11130146 chapter 2_-_matrix

BBMP1103 – Chapter 2 - Richard Ng (2008) of 47

BBMP 1103BBMP 1103

Matematik PengurusanMatematik Pengurusan

Chapter:2 Chapter:2 –– MatrixMatrix

Prepared by

Richard NgRichard Ng

BBMP1103 – Chapter 2 - Richard Ng (2008) Page 2 of 47

1. Introduction to Matrix

Matrix consists of data arranged in the form of rows andcolumns

Examples:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

251

⎥⎦

⎤⎢⎣

⎡5142

⎥⎦

⎤⎢⎣

⎡10

21

53

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

327510321

[ ]31⎥⎦

⎤⎢⎣

⎡23

1x2 Matrix 2x1 Matrix 2x2 Matrix

3x1 Matrix 2x3 Matrix 3x3 Matrix


2. Elements of a 3 x 3 Matrix

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

333231

232221

131211

aaaaaaaaa

a11 is located at row one and column onea12 is located at row one and column twoa13 is located at row one and column threea21 is located at row two and column onea22 is located at row two and column twoa32 is located at row three and column twoaij is located at row i and column j


3. Classifications of Matrices

⎥⎦

⎤⎢⎣

⎡23

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

102

⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢

⎣

⎡

−0112

Column Matrix2x1

Column Matrix3x1

Column Matrix4x1

[ ]21 [ ]102 − [ ]4013Row Matrix

1x2Row Matrix

1x3Row Matrix

1x4

Column Matrix (Column Vector): has only 1 column

Row Matrix (Row Vector): has only 1 row


⎥⎦

⎤⎢⎣

⎡0000

⎥⎦

⎤⎢⎣

⎡000000

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

000000000

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

000

000

⎥⎦

⎤⎢⎣

⎡4321

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

021843512

2x2 Matrix 3x3 Matrix

Zero Matrix (Null Matrix): all elements are zero

Square Matrix: number rows = number of columns


Main Diagonal

Elements located on the main diagonal in a squarematrix are elements which have the same row and column number. Examples:

In example (i): a11 = 1 and a22 = 4 hence the elements on the main diagonal are 1 and 4

In example (ii): a11 = 2, a22 = 4 dan a33 = 0 hence the elements on the main diagonal are 2, 4 dan 0

⎥⎦

⎤⎢⎣

⎡4321

(i) ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

021843512

(ii)⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

333231

232221

131211

aaaaaaaaa

(iii)


An Identity matrix is a square matrix where all the elements on the main diagonal = 1 and the rest = 0

⎥⎦

⎤⎢⎣

⎡1001

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

100010001

Identity Matrix

Example:


4. Matrix OperationsTranspose

Elements in a row elements in a columnElements in a column elements in a row

Example: 1

If ⎥⎦

⎤⎢⎣

⎡=

142301

A Then

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

142

301

AT

Example: 2

If⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

987654321

B Then⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

963852741

BT


Matrix Addition

Addition in a matrix involves elements at the same position and same type of matrix only

Example: 3

⎥⎦

⎤⎢⎣

⎡+⎥

⎦

⎤⎢⎣

⎡3102

4321

⎥⎦

⎤⎢⎣

⎡=

7423

Example: 4

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

231

412

013

201

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

244

613


Example: 5

Example: 6

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡−

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

231

412

013

201

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−−−

=222

211

⎥⎦

⎤⎢⎣

⎡−⎥

⎦

⎤⎢⎣

⎡3102

4321

⎥⎦

⎤⎢⎣

⎡−=

1221

Matrix Subtraction

Subtraction in a matrix involves elements at the same position and same type of matrix only


Scalar multiplication – when a matrix is multiplied by a scalar, every element must be multiplied with that scalar

Example: 7

⎥⎦

⎤⎢⎣

⎡=

4321

AIf Then ⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡=

8642

4321

22A

Example: 8

If⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

013412201

B Then⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

0391236603

013412201

33B

Matrix Multiplication


Multiplication of Matrices

Tips:

Matrix2 x 1

x Matrix1 x 2

Matrix2 x 2

Matrix2 x 3

x Matrix3 x 3

Matrix2 x 3

Multiplication of Matrices is only possible when the number of rows in the 1st matrix = number of columns in the 2nd

Matrix


⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡++++

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡10352

46031402

1021

4312

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡++++

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡9504

90320004

3102

3102

3102 2

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡++

=⎥⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡85

4432

12

4231

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

++++++++++++++++++

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

47532511412

004106104030200230092103192

011303232

102210131

Example: 9

Example: 10

Example: 11

Example: 12


a) For a 2 x 2 Matrix

⎥⎦

⎤⎢⎣

⎡3412

5. Determinant

Is defined only for square matricesIs used to find the invertible matrix

Finding determinant using cross multiplication

If A =

Hence the determinant of A = |A| = (2)(3) – (4)(1)

= 6 – 4= 2


b) For a 3 x 3 Matrix

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

024130312

402

231

If B =

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

024130312

Hence |B| =

= [(2)(3)(0)+(1)(1)(4)+(3)(0)(2)]–[(4)(3)(3)+(2)(1)(2)+(0)(0)(1)]

= [0 + 4 + 0]–[36 + 4 + 0]

= -36


If A = ⎥⎦

⎤⎢⎣

⎡ −4210

and B = ⎥⎦

⎤⎢⎣

⎡1021

Find:

a) |A| b) |B| c) |AB| d) |BA|

e) |A||B| f) |B||A|

Example: 13

Answer:

Hence |A|⎥⎦

⎤⎢⎣

⎡ −4210

a) A = = (0)(4) – (2)(-1)

= 0 – (-2)

= 2


b) B = ⎥⎦

⎤⎢⎣

⎡1021

c) AB = ⎥⎦

⎤⎢⎣

⎡ −4210

⎥⎦

⎤⎢⎣

⎡1021

= ⎥⎦

⎤⎢⎣

⎡++−++4402

)1(000

Hence |AB|

= ⎥⎦

⎤⎢⎣

⎡ −8210

Hence |B|= (1)(1) – (0)(2)

= 1 – 0

= 1

= ⎥⎦

⎤⎢⎣

⎡ −8210

= (0)(8) – (2)(-1)

= 0 – (-2) = 2


d) BA = ⎥⎦

⎤⎢⎣

⎡1021

⎥⎦

⎤⎢⎣

⎡ −4210

= ⎥⎦

⎤⎢⎣

⎡+++−+40208140

= ⎥⎦

⎤⎢⎣

⎡4274

= 2

= (4)(4) – (2)(7)

Hence |BA|

= 16 – 14

⎥⎦

⎤⎢⎣

⎡4274

=


e) From (a) dan (b):

|A| = 2 |B| = 1

Hence |A||B| = (2)(1) = 2

f) From (a) dan (b):

|A| = 2 |B| = 1

Hence |B||A| = (1)(2) = 2


Given P = ⎥⎦

⎤⎢⎣

⎡4321

Step: 1 – Find the matrix minor of P:

Minor of P:

⎥⎦

⎤⎢⎣

⎡=

4321

11a 4= ⎥⎦

⎤⎢⎣

⎡=

4321

12a 3=

⎥⎦

⎤⎢⎣

⎡=

4321

21a 2= ⎥⎦

⎤⎢⎣

⎡=

4321

22a 1=

⎥⎦

⎤⎢⎣

⎡=

2221

1211

aaaa

⎥⎦

⎤⎢⎣

⎡=

1234

Example: 14

Finding determinant using Cofactor Method


Step: 2 – Find the cofactor of P:

Element in a cofactor = element in minor x (-1)(i+j)

Hence cofactor of P ⎥⎦

⎤⎢⎣

⎡

−×−×−×−×

=++

++

)22()12(

)21()11(

)1(1)1(2)1(3)1(4

⎥⎦

⎤⎢⎣

⎡

−×−×−×−×

=)4()3(

)3()2(

)1(1)1(2)1(3)1(4

⎥⎦

⎤⎢⎣

⎡−

−=

1234


Step: 3 – Find the determinant:

Determinant = Product of elements in the original matrix and elements of the cofactor of a row

Original matrix P = ⎥⎦

⎤⎢⎣

⎡4321

Cofactor of P ⎥⎦

⎤⎢⎣

⎡−

−=

1234

Choose first row to find the determinant:

Determinant = (1)(4) + (2)(-3)

= 4 + (-6)

= -2⎥⎦

⎤⎢⎣

⎡4321

⎥⎦

⎤⎢⎣

⎡−

−1234


Example: 15

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

024130312

Step: 1 – Find the elements in the 2nd row of Minor Q:

Given that Q =

Using the second row to find the determinant, then:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

024130312

21a = (1)(0) – (2)(3) = 0 – 6 = - 6


= (2)(0) – (4)(3) = 0 – 12 = - 12⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

024130312

22a

= (2)(2) – (4)(1) = 4 – 4 = 0⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

024130312

23a

Step: 2 – Find elements in the 2nd row of cofactor Q:

6)1)(6()1)(6( )12(21 =−−=−−= +a

12)1)(12()1)(12( )22(22 −=−=−−= +a

0)1)(0()1)(0( )32(23 =−=−= +a


Step: 3 – Find the determinant:

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

024130312

6 -12 0

Determinant = (0)(6) + (3)(-12) + (1)(0)

= 0 + (-36) + (0)

= -36

Determinant = Product of elements in the original matrix and elements of the cofactor of a row


6. Inverse Matrix

⎥⎦

⎤⎢⎣

⎡3412

Let A =

Step: 1 – Find the determinant

= (2)(3) – (4)(1)

= 6 – 4

= 2

Finding the Inverse Matrix of the type 2x2

⎥⎦

⎤⎢⎣

⎡3412

|A| =

Example: 16


Step: 3 – Find the inverse matrix of A = A-1

A-1 = ⎥⎦

⎤⎢⎣

⎡−

−2413

x|A|

1

= ⎥⎦

⎤⎢⎣

⎡−

−2413

x21

⎥⎥

⎦

⎤

⎢⎢

⎣

⎡

−

−

1221

23

=

Step: 2 – Find the adjoint of A

⎥⎦

⎤⎢⎣

⎡3412

⎥⎦

⎤⎢⎣

⎡−

−2413 Elements of Main Diagonal

⇒Change positionElements Second Diagonal⇒Change sign

[ ]AintAdjox|A|

1A 1 =−


Example: 17

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

− 332020141

Let A =

Step: 1 – Find the determinant of A

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

− 332020141

|A| =324

201

−= [6+0+0] – [4+0+0]

= 6 – 4= 2

Finding the Inverse Matrix of the type 3x3


Step: 2 – Find the Minor for matrix A

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

− 332020141

a11 = = (2)(3) – (-3)(0) = 6 – 0 = 6

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

− 332020141

a12 = = (0)(3) – (2)(0) = 0 – 0 = 0

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

− 332020141

a13 = = (0)(-3) – (2)(2) = 0 – 4 = -4


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

− 332020141

a21 = = (4)(3) – (-3)(1) = 12 + 3 = 15

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

− 332020141

a22 = = (1)(3) – (2)(1) = 3 – 2 = 1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

− 332020141

a23 = = (1)(-3) – (2)(4) = -3 – 8 = -11


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

− 332020141

a31 = = (4)(0) – (2)(1) = 0 - 2 = -2

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

− 332020141

a32 = = (1)(0) – (0)(1) = 0 – 0 = 0

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

− 332020141

a33 = = (1)(2) – (0)(4) = 2 – 0 = 2


Matrix Minor A =⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

20211115406

Matrix cofactor A =⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

20211115406

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

+−+−+−+−+

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−

20211115406

=

Step: 3 – Find the cofactor of A


⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−−

21140102156


Adjoint A = [Cofactor A]T

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−

20211115406

A-1 = = =

Step: 5 – Find A-1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−−

21140102156

|A|1

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−−

21140102156

21

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−−

12112

0210

1215

3


7. Solving Simultaneous Equations

Example:18Find the value of x and y for the following equations:

x + y = 22x –y = 1

Step: 1 – Change the equations in the form AX = Bx + y = 22x –y = 1 ⎥

⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡− 1

21211

yx

Let A = ⎥⎦

⎤⎢⎣

⎡−1211

Hence |A| = (-1) – (2) = -3


Inverse Matrix Method


Step: 3 – Find the adjoint matrix A

⎥⎦

⎤⎢⎣

⎡−1211

⎥⎦

⎤⎢⎣

⎡−

−−1211

Step: 4 – Find the inverse matrix of A

A-1 = ⎥⎦

⎤⎢⎣

⎡−

−−1211

||1A = ⎥

⎦

⎤⎢⎣

⎡−

−−12111

3- =⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−31

32

31

31

Step: 5 – Multiply B with the inverse matrix

⎥⎦

⎤⎢⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−=⎥

⎦

⎤⎢⎣

⎡12

31

32

31

31

yx


Hence ⎥⎦

⎤⎢⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

+=⎥

⎦

⎤⎢⎣

⎡11

3333

31

34

31

32

yx

Thus x =1 and y = 1

x + y = 2 ……(a)2x –y = 1 ……(b)

(a) + (b) => 3x = 3Then x = 1

Substitute x = 1 into (a) Then 1 + y = 2and y = 1

Check:


x + 4y + z = 12y = 3

2x – 3y + 3z = 0

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

− 332020141

Step: 1 – Change the equations in the form AX = B

x + 4y + z = 12y = 3

2x – 3y + 3z = 0 ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

zyx

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

031


Dari ms 28: |A| = 2

Example: 19

Solve the following equations:


Step: 3 – Find the minor of A

From slide 32: Minor A = ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−−

20211115406

Step: 4 – Find the cofactor of A

From slide 32: Cofactor A = ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−

20211115406


From slide 33: Adjoint A =⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−

−−

21140102156


Step: 6 – Find the inverse matrix A-1

From slide 33: Inverse A =

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−−

12112

0210

12

153

Step: 7 – Solve the equations

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

031

AMatrixInverse

zyx


⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡−

=

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

++−

++

+⎟⎠⎞

⎜⎝⎛−+

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥⎥⎥⎥

⎦

⎤

⎢⎢⎢⎢⎢⎢

⎣

⎡

−

−−

=⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

2114211

2119

02332

0230

02453

031

12112

0210

12

153

zyx

Hence the value of x = 2119−

y = 211

and z =2114


Solve the equations:

5202

=−=+

yxyx

Step: 1 – Change the equation in the form AX = B

⎥⎦

⎤⎢⎣

⎡=⎥

⎦

⎤⎢⎣

⎡⎥⎦

⎤⎢⎣

⎡− 5

01221

yx

Step: 2 – Find |A|, |A1| dan |A2|

Let A = ⎥⎦

⎤⎢⎣

⎡−1221

|A| = (-1) – (4) = -5

Example: 20

Cramer’s Rule Method


Let A1 = ⎥⎦

⎤⎢⎣

⎡−1520

|A1| = (0) – (10) = -10

Let A2 = ⎥⎦

⎤⎢⎣

⎡5201

|A2| = (5) – (0) = 5

Step: 3 – Find the value of x and y

x =|A||A| 1 =

5-10- = 2

y =|A||A| 2 = = -1

5-5


Solve the equations:

02242332

=−=−+=+−

zxzyxzyx

Step: 1 – Change the equations in the form AX = B

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡=

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−

043

202121312

zyx

Example: 21


Step: 2 – Find |A|, |A1|, |A2| dan |A3|

⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−

202121312

Let A =

Hence |A| = [-8+2+0] – [12+0+2] = -6 –14 = -20

212

021−

Let A1 = ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−

−

200124313

043

021−

Hence |A1| = [-12+0+0] – [0+0+8] = -12 – 8 = -20


Let A2 = ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡

−−202141332

212

043

Hence |A2| = [-16+(-6)+0] – [24+0+(-6)] = -22 – 18 = -40

Let A3 = ⎥⎥⎥

⎦

⎤

⎢⎢⎢

⎣

⎡ −

002421312

212

021−

Hence |A3| = [0+(-8)+0] – [12+0+0] = -8 – 12 = -20


Step: 3 – Find the value of x, y and z

x =|A||A| 1 = = 1

2020

−−

y = = = 2|A||A| 2

2040

−−

z = = = 1|A||A| 3

2020

−−


End ofChapter

2

11130146 chapter 2_-_matrix

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