− = AdetAdet

Example :

Determine whether

=

21

53A is an inverse of

−

−=

31

52B .

=

−

−⋅

=

10

01

31

52

21

53AB . Thus, A is an inverse of B.

Example :

Let A and B be invertible matrices of the same size. Show that the product

AB is also invertible and ( ) 111 −−− = ABAB .

( )( ) ( ) IAAAIAABBAABAB ==== −−−−−− 111111 ,

( )( ) ( ) IBBIBBBAABABAB ==== −−−−−− 111111 .

Example :

Given

=

21

01A and

=

11

12B . Find ( ) 1−

AB , 11 −−BA and 11 −−

AB .

=

34

12AB ,

−

−=−

21

111B ,

−=−

11

02

2

11A .

( )

−

−=−

24

13

2

11AB ,

−

−=−−

32

22

2

111BA ,

−

−=−−

24

13

2

111AB .

Observe that ( ) 111 −−− = ABAB but ( ) 111 −−− = BAAB .

23

Example :

Show that if A is an invertible matrix, then TA is also invertible and

( ) ( )TTAA

11 −−= .

( ) ( ) IIAAAATTTT === −− 11

( ) ( ) IIAAAATTTT

=== −− 11 .

Example :

If A is invertible, show that kA is invertible when 0≠k , with inverse 11 −− Ak .

Since 0≠k , k

k11 =− exists. Then ( )( ) ( )( ) .IIAAkkAkkA =⋅== −−−− 11111

Hence, 11 −− Ak is the inverse of kA.

Example :

Given

−

−=

21

12A . Determine whether

11 −−= AA .

( )( ) ( )( ) 31122 =−−−=A , 3

13 11

==⇒ −−A .

=−

3

2

3

13

1

3

2

1A , 3

1

3

1

3

1

3

2

3

21 =

−

=−A .

Thus, 11 −−

= AA .

24

System of linear Equations A linear system of m linear equations in n unknowns ( 1x , 2x , 3x , 4x , nx ,L ):

11212111 bxaxaxa nn =+++ L

M

L

22222121 bxaxaxa nn =+++ ( )∗

mnmnmm bxaxaxa =+++ L2211

If 021 ==== mbbb L the system is homogeneous. Note that

0321 ===== nxxxx L is always a solution of a homogeneous system.

This is known as trivial solution.

If not all mbbb , , , 21 L equal to zero then the system is non-homogeneous.

We can write the system ( )∗ in the form

BX =A

where

=

mnmjm

iniji

nj

aaa

aaa

aaa

A

LL

MOMM

LL

MMOM

LL

1

1

1111

;

=

nx

x

x

M

M

2

1

X ;

=

mb

b

b

M

M

2

1

B

A is called coefficient matrix.

[ ]

=

mmnmjm

iiniji

nj

baaa

baaa

baaa

A

LL

MMOMM

MMMOM

LL

1

1

11111

B is called augmented matrix.

25

Solving a system of linear equations by using Inverse of a matrix

If BX =A with 0≠A then 1−A is exist and the unique solution of X can

be found by using 1−A with the following procedures:

Step 1:

Write down the system of linear equations in the form of BX =A . Then

find 1−A .

Step 2:

Multiply both sides of the system of linear equation BX =A with 1−A i.e

BX 11 −− = AAA

BX 1−= AI

BX 1−= A Step 3:

Find B1−A to get the solution of X.

Example :

Solve the following system of linear equations by using Inverse of a matrix.

343

125

2 2

321

32

321

=++

=−

=++

xxx

xx

xxx

Step 1:

The system of linear equations can be written as BX =A .

i.e

=

−

3

1

2

431

250

112

3

2

1

x

x

x

Let

−=

431

250

112

A , 1−A can be found as shown in example 32.

−−

−

−−

=−

1055

472

7126

45

11A .

26

Step 2:

Multiply both sides of the system of linear equation BX =A with 1−A i.e

−−

−

−−

=

−

−−

−

−−

3

1

2

1055

472

7126

45

1

431

250

112

1055

472

7126

45

1

3

2

1

x

x

x

−−

−

−−

=

3

1

2

1055

472

7126

45

1

100

010

001

3

2

1

x

x

x

−−

−

−−

=

3

1

2

1055

472

7126

45

1

3

2

1

x

x

x

Step 3:

=

3

13

13

2

3

2

1

x

x

x

.

27

Elementary Row Operations

Let A be an m x n matrix. The following operations are called elementary

row operations:

(1): Interchange the ith row and the jth row: Ri ↔Rj.

(2): Multiply the ith row by a nonzero scalar k: Ri →kRi, k ≠ 0.

(3): Replace the ith row by k times the jth row plus the ith row:

Ri → Ri + kRj.

Example:

Performs the operations R1 ↔ R2, R3 → -7R3 and R2 → R2 -3R1 to the

identity matrix

=

100

010

001

3I .

Solution:

100

010

001

100

001

010

100

010

001

− 700

010

001

100

010

001

−

100

013

001

R1 ↔ R2

R3 → -7R3

R2 → R2 -3R1

Changes in row 1

and row 2

Changes in row 3.

Changes in row 2.

28

Elementary Matrix

Identity matrix

Eij: Interchange the ith row and the jth row in Identity matrix

I: Ri ↔Rj.

Ei(k): Multiply the ith row by a nonzero scalar k in Identity

matrix I: Ri →kRi, k ≠ 0.

Eij(k): Replace the ith row by k times the jth row plus the ith row

in Identity matrix I: Ri → Ri + kRj.

100

010

001

100

010

501

Questions:

Which of the following are elementary matrices?

0010

0100

1000

0001

,

200

110

011

,

10

05.

Single elementary

row operation

Elementary

Matrix

( )513E or 311 5RRR +→

29

Theorem

If the elementary matrix E results from performing a certain row

operation on Im and if A is an m x n matrix, then the product EA is

the matrix that results when the same row operation is performed

on A.

Example:

(a) Find an elementary matrix, E by using E31(3). and

(b) Perform the same row operation as given in (a) to

−=

0441

6312

3201

A . Compare the answer with EA.

Solution:

(a) EIRRR =

→

= +→

103

010

001

100

010

001

1333

(b) Same row operation is performed on A:

− →

−=+→

91044

6312

3201

0441

6312

3201133

3RRRA

E multiple by A:

−=

−

=

91044

6312

3201

0441

6312

3201

103

010

001

EA

A single elementary row operation

is performed on I.

Elementary

Matrix

Same result

30

Row equivalent

A is said to be row equivalent to a matrix B:

- If B can be obtained from A by performing a finite sequence

of elementary row operations successively to A.

- If and only if there exist elementary matrices E1, E2, . . . , Es

such that Es. . . E2E1A = B.

Example:

122

511

121

RRR

A

−→

=

−

610

1211AE=

−

−

22 RR −→

610

12112 )AE(E=

−

−

2 211 R-RR →

B)AEE(E ==

−

610

1101123

122

10

01

RRR

I

−→↓

=

111

01E=

−

22 RR

I

−→↓

210

01E=

−

211 2 R-RR

I

→↓

310

21E=

−

A is row

equivalent

to B.

31

Theorem

Every elementary matrix is invertible, and the inverse is also an

elementary matrix. More precisely:

E-1

is the elementary matrix obtained from I by the inverse of the row

operation that produced E from I.

(a) Eij -1 = Eij. : Ri ↔Rj.

(b) Ei(k)-1 = Ei(

k

1 ):Ri →k

1 Ri, k ≠ 0

(c) Eij(k)-1 = Eij(-k), i ≠ j, k ≠ 0: Ri → Ri - kRj.

Example:

100

010

001

100

010

001

100

010

001

100

010

001

010

100

001

010

100

001

100

030

001

100

00

001

3

1

=

−

010

100

001

010

100

0011

=

−

100

00

001

100

030

001

3

1

1

E23

123−

E = E23 E2(3) E2(3)

-1=E2(1/3)

32

Matrix Inverse by Elementary Operations

Theorem: If an nxn matrix can be reduced to the identity matrix, then it is

invertible.

Theorem: If A is an nxn invertible matrix, then A can be reduced to the

identity matrix In by elementary row operations. It follows that Es . . . E1A = I where E1, . . . , Es are the row elementary matrices

corresponding to the row operations and A-1

= Es . . . E1.

How to find the inverse of an invertible matrix?

1. By using Elementary Row Operations.

2. By using Classical Adjoint.

By using Elementary Row Operations:

( )IA ( )BI

⇒ BA =−1

( )IA : augmented matrix

Perform elementary

operations

33

Example:

The matrix

=

100

420

011

A is nonsingular. Find its inverse by using

elementary operations.

Solution:

( )

− →

=−→

100

410

001

100

020

011

100

010

001

100

420

011322 4

3RRR

I/A

−

−

→

− →−→→

100

20

21

100

010

001

100

20

001

100

010

011

2

12

1

2

12 21122 RRR/RR

−

−

=∴ −

100

20

21

2

12

1

1A

34

By using Classical Adjoint

=

nnnn

n

n

aaa

aaa

aaa

A

K

MMMM

L

L

21

22221

11211

adj A =

nnnn

n

n

AAA

AAA

AAA

L

MLMM

L

L

21

22212

12111

, Aij =(-1)i+j

Mij, where Mij is the minor of |A|,

that is, a determinant obtained by deleting the i-th row and j-th column of |A|.

adjAAdet

A11 =−

cofactor

35

Example:

The matrix

=

100

420

011

A is nonsingular. Find its inverse by using classical

adjoint.

Solution:

1. Find the determinant of A, det A = 2.

2. Find the cofactors ijA .

210

4211 ==A , 1

10

0121 −=−=A , 4

42

0131 ==A

010

4012 =−=A , 1

10

0122 ==A , 4

40

0132 −=−=A

000

2013 ==A , 0

00

1123 =−=A , 2

20

1133 ==A

3.

−

−

=

200

410

412

Aadj

4. adjAAdet

A11 =−

−

−

=

200

410

412

2

1

−

−

=

100

20

21

2

12

1

36

Remarks

To solve a system of n equations in n unknowns by Cramer’s Rule, it is

necessary to evaluate 1+n determinants of nn × matrices. For system with

more than three equations, Gaussian elimination is far more efficient, since

it is only necessary to reduce one ( )1+× nn augmented matrix.

Gaussian Elimination

Gaussian Elimination is a method of solving system of linear equations.

Elementary Operations

The basic method for solving a system of linear equations is to replace the

given system by a new system that has the same solution set but which is

easier to solve.

Elementary Operations for Linear Equations

(1) Multiply an equation by a nonzero number.

(2) Interchange two equations.

(3) Add a multiple of one equation to another.

Since the rows of the augmented matrix correspond to the equations in the

associated system, these operations correspond to the following elementary

row operations which are performed on the augmented matrix.

Elementary Row Operations

(1) Multiply a row through by nonzero constant: ii RR α→ , { }0\R∈α .

(2) Interchange two rows: ji RR ↔ .

(3) Addition of a constant multiple of one row to another row:

jii RRR α+→ .

Reduced Row-Echelon form

A matrix in reduced row-echelon form must contain following properties.

(1) If a row does not consist entirely of zeros, then the first nonzero

number in row is a 1, called a “leading 1” or pivot.

(2) Any rows consisting entirely of zeros are grouped together at the

bottom of the matrix.

(3) If any two successive rows that do not consist entirely of zeros, the

leading 1 in the lower row occurs farther to the right than the leading

1 in the higher row.

(4) Each column that contains a leading 1 has zeros everywhere else.

37

A matrix having properties (1), (2) and (3)(but not necessarily 4) is said to

be in row-echelon form.

Example :

100

010

001

in reduced row-echelon form.

000

010

011

in row-echelon form.

Example :

Which of the following matrices are said to be in (i) row-echelon form (r.e.f)

(ii) reduced row-echelon form (r.r.e.f)?

=

100

510

821

A

=

1000

0000

0010

0001

B nIC =

=

100

001

011

D

=

100

001

010

E

A - r.e.f.

C – r.e.f / r.r.e.f

B , D and E - neither.

Procedure to reduce a matrix to reduced row-echelon form/ row-echelon

form: (Study this together with example 43 below)

(1) Locate the leftmost column that does not consist entirely of zeros.

(2) Interchange the top row with another row, if necessary, to bring a

nonzero entry to the top of the column found in (1).

(3) If the entry that is now at the top of the column found in (1) is a, then

multiply the first row by a

1 in order to get a “leading 1”.

(4) Add suitable multiples of the top row to the rows below so that all

entries below the “leading 1” become zeros.

(5) Ignore the rows already containing “leading 1”. Locate the leftmost

column that does not consist entirely of zeros. Repeat (2), (3), (4) and

(5) until the entire matrix in reduced row-echelon form/ row-echelon.

38

The procedure for reducing a matrix to reduced row-echelon form is called

Gauss-Jordan elimination.

If we use the procedure to reduce a matrix up to a row-echelon form is called

Gaussian elimination.

Example :

Reduce this matrix into reduced row-echelon form.

=

1101

2022

2010

A

(1)

1101

2022

2010

(2) 21 RR ↔ (interchange row 1 and row 2)

1101

2010

2022

(3) 112

1RR →

1101

2010

1011

(4) 133 RRR −→

− 0110

2010

1011

Leftmost nonzero column

39

(5)

− 0110

2010

1011

Repeat step (2), (3), (4) and (5).

−

2100

2010

1001

is the reduced row-echelon form for matrix A.

Example :

Reduce the following matrix into row-echelon form and reduced row-

echelon form.

=

333

112

121

A

333

112

121

−

−−

030

130

121

− 0303

110

121

1003

110

121

1003

110

3

101

Leftmost nonzero column

122 2RRR −→

133 3RRR −→

223

1RR −→

233 3RRR +→211 2RRR −→

3113

1RRR −→

3223

1RRR −→

Row-echelon form

40

100

010

001

How to use Gauss Elimination/Gauss-Jordan Elimination to solve a

system of linear equations? (Study the following example)

Reduced row-echelon form

41

Example :

Solve the following system of linear equation by using Gauss-Jordan

Elimination method.

3333

12

22

=++

=++

=++

zyx

zyx

zyx

Above system is in standard form.

The augmented matrix is

=

3333

1112

2121

A .

Reduce the matrix A into Reduced row-echelon form.

3333

1112

2121

−−

−−−

3030

3130

2121

−− 3030

13

110

2121

0100

13

110

2121

0100

13

110

03

101

0100

1010

0001

0 ,1 ,0 ===∴ zyx .

122 2RRR −→

133 3RRR −→

223

1RR −→

233 3RRR +→211 2RRR −→

3113

1RRR −→

3223

1RRR −→

Row-echelon form

42

Existence and properties of solution of Linear System

After we apply the Gauss elimination / Gauss–Jordon elimination to the

augmented matrix [ ]BA .

1. If there are zero rows in the row echelon form and there is a leading 1 in

the B column, the system is inconsistent (⇒ no solution).

For example,

1000

3010

2001

2. If there are one or more zero rows in the augmented matrix, then the

system is having infinitely many solutions.

For example,

0000

3010

2001

0000

0000

2001

3. If there is NO zero row in the augmented matrix, then the system is

having precisely one solution.

For example,

1100

3010

2001

The equation is

inconsistent

Infinitely many

solutions

Infinitely many

solutions

Precisely one

solution

43

Example : (Gauss elimination if no solution exists)

Solve the system

6426

02

323

321

321

321

=++

=++

=++

xxx

xxx

xxx

( ) 133

122

2

3

2

6

0

3

426

112

123

RRR

RRR

−+→

−+→

( ) 233 60

2

3

2203

1

3

10

123

RRR −+→

−

−

−

−−

12

2

3

0003

1

3

10

123

323 321 =++ xxx

23

1

3

132 −=+− xx

120 =

This shows that the system has no solution.

44

Example : (Gauss elimination if infinitely many solutions exist)

Solve the linear system of three equations in four unknowns

1242303021

7245515160

0805020203

4321

4321

4321

.x.x.x.x.

.x.x.x.x.

.x.x.x.x.

=+−−

=−++

=−++

133

122

40

20

12

72

08

42303021

45515160

05020203

R).(RR

R).(RR

.

.

.

....

....

....

−+→

−+→

−−

−

−

23311

11

08

4411110

4411110

05020203

RRR.

.

.

...

...

....

+→

−−−

−

−

−

−

0

11

08

0000

4411110

05020203

.

.

...

....

Back substitution: 432 41 xxx +−=

41 2 xx −=

Therefore we have infinitely many solutions. If we choose a value of 3x and

a value of 4x then the corresponding values of 1x and 2x are uniquely

determined.

45

Example : (Gauss elimination if a unique solution exists)

Solve the following system of linear equation by using the Gauss elimination

method

632 321 =++ xxx

14232 321 =+− xxx

23 321 −=−+ xxx

The system can be written as

BX =A i.e

−

=

−

−

2

14

6

113

232

321

3

2

1

x

x

x

−−

−

2

14

6

113

232

321

−−−

−−

20

2

6

1050

470

321

−−

4

2

6

210

470

321

−− 4

2

6

470

210

321

30

4

6

1000

210

321

3

4

6

100

210

321

~ row echelon form

( ) 122 2 RRR −+→

( ) 133 3 RRR −+→

335

1RR −→ 32 RR ↔

233 7RRR +→ 33

10

1RR →

46

=

=+

=++

⇒

−=−+

=+−

=++

3

42

632

23

14232

632

3

32

321

321

321

321

x

xx

xxx

xxx

xxx

xxx

Back substitution,

33 =x

224 32 −=−= xx

1236 231 =−−= xxx

The solution for X ( ) ( )321321 ,,x,x,x −==

47

Gauss-Jordan elimination also can be use to find the inverse of a matrix?

The algorithm is as follows.

[ ]nIA [ ]BIn then BA =−1 .

Example :

Find the inverse of

−

=

220

720

211

A .

[ ]nIA

−−

−

110

010

001

500

720

211

−

−−−

20200

0500

001

100

5310

211

..

..

A finite series of elementary row

operations

33

22

11

20

50

R.R

R.R

RR

→

→

−→

322

311

53

2

R.RR

RRR

−→

+→

233 RRR −→

48

−

−

−−−

20200

70200

40401

100

010

011

..

..

..

−

−

−

20200

70200

30201

..

..

..

In

Hence

−

−

−

=−

20200

70200

302011

..

..

..

A .

Exercise

What is the different within Gauss Elimination and Gauss-Jordan

Elimination?

Find the answer from lecture notes.

211 RRR +→

49

Linear Dependence and Independence, Rank of a matrix

Let nR be the collection of all ordered n-tuples row vectors, that is

( ){ }niRxxxxR inn , 2, ,1 , , , , 21 LL =∈= .

Definition

Let nk RAAA ∈ , , , 21 L . Any row vector in nR of the form

kk AAA α++α+α L2211

where iα , ki , ,3 ,2 ,1 L= are scalar in R, is called a linear combination of

vectors kAAA , , , 21 L . A row vector A is a linear combination of

kAAA , , , 21 L if and only if there exists scalar Rk ∈ααα , , , 21 L such that

kk AAAA α++α+α= L2211 .

Definition

The vectors nk RAAA ∈ , , , 21 L are said to be linearly independent if

02211 =α++α+α kk AAA L implies 0 21 =α==α=α kL .

Definition

The vectors nk RAAA ∈ , , , 21 L are said to be linearly dependent if there

exists scalars Rk ∈ααα , , , 21 L , not all equal to zero, such that

02211 =α++α+α kk AAA L .

Theorem

(1) A finite set of vectors that contains the zero vectors is linearly

dependent.

(2) Two vectors are linearly dependent if and only if either vector is a

scalar multiple of the other.

Example :

Show that any nonzero vector v is, by itself, linearly independent.

Suppose 0=kv , but 0≠v . Then 0=k . Hence v is linearly independent.

50

Example :

Determine whether u and v are linearly dependent.

(a) ( )5 ,3=u , ( )10 , 6=v (b) ( )4 ,3=u , ( )10 , 6=v

Two vectors u and v are dependent if and only if one is a multiple of the

other.

(a) Yes; uv 2= (b) No; neither is a multiple of the other.

Example :

Determine whether the vectors ( )1 ,2 ,1 − , ( )0 2, ,2 and ( )3 ,4 ,5 − are linearly

dependent.

Method 1:

Set a linear combination of the vectors equal to the zero vector using

unknown scalars 1α , 2α and 3α :

( ) ( ) ( ) ( )0 0, ,03 ,4 ,50 2, ,21 ,2 ,1 321 =−α+α+−α

Then, ( ) ( )0 0, 0,3 ,422 ,52 31321321 =α+αα−α+α−α+α+α

Set corresponding components equal to each other to obtain the equivalent

homogeneous system, and reduce to echelon form:

03

0422

052

31

321

321

=α+α

=α−α+α−

=α+α+α

or

022

066

052

32

32

321

=α−α−

=α+α

=α+α+α

or

0

052

32

321

=α+α

=α+α+α

The system, in echelon form, has only two nonzero equations in the three

unknowns; hence the system has a nonzero solution. Thus the original

vectors are linearly dependent.

51

Method 2:

Form the matrix whose rows are the given vectors, and reduce to echelon

form using the elementary row operations:

−

−

345

022

121

to

−

−

−

260

260

121

to

−

−

0003

110

121

Since the echelon matrix has a zero row, the vectors are dependent.

Example :

Determine whether ( )3- 2, 1, , ( )2 3,- 1, and ( )5 1,- 2, are linearly

independent.

Form the matrix whose rows are the given vectors, and reduce to echelon

form using the elementary row operations:

−

−

−

512

231

321

to

−

−

−

1150

550

321

to … to

−

−

100

110

321

Since the echelon matrix has no zero rows, the vectors are independent.

Definition

The maximum number of linearly independent row vectors of a matrix

[ ]ijaA = is called the rank of A and is denoted by ( )Arank or ( )Aρ .

Theorem

(1) If A is any matrix, then ( ) ( )ArankArank T = .

(2) If A is an nm × matrix in row echelon form, then ( )Arank is the

number of nonzero rows of A.

(3) Row equivalent matrices have the same rank.

52

Example :

Given

−

−

−

=

512

231

321

A . Find ( )Arank .

From example 54, number of nonzero rows of row equivalent matrix for A is

3. Thus ( ) 3=Arank .

53

Theorem (Fundamental Theorem of Linear Systems)

A linear system of m equation in n unknowns ( 1x , 2x , 3x , 4x , nx ,L ):

11212111 bxaxaxa nn =+++ L

M

L

22222121 bxaxaxa nn =+++ ( )∗

mnmnmm bxaxaxa =+++ L2211

has solution if and only if the coefficient matrix A and the augmented matrix ~

A have the same rank; that is, ( )

=

~

ArankArank . Moreover,

(1) If ( ) nArankArank =

=

~

, then the system ( )∗ has precisely one

solution.

(2) If ( ) nArankArank <

=

~

, then the system ( )∗ has infinitely many

solutions.

(3) If ( )

<

~

ArankArank , then the system (*) has no solution.

Theorem

Let A be an nn × matrix. Then the following are equivalent:

(1) A is invertible,

(2) ( ) nArank = ,

(3) BX =A has exactly one solution for every 1×n matrix B,

(4) 0=XA has only the trivial solution.

54

Solving a system of linear equations by using Determinant (or Cramer’s

Rule)

Cramer’s Rule

Cramer’s Rule is useful in solving a system of linear equation of n equations

in n unknowns.

Theorem (Cramer’s Rule)

If ~~bx =A is a system of n linear equations in n unknowns such that

( ) 0≠Adet , then the system has a unique solution. This solution is

( )( )Adet

Adetx i

i = , ni , 3, 2, ,1 L=

where iA , ni , 3, 2, ,1 L= is the matrix obtained by replacing the entries in

the ith column of A by the entries in the matrix

=

nb

b

b

M

2

1

~b .

55

Example :

Solve by determinant (or Cramer’s Rule): 52 =− yx

0321 =++ xy .

First arrange the system in standard form:

123

52

−=+

=−

yx

yx

The coefficient matrix is

−=

23

12A .

The determinant of coefficient matrix is ( ) 723

12=

−=Adet .

Since ( ) 0≠Adet , the system has a unique solution,

( ) 921

151 =

−

−=Adet , ( ) 17

13

522 −=

−=Adet .

Thus the unique solution of the system is

( )( ) 7

91 ==Adet

Adetx ,

( )( ) 7

172 −==Adet

Adety .

56

Example :

Solve by determinant (or Cramer’s Rule):

yxz

yzx

zxy

213

83

23

−=−

−=+

=+

First arrange the system in standard form:

132

83

032

=++−

=++

=−+

zyx

zyx

zyx

The coefficient matrix is

−

−

=

321

113

132

A .

The determinant of coefficient matrix is ( ) 35−=Adet .

Since ( ) 0≠Adet , the system has a unique solution,

( ) 84

321

118

130

1 −=

−

=Adet , ( ) 35

311

183

102

2 =

−

−

=Adet ,

( ) 63

121

813

032

3 −=

−

=Adet .

Thus the unique solution of the system is

( )( ) 35

841 ==Adet

Adetx ,

( )( )

12 −==Adet

Adety ,

( )( ) 35

633 ==Adet

Adetz .

57

Further Matrices

Partitioning

Any matrix A may be partitioned into a number of smaller matrices

called blocks by vertical lines that extend from bottom to top, and

horizontal lines that extend from left to right.

Example:

=

−

=

−

=

3231

2221

1211

1

0

7

3

64

43

21

01

164

043

721

301

AA

AA

AA

A

where the blocks are

( ) ( ) ( ) ( )1640

7

43

21301 323122211211 ==

=

=−== A,A,A,A,A,A .

Clearly the partition is NOT unique. We could also have

−

=

1

0

7

3

6

4

2

0

4

3

1

1

A =

232221

131211

AAA

AAA

58

Some properties: Example:

If A and B are partitioned as

=

mnmm

n

AAA

AAA

A

L

MMM

L

21

11211

and

=

pqpp

q

BBB

BBB

B

L

MMM

L

21

11211

then

(i)

ααα

ααα

=α

mnmm

n

AAA

AAA

A

L

MMM

L

21

11211

(ii) If A and B are matrices of the same size,

i.e., m = p and n = q and each Aij block is

of the same form as the corresponding Bij

block, then

+++

+++

=+

mnmnmmmm

nn

BABABA

BABABA

BA

L

MMM

L

2211

1112121111

(iii) If n = p and we denote AB = C, then

,BACn

k

kjikij ∑=

=1

provided that the number of columns in

each Aik is the same as the number of rows

in the corresponding Bkj.

If A and B are partitioned as

=

=2221

1211

323

112

401

AA

AAA and

=

=2221

1211

103

221

100

BB

BBB then

(i)

=

=

151015

5510

2005

55

555

2221

1211

AA

AAA

(ii)

++

++=+

22222121

12121111

BABA

BABABA

=

426

333

501

(iii)

=

2221

1211

2221

1211

BB

BB

AA

AAAB

++

++=

2222122121221121

2212121121121111

BABABABA

BABABABA

( )031

4

21

00

12

0121121111

+

=+ BABA

=

24

012

( )

=

+

=+

5

51

1

4

2

1

12

0122121211 BABA

( ) ( )( ) ( )41103321

002321221121 =+

=+ BABA

( ) ( )( ) ( )10132

12322221221 =+

=+ BABA

=

10411

524

5012

AB

59

Suppose that an n x n matrix A can be partitioned into the block

diagonal form

mA

A

A

00

0

0

00

2

1

L

OM

M

L

where A1, . . . , Am are all square matrices. (not necessarily all of

the same size)

Then

(i) A is invertible if and only if A1, . . . , Am are invertible.

(ii) assuming that A1, . . . , Am are invertible, then

=

−

−

−

−

1

1

2

1

1

1

00

0

0

00

mA

A

A

A

L

OM

M

L

Example:

Find entry(1,2) in A2, where A =

−−

−

−

211

011

011

.

Solution:

By partitioning: By original way:

=

−−

−

−

2212

1211

211

011

011

AA

AA

−

−

−

=

444

022

0222

A

−

−=

−

−

−

−=

22

22

11

11

11

111111AA 212 −=a

212 −=a .

60

Example:

Find entry(1,2) in A2, where A =

−−

−

−

211

111

111

.

Solution:

By partitioning: By original way:

=

−−

−

−

2212

1211

211

111

111

AA

AA

−−

−−

=

400

211

2332

A

−

−=

−

−

−

−=

22

22

11

11

11

111111AA 312 −=a

212 −=a . Is this TRUE??

61

Eigenvalue and Eigenvector

Let A be an nn × matrix. A number λ is said to be an eigenvalue of

A if there exists a nonzero solution vector X of the linear system

AX = λX.

The solution vector X is said to be an eigenvector corresponding to

the eigenvalue λ.

From the definition, λλλλ = 0 can be an eigenvalue but the zero

vector is NOT an eigenvector for any matrix.

Example:

Is

−1

1 an eigenvector of the matrix

=

34

21A ?

Solution:

−⋅−=

−

1

11

1

1

34

21

Hence,

−1

1 is an eigenvector of A associated to the eigenvalue 1− .

A X = λ X

62

How to find eigenvalues of an nn× matrix A? Solve characteristic equation of A to get eigenvalues.

What is characteristic equation of A?

We rewrite xAx λ= as IxAx λ= , or equivalently,

( ) 0=−λ xAI .

( ) 0=−λ AIdet : Characteristic equation of A.

When ( )AIdet −λ is expanded, it is a polynomial in λ called

characteristic polynomial of A. This characteristic polynomial

has the form: λn + cn-1λ

n-1 +. . . + c1λ + c0 .

Example:

Find the eigenvalues of

−

−

=

6116

100

010

A .

Solution:

Characteristic polynomial:

( ) 6116

6116

10

0123 −λ+λ−λ=

−λ−−

λ

λ

=−λ detAIdet .

Characteristic equation:

( ) 0=−λ AIdet

06116 23 =−λ+λ−λ

( )( )( ) 0321 =−λ−λ−λ

321 ,,=λ

The eigenvalues of A are 1, 2 and 3.

Nonzero ( ) 0=−λ⇔ AIdet

63

Eigenvalues of triangular matrices

Theorem:

If A is an n x n triangular matrix (upper triangular, lower

triangular, or diagonal), then the eigenvalues of A are the entries

on the main diagonal of A.

Example:

Find the eigenvalues of

=

527

029

001

A (lower triangular matrix).

Solution:

By inspection, the eigenvalues are 1, 2 and 5.

How to find eigenvectors of an nn× matrix A? The eigenvectors of A corresponding to an eigenvalue λ are the

nonzero vectors x that satisfy xAx λ= .

Study the following 3 examples:

Example: (All the eigenvalues are different)

Example: (Some of the eigenvalues are the same)

Example: (All the eigenvalues are the same)

64

Example: (All the eigenvalues are different)

Determine the eigenvalues and eigenvectors for the matrix A.

−

−

−

=

110

121

211

A .

Solution:

The characteristic equation, |A - λI| = 0

0

110

121

211

=

λ−−

λ−−

−λ−

(1 – λ)[–(2 – λ)(1 + λ) – 1] + [(–1 – λ) + 2] = 0

(1 – λ)[λ2 - λ – 3] + (1 – λ) = 0

(1 – λ)[λ2 – λ – 2] = 0

(1 – λ)(λ + 1)(λ – 2) = 0

Hence, the eigenvalues are λ1 = 2, λ2 = 1 and λ3 = –1.

For λ = λ1 = 2, we have

A - λI =

−

−

−−

=−

310

101

211

2IA

Let e1 be the eigenvector of λ = 2. Then

(A – 2I)e1 = 0

0

310

101

211

3

2

1

=

−

−

−−

x

x

x

that is,

–x1 + x2 – 2x3 = 0

–x1 + 0x2 + x3 = 0

0x1 + x2 – 3x3 = 0

⇒ x1 = x3

x2 = 3x3

65

Thus the eigenvector e1 corresponding to the eigenvalue λ = 2 is

α=

1

3

1

1e ,

where α is an arbitrary nonzero scalar.

For λ = λ2 = 1, we have

(A – (1)I)e2 = 0

0

210

111

210

3

2

1

=

−

−

−

x

x

x

that is,

x2 – 2x3 = 0

–x1 + x2 + x3 = 0

x2 – 2x3 = 0

⇒ x2 = 2x3

x1 = x2 + x3 = 3 x3

Thus the eigenvector e2 corresponding to the eigenvalue λ = 1 is

β=

1

2

3

2e

where β is an arbitrary nonzero scalar.

For λ = λ3 = –1, we obtain

(A – (–1)I)e3 = 0

0

010

131

212

3

2

1

=

−

−

x

x

x

that is,

2x1 + x2 – 2x3 = 0

–x1 + 3x2 + x3 = 0

x2 = 0

⇒ x2 = 0 , x1 = x3

Thus the eigenvector e3 corresponding to the eigenvalue λ = –1 is

γ=

1

0

1

3e .

66

Example: (Some of the eigenvalues are the same)

Determine the eigenvalues and eigenvectors for the matrix

−−−

=

1144

814

1685

A .

Solution:

The characteristic equation, |A - λI| = 0

0

1144

814

1685

=

λ−−−−

λ−

λ−

(λ + 3)[λ2 + 2λ – 3] = 0

(λ + 3)2(λ– 1) = 0

Hence, the eigenvalues are λ1 = –3, λ2 = –3 and λ3 = 1.

For λ = -3 we have

(A - λ1I)e1 = 0

that is,

–8x1 – 8x2 – 16x3 = 0

–4x1 - 4x2 – 8x3 = 0

4x1 + 4x2 + 8x3 = 0

which form a single equation

⇒ x1 + x2 + 2x3 = 0

We are free to choose any two of the components x1, x2, and x3 at will, with

the remaining one determined by the above equation. Suppose we set, x2 = α

and x3 =β; then

−

β+

−

α=

β

α

β−α−

=

1

0

2

0

1

12

X

Thus the two linearly independent eigenvectors e1 and e2 corresponding to

the eigenvalue λ = -3 are

−

=

0

1

1

1e and

−

=

1

0

2

2e .

For λ = 1, the eigenvector is

−

=

1

1

2

3e .

67

Example: (All the eigenvalues are the same)

Determine the eigenvalues and eigenvectors for the matrix A.

−=

71

43A .

Sol: The characteristic equation, |A - λI| = 0

( ) 0571

43 2 =−λ=λ−−

λ−

Hence, the eigenvalue is λ1 = λ2 = 5.

(A –λ1I)e = 0

–2x1 + 4x2 = 0

–x1 + 2x2 = 0

⇒ x1 = 2x2

Thus , we find the single eigenvector

=

1

2e .

68

Eigenvalues and invertibility

Theorem: A square matrix A is invertible if and only if λ = 0 is

NOT an eigenvalue of A.

Example:

Is matrix

10

10 invertible?

Solution:

From the above theorem, we see that

The eigenvalues are 0 and 1. Since we

have 0=λ , then A is not invertible.

OR,

From the formula

If

=

10

10A , then

( )( ) ( )( )

−

−=−

00

11

0110

11A

Given a 22 × matrix

=

dc

baA , the

inverse is

−

−

−=−

ac

bd

bcadA

11

0

1

69

Some useful properties of eigenvalues

1. The sum of the eigenvalues A is ∑∑==

==λn

i

ii

n

i

i aAtrace11

.

Example:

Given a square matrix

=

300

520

421

A , find the sum of eigenvalues and

compare with trace A.

Solution:

=

300

520

421

A is a upper triangular matrix.

∴ Eigenvalues are 1, 2 and 3.

Sum of eigenvalues = 1+2+3=6

trace A = 1+2+3 (the values in the diagonal of A)

2. The product of the eigenvalues of A is Adetn

i

i =λ∏=1

, where

det A denotes the determinant of the matrix A.

Example:

Given a square matrix

=

300

520

421

A , find the product eigenvalues and

compare with det A.

Solution:

Eigenvalues are 1, 2 and 3.

( )( )( ) 63213

1

==λ∏=i

i .

( )( )( ) 6321 ==Adet .

Same answer

Same answer

70

3. Suppose the eigenvalues for A are 1λ , nλλ , ,2 K . Then the

eigenvalues of the inverse matrix A-1

, provided it exists, are

.,...,,nλλλ

111

21

Example:

Given a square matrix

=

300

520

421

A , find the eigenvalues of

(i) A

(ii) A-1

.

Solution:

(i) Eigenvalues of A are 1, 2 and 3.

(ii)

−

−

=−

31

65

21

31

1

00

0

11

A

This an upper triangular matrix,

∴ eigenvalues are 1, 1/2 and 1/3.

By using elementary operations

method

Or

By using classical adjoint method.

71

4. The eigenvalues of the tranposed matrix AT are λ1, λ2, . ., λn

as for the matrix A.

Example:

Given a square matrix

=

300

520

421

A , compare the eigenvalues for A and

TA .

Solution:

Eigenvalues for A are 1, 2 and 3.

=

354

022

001T

A

Eigenvalues are 1, 2 and 3.

5. Suppose the eigenvalues for A are 1λ , nλλ , ,2 K . If k is a

scalar then the eigenvalues of kA are kλ1, kλ2, . ., kλn.

Example:

Given a square matrix

=

300

520

421

A , find the eigenvalues for 2A.

Solution:

=

=

600

1040

842

300

520

421

22A .

Eigenvalues for 2A are 2, 4 and 6 (which are 2(1), 2(2), 2(3)).

Same answer

1, 2 and 3 are eigenvalues of A.

72

6. Suppose the eigenvalues for A are 1λ , nλλ , ,2 K . If k is a

scalar and I the identity matrix then the eigenvalues of A ±±±± kI

are respectively λ1 ± k, λ2 ± k, . . . , λn ± k.

Example:

Given a square matrix

=

300

520

421

A , find the eigenvalues for IA 3+ .

Solution:

=

+

=+

600

550

424

100

010

001

3

300

520

421

3IA .

The eigenvalues for IA 3+ are 4, 5 and 6 (which are 1+3, 2+3 and 3+3).

7. Suppose the eigenvalues for A are 1λ , nλλ , ,2 K . If k is a

positive integer then the eigenvalues of Ak are k

nkk

,,, λλλ K21 .

Example:

Given a square matrix

=

300

520

421

A , find the eigenvalues for 2A .

Solution:

=

900

2540

26612

A

The eigenvalues for 2A are 1, 4 and 9 (which are 12, 2

2 and 3

2).

1, 2 and 3 are eigenvalues of A.

73

Similar Matrices Suppose A and B are nn × matrices. Then A is said to be similar to

B, denote A ∼∼∼∼ B, if there exists a nonsingular nn × matrix P such

that P-1

AP = B.

(The matrix P is not unique.)

Example:

Suppose

−

−=

26

37A and

=

40

01B . Show that BAPP =−1 where

(a)

=

12

11P

(b)

=

24

22P

Solution:

(a)

−−

=

−−

12

11

26

37

12

111

1 APP

B=

=

−−

−

−=

40

01

12

11

26

37

12

11

(b)

−−

=

−−

24

22

26

37

24

221

1 APP

B=

=

−−

−

−=

40

01

24

22

26

37

121

21

21

74

We have the following observations:

1. A ∼∼∼∼ A.

2. A ∼∼∼∼ B ⇒ B ∼∼∼∼ A

3. A ∼∼∼∼ B, B ∼∼∼∼ C ⇒ A ∼∼∼∼ C.

Theorem:

If A ∼∼∼∼ B, then

(i) the characteristic polynomial of A is equal to the

characteristic polynomial of B.

(ii) A and B have the same set of eigenvalues.

Example:

Suppose

−

−=

26

37A and

=

40

01B , find

(a) the characteristic polynomial of A and B respectively.

(b) the eigenvalues of A and B respectively.

Solution:

(a) Characteristic polynomial of A = AI −λ = 4526

37 2 +λ−λ=+λ

−−λ.

Characteristic polynomial of B = BI −λ = 4540

01 2 +λ−λ=−λ

−λ.

(b) Eigenvalues for A:

0=−λ AI

0452 =+λ−λ =λ 1 and 4.

Eigenvalues for B:

0=−λ BI

0452 =+λ−λ =λ 1 and 4.

From (a) and (b), matrices A and B are having the same characteristic

polynomial and eigenvalues. And from previous example, we know that A ∼∼∼∼

B.

75

Diagonalization of Matrices A square matrix A is called diagonalizable if there is an invertible

matrix P such that P-1

AP is a diagonal matrix; the matrix P is said

to diagonalize A. We may say that A ~ a diagonal matrix.

(Useful in finding nA )

APPB1−=

1−= PBPA

( )nn PBPA 1−=

( )( )( ) ( )1−−−−= BPBBPBAn PPPPPP K111

1−= PPBA

nn

Diagonal matrix

IPP =−1

nB can be found easily For example,

If

=

30

02B is diagonal

matrix then

=

=

n

nn

nB

30

02

30

02

An is a product of 3 matrices: P, B

n and P

-1.

76

How to find a matrix P in Diagonalization of

Matrices?

Diagonalization algorithm Given an n x n matrix A:

(1) Find the eigenvalues of A.

(2) Find (if possible) n linearly independent eigenvectors p1, p2, . . . , pn.

(3) Form P = (p1, p2, . . . , pn) - the matrix with the pi as

columns.

(4) Then P-1

AP is diagonal, the diagonal entries being the

eigenvalues corresponding to p1, p2, . . . , pn

respectively.

Linearly Independent: If 1v , 2v , …, nv are n vectors, then the vector equation

02211 =+++ nnvcvcvc K has at least one solution, namely 01 =c ,

02 =c , … and 0=nc . If this is the only solution, then 1v , 2v , …,

nv are linearly independent.

77

Example:

Diagonalize the matrix

−−−

=

1144

814

1685

A .

Sol:

The eigenvalues are found to be λ = -3 and λ = 1, with eigenvectors

−

=

−

=

−

=

1

1

2

1

0

2

0

1

1

321 p,p,p . These vectors are linearly independent.

So,

−

−−

==

110

101

221

321 )p,p,p(P

is invertible and will diagonalize A. In fact,

−−

=−

211

311

2011

P and

−

−

=−

100

030

0031APP .

78

Theorem (Distinct Eigenvalues, Linear Independent Eigenvectors)

If an n x n matrix A has distinct eigenvalues λ1, . . . , λn, then the

corresponding eigenvectors e1, . . . , en are linearly independent.

Theorem

Let A be an n x n matrix.

1. A is diagonalizable if and only if it has n linearly independent

eigenvectors.

2. If A has n linearly independent eigenvectors e1, . . . , en and

we make these the columns of Q, so that Q = [e1, . . . , en],

then Q-1

AQ = D is diagonal, with the eigenvalues of A as the

entries on the main diagonal.

Example:

Diagonalize the matrix

−

=

253

021

001

A .

Sol:

The characteristic polynomial of A is

221

253

021

001

))((IA −λ−λ−=

λ−−

λ−

λ−

=λ−

so the characteristic equation is (λ-1)(λ-2)2 = 0.

Thus the eigenvalues of A are λ = 1 and λ = 2.

The eigenvectors of respective eigenvalue are

−=

1

81

81

1p and

=

1

0

0

2p .

Since A is a 3 x 3 matrix and there are only two basis vectors, A is NOT

diagonalizable.

From the above example, we know that not every matrix is diagonalizable.

79

Theorem

Let A be any nn × real matrix. Then the followings are equivalent:

1. A is nonsingular.

2. Rank(A) = n.

3. All the rows(column) are linearly independent.

4. |A| ≠ 0.

5. The eigenvalues of A are not zero.

Theorem

Every symmetric matrix is diagonalizable.

Example:

Find a matrix that diagonalizes

=

422

242

224

A .

Sol:

The characteristic equation of A is (λ - 8)(λ - 2)2 = 0.

For λ = 2, we have the corresponding eigenvectors,

−

=

0

1

1

1u and

−

=

1

0

1

2u .

For λ = 8, we have the corresponding eigenvectors, .u

=

1

1

1

3

Let

−−

=

110

101

111

P . Then

=−

800

020

0021APP .

80

Applications

(1) Homogeneous systems of linear first order differential equations with

constant coefficients.

(2) Non-homogeneous systems of linear first order differential equations

with constant coefficients.

Homogeneous systems of linear first order differential equations with

constant coefficients:

Consider a system of linear first-order differential equations with constant

coefficients

x1′= a11x1 + a12x2 + . . . + a1nxn

x2' = a21x1 + a22x2 + . . . + a2nxn

.

.

.

xn' = an1x1 + an2x2 + . . . + annxn

with initial conditions

x1(0) = c1 , x2(0) = c2 , . . . , xn(0) = cn ,

where aij ’s are constants. Here we have n unknown functions xj = xj(t) in

one variable t and n differential equations.

Let A

=

nnnn

n

n

aaa

aaa

aaa

L

MM

L

L

21

22221

11211

, ,2

1

==

)t(x

)t(x

)t(x

)t(XX

n

M

and 2

1

.

)t('x

)t('x

)t('x

)t('X'X

n

==M

81

Then the system can be represented in matrix equation:

X' = AX (1)

The solution of such a system of differential equations is generally tedious

but is simple if A is diagonal.

We begin the solution of (1) by making a linear change of variables from

nxxx , , , 21 L to y1 ,y2,…,yn :

YQ X = (2)

where Q is an invertible matrix. Thus, YQX ′=′ . Putting (2) into (1) gives

YAQ YQ =′

YAQQY -1 =′

So the idea is to find such a matrix Q such that

DAQ Q 1 =−

where D is a diagonal matrix.

Example:

Find a solution to the system

211 3xxx +=′

212 22 xxx +=′

where x1 and x2 are functions of t and x1(0) = 0, x2(0) = 5.

Solution:

Let

=

22

31A ,

=

2

1

x

xX ,

=

'x

'x

2

1X' and X(0) = C =

5

0

.

Then this system can be written as

( ) ( )tt AXX =′ , X(0) = C =

5

0

.

82

Solving for the eigenvalues and eigenvectors , we obtain

=

1

11q and

−=

2

32q are eigenvectors corresponding to the eigenvalues 4 and

1− , respectively.

Hence with

−==

21

31),( 21 qqQ ,

−=−

10

04AQQ 1

= D.

Now consider new functions y1 = y1(t) and y2.= y2(t). Let YQ X = ,

where

=

2

1

y

yY

ie,

−=

2

1

2

1

21

31

y

y

x

x that is ,

212

211

2

3

yyx

yyx

−=

+=

Then, from QYX = , we have YQX ′=′ , substitute into AXX =′ , we

obtain

YAQ YQ =′

DYYAQQY ==′ 1- .

This means that

−=

2

1

2

1

10

04

y

y

'y

'y so

22

11 4

y'y

y'y

−=

=

Solving the two first order linear differential equations, we obtain

( ) tAety

41 = and ( ) t

Bety−=2 ,

where A and B are constants.

Then,

−

+=

−=

−=

−

−

− tt

tt

t

t

BeAe

BeAe

Be

Ae

y

y

x

x

2

3

21

31

21

31

4

44

2

1

2

1

83

so the general solution is

x1(t) =Ae4t

+ 3Be-t

x2(t) = Ae4t

– 2Be-t

Finally, the requirement in this example that x1(0) = 0, x2(0) = 5

determines the constants c and d:

0 = x1(0) = Ae0 + 3Be

0 = A + 3B

5 = x2(0) = Ae0 – 2Be

0 = A – 2B

These equations give A = 3 and 1−=B , so

x1(x) = 3e4t − 3e

-t

x2(x) = 3e4t + 2e

-t.

Example:

Find the general solution to the system

x1' = 5x1 + 8x2 + 16x3

x2' = 4x1 + x2 + 8x3

x3' = -4x1 - 4x2 - 11x3

Then find a solution satisfying the boundary conditions

x1(0) = x2(0) = x3(0) = 1.

Solution:

The system has the form X’ = AX, where

−−−

=

1144

814

1685

A

The eigenvalus are –3, 3− and 1 and the corresponding eigenvectors are,

respectively,

−

=

0

1

1

1q ,

−

=

1

0

2

2q ,

−

=

1

1

2

3q

84

Let ( )

−

−−

==

110

101

221

, , 321 qqqQ . Then Q diagonalizes A with

Q-1

AQ = D=

−

−

100

030

003

, a diagonal matrix.

Now let QYX = so 'QYX =′ .

From the equation X’ = AX, we have

AQYYQ =′

DYAQYQY' == −1

i.e: 11 3yy −=′ , 22 3yy −=′ and 33 yy =′ .

Hence we have

tecy

311

−=

tecy

322

−=

tecy 33 = , where c1, c2, c3 are arbitrary constants.

Hence , QYX =

=

−

−−

110

101

221

3

2

1

y

y

y

=

−

+

+−−

−

−

−−

tt

tt

ttt

ecec

ecec

ececec

33

2

33

1

33

23

1 22

So the general solution is

x1(t) = c4e-3t

+ 2c3et, where ( )214 2ccc +−=

x2(t) = c1e-3t

+ c3et

x3(t) = c2e-3t

- c3et

85

The boundary conditions, x1(0) = x2(0) = x3(0) = 1 determine the constants

ci.

( )

−

+

+

==

32

31

34 2

0

1

1

1

cc

cc

cc

X

c4 = -(c1 + 2c2)

c4 + 2c3 = 1 c1 = -3

⇒ c1 + c3 = 1 ⇒ c2 = 5

c2 - c3 = 1 c3 = 4

c4 = -7

Hence,

( ) tteetx 87 3

1 +−= −

( ) tteetx 43 3

2 +−= −

( ) tteetx 45 3

3 −= −

86

Non-homogeneous System of linear first order differential equations

with constant coefficients:

Consider a system of linear first-order differential equations with constant

coefficients

( )tfxaxaxax nn 112121111 ++++=′ K

( )tfxaxaxax nn 222221212 ++++=′ K

.

.

.

( )tfxaxaxax nnnnnnn ++++=′ K2211

with initial conditions

( ) 11 0 cx = , ( ) 22 0 cx = , . . . , ( ) nn cx =0 ,

where aij ’s are constants. Here we have n functions ( )txx jj = in one

variable t and n differential equations.

Let A

=

nnanana

naaa

naaa

L

MM

L

L

21

22221

11211

, 2

1

,

)t(x

)t(x

)t(x

)t(XX

n

==M

2

1

,

)t('x

)t('x

)t('x

)t('X'X

n

==M

and

=

)t(f

)t(f

)t(f

)t(F

n

M

2

1

Then the system can be represented in matrix equation:

X' = AX + F(t) ; (6)

X(0) =

=

nc

.

.

c

c

C

2

1

The system is said to be homogeneous , if F(t) = 0 .

87

Suppose that the coefficient matrix A is diagonalizable, so there is an

invertible matrix Q such that AQQD1−= is diagonal.

We choose Q to be a matrix whose columns are n linearly independent

column eigenvectors ,corresponding to the eigenvalues λ1, λ2, . . . , λn of A respectively. Thus,

D = diag(λ1, λ2, . . . , λn)

Step 1:

Let QYX = , where

=

y

.

.

y

y

Y

2

1

Hence, Y' QX =′ , since Q is a constant matrix. Substituting for X and X’ in

the equation (6) X' = AX + F(t) , we obtain

( )tFAQYYQ +=′ ; X(0) = CQY =)0(

( )tFQYAQQY11' −+

−= ; ( ) CQY

10 −=

We can rewrite the above equation as:

( )tGDYY ' += ;

where )(1)( tFQtG−= = (g1(t),…,gn(t))

T . Let CQK

1−= = (d1,…,dn)T.

This is a system of linear differential equations in nyyy , , , 21 L , the entries

of Y, which has the very simple form

( )tgyy 1111 +λ=′ , ( ) 11 0 dy =

( )tgyy 2222 +λ=′ , ( ) 22 0 dy =

….

88

( )tgyy nnnn +λ=′ , ( ) nn dy =0

Step 2: Solve for Y

Each of these equations is a first-order linear differential equation, it can be

solved by using the integrating factor method.

Step 3

Finally, to get X, we use X= Q Y.

========================

**The Integrating Method for solving the differential equation of the

following form :

( )xqydx

dy=λ+

where λ is a constant. We multiply the above equation by the integrating

factor

∫λdxe

Then we obtain

∫λdxe ( )xqy

dx

dy=

λ+ ∫λdxe

∫λdxe

dx

dy + λ y ∫ dx

eλ

=q(x) ∫ dx

eλ

ie. (dx

dy ∫ dx

eλ

) = q(x) ∫ dx

eλ

Hence

y ∫ dx

eλ

= ∫ q(x) ∫ dx

eλ

dx + C

We can then find y.

=====================

89

Example:

Solve the system

1

x′ = 6x1 + x2 + 6t

2

x′ = 4x1 +3x2 −10t +4

Solution:

The system has the form

( )tFAXX +=′ ,

where

=

34

16A and ( )

+−=

410

6

t

ttF .

The characteristic equation of A is

01492 =+λ−λ=−λ AI .

The eigenvalues and eigenvectors are found to be

−==λ

4

12 11 e, and

==λ

1

17 22 e, .

Let ( )

−==

14

1121 eeQ . Then Q diagonalizes A with

==−

70

021 DAQQ .

Now let QYX = , so 'QYX =′ .

From the equation ( )tFAXX +=′ , we have

( )

( )

( )tFQDY

tFQAQYQY

tFAQYQY

1

11'

'

−+=

−+−=

+=

Let

( ) ( )

+

−=

+−

−=

−=

27

28

5

2

410

6

14

11

5

1

1

t

t

t

t

tFQtG

So g1(t) = 5

2(8t −2) , g2 (t) =

5

2( 7t + 2).

90

Solving

( )( )tgy'y

tgy'y

222

111

7

2

+=

+=

by the integrating factor method, we get

y1 = C1e2 t

−5

2 (4t +1 ) and

y2 = C2 e7 t

− 5

2t −

35

6 .

To get x1 and x2, we use X = QY .

To get X, use

X=

2

1

x

x

+−

+=

−=

=

21

21

2

1

4

14

11

yy

yy

y

y

QY

So, the general solution is

( )

( )7

1064

7

42

7

2

2

12

7

2

2

11

+++−=

−−+=

teCeCtx

teCeCtx

tt

tt

91

Example:

Solve the system

( )( ) 3032

20344

2

2

212

1

2

211

=−+−+=

=−−+=

x;ttxxx

x;txxx

'

'

Solution:

Let

=

11

41A , ( )

−+−

−−=322

324

tt

ttF and

=

3

2C .

The eigenvalues and eigenvectors are found to be

==λ

1

23 11 e, and

−=−=λ

1

21 22 e, .

Let ( )

−==

11

2221 eeQ . Then

−=−

21

41

21

41

1Q and

−==−

10

031 DAQQ .

Let QYX = . Then 'QYX =′ .

From the equation ( )tFAXX +=′ , ( ) CX =0

QY’=AQY + F(t) ,

Y’=Q−1

AQY +Q−1

F(t)

( )tGDY'Y += ,

where

G ( ) ( )

−+

−+−== −

4

3

2

14

9

2

3

2

2

1

tt

tt

tFQt and ( )

=−=

1

210 CQY .

Therefore,

( )

( ) 104

3

2

1

204

9

2

33

22

22

12

11

=−++−=

=−+−+=

y;tty'y

y;tty'y

Solving these equations by integrating factor method, we have

92

( )

( )4

3

2

1

4

7

4

3

2

1

4

5

2

2

23

1

−+=

++=

−tety

tety

t

t

Hence,

( )

+

−=

=

21

212

yy

yy

QYX

So, the solution is

( ) ( )

( ) tt

tt

eetyytx

eeyytx

−

−

++=+=

−+=−=

4

7

4

5

2

7

2

532

32

212

3

211

Example:

Find a solution to the system

4912

21

+−=′

+−=′

xx

tsinxx

subject to the initial conditions ( ) 001

=x , and ( ) 102

=x .

Solution:

We have

+

−

−=

′

′

409

10

2

1

2

1tsin

x

x

x

x, with

( )( )

=

=

1

0

0

0

2

1

x

xc .

Eigenvalues are 3321

−=λ=λ , , with eigenvectors

=

−=

3

1

3

121

e,e .

Therefore

=

−=

−−

61

21

61

21

1

33

11Q,Q , and as before , we obtain

( ) ( )10

9

6060

61

9

4

10180180

61 33

2

33

1

tsineetx,

tcoseetx

tttt

+−=+−−−

=∴−−

93

Appendix:

A few remarks on the Gauss-Jordan elimination method (by using

elementary row operations)

1. Suppose in solving a system of linear equations, we reduced the

augumented coefficient matrix to its reduced row echelon form as

follows:

710000

503100

304021

Then x2 ,x4 are the free variables and x1, x3, x5 are the basic variables.

(They correspond to the column with the SPECIAL 1). Thus we set

x2 = s

x4 = t

and then solve for x1, x3, x5 in terms of x2, x4 as follows :

x1 =3-2s-4t

x2 =s

x3 =5-3t

x4 =t

x5 =7

or

=

5

4

3

2

1

x

x

x

x

x

X =

7

0

5

0

3

+s

−

0

0

0

1

2

+t

−

−

0

1

3

0

4

=

2. Suppose the reduced row echelon form of the augumented cofficient

matrix of a system of linear equations is as follows:

00000

90100

70001

Then

x1= 7

x3 =9

94

There is no restriction on x2 ,x4 , so they take any values, that is , they are

free.So ,we put

x2 = s

x4 = t

We can then write the solutions as follows:

+

+

=

=

1

0

0

0

0

0

1

0

0

9

0

7

4

3

2

1

ts

x

x

x

x

X ,

where s and t are parameters.

~END~