Unit2 Matriks MATH2(D) Ikbn (Student Version)

7/29/2019 Unit2 Matriks MATH2(D) Ikbn (Student Version)

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UNIT 2

MATRICES

2.1 Introduction

Matrices and Linear Equations made their appearance long time ago. The word

matrix was first introduced by Sylvester(1850). Many of the symbols and

operations that we used today were formalized by Cayley(1841). In the present

day, we use matrices to solve many complex problems of very great importance in

Medical Diagnostic Systems, Public Transport System, Communication Network

and others. In this topic students will be exposed to the operations of matrices,

determine the inverse matrices and hence solve the system of linear equation by

using the matrices.

Objectives

At the end of the topic, students should be able to:

State the size of matrices according to the number of rows and columns.

Perform the algebraic operation on matrices for 2 and 3 rows matrices.

Recognize matrices according to their types .

Determine the inverse matrix by determinant and adjoin method for 3 by 3

matrices.

Solve the system of linear equation using the inverse matrix method and the

Cramers rule for 3 by 3 matrices.

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2.2 Definition of Matrices

2.2.1 Size and Symbol

Matrices have columns and rows.

Let

=43

21A

A symbol of a matrix can be written as :

( ) or [ ]

2.2.2 Element of Matrices

Each number in a matrix is called the element of the matrix.

If

=43

21A ,

so numbers like 1, 2, 3, and 4 are the elements of matrix A.

2.2.3 Size of a Matrix

The size of a matrix is determined by the number of its rows and columns.

If a matrix has m rows and n columns, so the size of the matrix is (m by n) or can

also be written as ( nm ).

If

=

112

231B .

The size of matrix B is (2 3) or we can say B is a (2 by 3) matrix.

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row 1

row 2

Column 1 Column 2


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Example 2.1

Given matrix

=

223

124

312

A .

(a) State the size of matrixA.

(b) Determine the elements of 12A and 33A .

2.3 Operation and Types of Matrices

2.3.1 Addition of Matrices

Addition of matrices can be done if the matrices involve are of equal size.

Example 2.2

If

=

=

21

22

32

41BA .

size : 2 by 2 size : 2 by 2

It means matrixA and matrix B can be added. How to do it?

Add the elements of the same row and column.

Solution:

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3.1.2 Subtraction of Matrices

The same condition applies to the subtraction of matrices.

Subtract the same elements of rows and columns.

Example 2.3

=

=

21

22,

32

41BA

Solution

2.3.3 Multiplication of a Matrix with a Constant

Example 2.4

If given

=13

21C , determine 2C.

Solution:

2.3.4 Multiplication of Matrices

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Two matrices can be multiplied if the number of column for the first matrices and

the number of rows for the second matrices are equal.

If A = m byp matrix and B =p by n matrix, canA be multiplied with B?

Let us check!

The number of columns ofA =p and the number of rows of B =p.

EQUAL so A and B can be multiplied.

Can you guess the size of the product?

matrix)(matrix)(matrix)( pmnppm =

So the product ofAB has m rows andp columns.

Let

=

232221

131211

aaa

aaaA and

=

3231

2221

1211

bb

bb

bb

B .

( 2 by 3) (3 by 2)

Can we get the product ofAB?

Let us check the number of columns of matrix A and the number of rows of

matrix B.

The size of matrixA = 2 by 3 so the number of columns = 3.

The size of matrix B = 3 by 2 so the number of rows = 3.

EQUAL!We can get the product. Can you guess the size of the product?

Of course, it is a 2 by 2 matrix.

How to get the product ofAB?

++++++++=

=

322322221221322321221121

321322121211311321121111

3231

2221

1211

232221

131211

babababababa

babababababa

bb

bb

bb

aaa

aaaAB

Note : Make the first row of matrix A has been multiplied by every column

of matrix B. Then, only you move to the second row of matrix A.

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Example 2.5

Given

=

=

3

2and

13

21BA . FindAB.

Solution:

Example 2.6

Given

=

=

4251and

2301 BA . FindAB.

Solution:

Check whetherA and B can be multiplied. How?

Size ofA = 2 by 2, number of columns = 2.

Size of B = 2 by 2, number of rows = 2.

EQUAL, it meansA and B can be multiplied.

Can you guess the size of the product,AB?

Of course, a 2 by 2 matrix.

Example 2.7

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Given

=

=

112

031

401

and113

021BA . FindAB.

Solution:

Characteristics of matrix multiplication

(i) (mAB) = m(AB)

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(ii) A(BC) = (AB)C

(iii) (A + B)C=AC+ BC

(iv) C(A + B) = CA + CB

(v)TTT

ABAB =)(

Note : T is the symbol of a transpose matrix.

If

=

=

42

31then

43

21TAA .

If

=

=

963852

741

then987654

321T

BB .

Can you see what happened here?

The process of transposition changes the first row to be the first column, the

second row to be the second column and so on.

2.3.5 Types of Matrices

Here are a couple of examples of different types of matrices:

A. Symmetric

653

502

321

12

21.

B. Diagonal

600

040

001

40

01.

C. Upper Triangular

600

570

321

10

21.

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D. Lower Triangular

353

072

001

12

01.

E. Zero

000

000

000

00

00.

F. Identity

100

010

001

10

01.

Practice 2.1

1. Solve the equation of the following matrices.

=

+

54

31

52

21

y

x.

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2. Given

=

=

=

01

10

21

and112

101,

421

112CBA .

Calculate A + B and B + C.

3. Given

=

=

=

10

01and

30

12,13

21CBA .

Calculate

(a) 3A .

(b) 2B + 3C.

(c) BC.

(d) (BC) T .

2.4 Determinants, Minor, Adjoin, and Inverse Matrix

2.4.1 Determinants

Determinants play an important role in finding the inverse of a matrix and also in

solving system of linear equations. In the following we assume we have a square

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matrix (m rows = m columns). The determinant of a matrix A will be donated by

det(A) or A .

Determinant of a 3 by 3 matrix.

Assume

=

333231

232221

131211

aaa

aaa

aaa

A .

det(A) = A

3231

2221

13

3331

2321

12

3332

2322

11

333231

232221

131211

aa

aaa

aa

aaa

aa

aaa

aaa

aaa

aaa

+=

=

Example 2.8

Given

=

543

332

111

A . Find det(A).

2.4.2 Minor and Cofactor

A. Minor for a 3 by 3 matrix.

Assume

=

333231

232221

131211

aaa

aaa

aaa

A .

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Minor for3332

2322

11 isaa

aaa .

Minor for3332

2321

12 isaa

aaa .

Minor for3231

2221

13 isaa

aaa .

By now, you can figure out how to determine the minor of an element. Minor is

formed by excluding the row and column of the element involved.

Example 2.9

Given

=

543332

111

A . Determine the minor for the first rows elements.

B. Cofactor of a 3 by 3 matrix

Cofactor for elements ija is defined as :

ij

ji

ijA +

= )1( . ij is the minor for the element of row i and columnj.

Example 2.10

Given

=

543

332

111

A .

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Cofactor matrix of

333231

232221

131211

AAA

AAA

AAA

A = .

11)1(1.)1(

11)1(1)1(

00)1(0)1(

11.)1(1)1(

22)1(2)1(

11)1(1)1(

1)1()1()1()1(

11)1(1)1(

33)1(3)1(

63333

523

32

413

31

532

23

422

22

312

21

431

13

321

12

211

11

===

===

===

===

===

===

===

===

===

+

+

+

+

+

+

+

+

+

A

A

A

A

A

A

A

A

A

So the cofactor matrix of A is defined as :

Cofactor

=

110

121

113

A .

2.4.3 Adjoin Matrix

Adjoin matrix is defined as the transpose of the cofactor matrix.

T

ijAA ][)(adj =

Example 2.11

=543

332

111

A and cofactor

=110

121

113

A , so adjoin

=

111

121

013

A .

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2.4.4 Inverse Matrix

Assuming we have a square matrix A, which is non-singular ( det(A) 0 ), then

there exists matrix 1A which is called the inverse of A, such that this property

holds :

IAAAA == 11 whereIis the identity matrix.

The inverse of any square matrix, size of n by n is given by :

)det(

)(1

A

AadjA = .

Example 2.12

Given

=

543

332

111

A . We have found earlier that adjoin

=

111

121

013

A .

Det(A) = 1 .

So)det(

)(1

A

AadjA = =

1111

121

013

=

111

121

013

.

If we want to make sure the inverse matrix is indeed the inverse of matrix A, then

we have to multiply matrixA with the inverse matrix, 1A .

By definition, IAAAA == 11 whereIis the identity matrix.

So,

=

543

332

111

A and

=

111

121

0131A .

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=

111

121

013

543

332

1111AA

=

++++

++

540583549330362536

110121113

=

100

010

001

.

The product is an identity matrix so the inverse matrix that we got is correct.

Practice 2.2

1. Given

=

341

431

321

A . Find the 1A .

2. Given

=

113

232

321

A .

(a) By following the four steps, find the 1A .

(b) Show that the inverse matrix in (a) is correct.

Solution

2.5 The System of Linear Equations

Solving the system of linear equations by using matrices.

A system of linear equations is a set of equations with n equations and n

unknowns, is of the form of

........

...

...

.......

.......

2211

22222121

11212111

nnnnnn

nn

nn

bxaxaxa

bxaxaxa

bxaxaxa

=+++

=+++

=+++

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The unknowns are denoted by nxxx .......,,, 21 and the coefficient (as and bs)

are assumed to be given. In matrix form the system of equations above can be

written as :

=

nnnnnn

n

n

b

b

b

x

x

x

aaa

aaa

aaa

::

........

::::

........

........

2

1

2

1

21

22221

11211

A X B

A simplified way of writing, AX= B.

Example 2.13

Express the system of linear equations given below in matrix form.

(a).53

32

==+

yx

yx(b)

.323

223

12

=++

=+

=+

zyx

zyx

zyx

Solution:

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2.5.1 Inverse Matrix MethodThe inverse matrix method uses the inverse of a matrix to help solve a system of

linear equations, such like the above,AX= B.

Multiply both sides with 1A .

BAXAA

BAAXA

BAX

)()(

)()(

11

11

=

=

=

So it means BAX 1=

By calculating the inverse of a matrix and multiplying this by matrix, B we can

find the solution to the system of equations directly.

Looking at three equations we might have that

wvzuytx

srzqypx

dczbyax

=++

=++

=++

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Written in matrix form would look like

=

w

s

d

z

y

x

vut

rqp

cba

.

By rearranging we would get that the solution would look like

=

w

s

d

vut

rqp

cba

z

y

x1

.

Example 2.14

Solve the given the system of linear equations,

.3543

1332

2

=++

=++

=++

zyx

zyx

zyx

Solution

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2.5.2 Cramers Rule

Cramers rule uses a method of determinants to solve system of equations.

Let the system of equations be written asAX= B.

=

3

2

1

333231

232221

131211

b

b

b

z

y

x

aaa

aaa

aaa

.

Here,

=

333231

232221

131211

aaa

aaa

aaa

A .

A

Ax

1= where

=

33323

23222

13121

1

aab

aab

aab

A . Note : column

31

21

11

a

a

a

is replaced by

3

2

1

b

b

b

.

A

Ay

2= where

=

33331

23221

13111

2

aba

aba

aba

A . Note : column

32

22

12

a

a

a

is replaced by

3

2

1

b

b

b

.

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A

Az

3= where

=

33231

22221

11211

3

baa

baa

baa

A . Note : column

33

23

13

a

a

a

is replaced by

3

2

1

b

b

b

.

Example 2.15

Solve the system of linear equations below by using the Cramers rule.

.32

42

132

=+

=+

=+

zyx

zyx

zyx

Solution:

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Practice 3.3

1. Solve the system of linear equations given by the inverse matrix method.

.343

234

133

=++

=++

=++

zyx

zyx

zyx

Solution

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2. Solve the given system by using the Cramers rule.

.223

12

12

=++

=+

=++

zyx

zy

zyx

Solution

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EXERCISE

1. Find the values of a, b, c and d if

(a)

=

1842

436

124

422

d

c

b

a.

(b)

=

+ b

c

bd

a

21

23

43

22 .

2. Given

=

=

=22

43and

11

21

12

,

12

04

13

CBA .

Calculate,

(a) A + 2B.

(b) BA.

(c) TBC)( .

3. Find the determinants of the followings.

(a)150

232

403

. (b)141

213

342

.

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4. Given

=

cossin

sincosB .

Calculate,

(a) 2B . (b)3B .

5. Given

=

=

rq

pBA

0

30

055

,

010

210

121

.

Calculate,

(a) AB. (b) BA.

(c) Hence, determine the values ofp, q and r ifAB = BA.

6. Find the inverse matrices of the given matrices below.

(a)

=

524

012

321

A . (b)

=

301

021

112

B .

(c)

=1125412

613

C . (d)

=112212

221

D .

7. Solve the system of linear equations given by using

(i) the inverse method

(ii) the Cramers rule

(a)11273

543

32

=+=+

=+

zyx

zyx

zyx

. (b)043

1

32

=+=+

=+

zyx

zx

zyx

.

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(c)133

32

2253

=+

=+

=+

zyx

zyx

zyx

. (d)1363

632

2

=++

=++

=++

zyx

zyx

zyx

.

Answers to exercise:

1. (a) a = -4, b = -4, c = 1, d = -0.5(b) a = -3, b = 14, c = -1.33, d = 0.5

2. (a)

34

46

17

(b) cannot multiply (c)

6810

578

3. (a) 1 (b) -5

4. (a)

2cos2sin

2sin2cos(b)

3cos3sin

3sin3cos

5. (a)

++

++

p

rpq

rpq

30

2230

2115

(b)

+

+

qrq

p

20

630

15155

(c)p = 8, q = 4, r = -1

6. (a)

568

6710

345

(b)

43.014.029.0

14.071.043.0

29.043.086.0

(c)

111

032

213

(d)

110

67.0167.0

67.0033.0

7. (a)x = 1,y = 2,z = 3(b)x = 3,y = 1,z = -2(c)x = 2,y = 6,z = 13(d)x = 1,y = -2,z = 3

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Activity

1. The manufacture of an automobile requires painting, drying, and polishing. The

Parit Motor Company produces three types of cars : the Bengkok, the Jambol, and

the Botak. Each Bengkok requires 8 hours for painting, 2 hours for drying, and 1

hour for polishing. A Jambol needs 10 hours for painting, 3 hours for drying, and

two hours for polishing. It takes 16 hours of painting, 5 hours of drying, and 3

hours of polishing to prepare a Botak. The company uses 240 hours for painting,

69 hours for drying, and 41 hours for polishing in a given month. As the General

Manager, you have to determine how many of each type of car are produced.

You have to submit the report to your Board of Directors three days from now.

2. Mahkota Sdn Bhd produces luggage of three types : economy, standard and

deluxe. The company produces 1000 pieces of luggage at a cost of RM 20, RM

25, and RM 30 for the economy, standard, and deluxe luggage respectively.

Mahkota Sdn Bhd has a budget of RM 20,700. Each economy luggage requires 6

hours of labor, each standard luggage requires 10 hours of labor, and each deluxe

model requires 20 hours of labor. Mahkota Sdn Bhd has a maximum of 6800

hours of labor available. However, Mahkota sells all the luggage, consumes the

entire budget, and uses all the available labor.

Considering facts given, as the General Manager of Mahkota Sdn Bhd, you have

to determine how many of each type of luggage should be produced. Please do itas fast as you can, because you have to report it in the next meeting which is two

days from now.

Answers

1. Bengkok = 10, Jambol = 8, Botak = 5

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2. Economy = 900, Standard = 60, Deluxe = 40

Unit2 Matriks MATH2(D) Ikbn (Student Version)

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