3. Math for XII Science Trial Editionx

8/3/2019 3. Math for XII Science Trial Editionx

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Mathematics for Science XII Grade 1

XII GRADE

Syaiful Yazan

SMA NEGERI 1 MARTAPURAJl. A. Yani 59 A 0511-4721272 Martapura

School Web: www.sman1mtp.sch.idPrivate Web: www.msyna.comE-mail: [email protected]


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Mathematics for ScienceXII GRADE1st Edition

Syaiful Yazan

SMA Negeri 1 MartapuraMartapura, Kalimantan SelatanIndonesiaSchool Phone 0511-4721272School Web: www.sman1mtp.sch.idPrivate Web: www.msyna.comE-mail: [email protected]

First PublishedPrinted by Mitra PressMartapura, Kalimantan SelatanIndonesia, 2010

Copyright 2010 by The Author.All rights reserved. No part of this book may bereproduced in any form, electronic or mechanical,including photocopy, recording, or any information

storage and retrieval system, without permission fromthe Author.

Covert Art: msyna


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PREFACEby Drs. H. Busra

(The Head Master of SMA Negeri 1 Martapura)

All gratefull for Allah, on His willing this international adabted learning material of

mathematics for science XI grade can be completed fluently on time. This learningmaterial is used in mathematics learning process for science XI grade at SMANegeri 1 Martapura (State Senior Hight School 1 Martapura), Banjar regency, as aPionering Senior High School for International Level (Rintisan SMA BertarafInternasional, RSMABI).

This learning material is expected as one of main references and can enhance theknowldege for students and teachers as the effort to increase the services andquality of mathematics instruction. The students can learn this learning material bytheirselves, whether inside or outside of schools area.

By learning this material, it is also expected can motivate the effort of masteringEnglish better, especially related to the adapted international material in interactivelearning process.

We thank to the commitees and the participants of the workshop of making theadapted international learning material. Of course, this mathematics learningmaterial is need to be repaired and revised at the next time accordance with thedemand of curriculums development. Therefore, suggestions to improve thislearning material is expected.

We thank also to the Government, in this case Direktorat Pembinaan SMA,Direktorat Jenderal Manajemen Pendidikan Dasar dan Menengah, considering thefunding of making this learning material is from APBN through the blockgrant ofRSMABI for the financial year 2009/2010. Hopely, this work is a part of our deedsthat is useful for the development of our education quality and becames charity inAllah SWTs worship. Amen.

Martapura, Januari, 2011The Head Master of SMA Negeri 1 Martapura

Drs. H. BusraPembinaNIP 19571021 198511 1 001


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INTRODUCTION

All gratefull for Allah, on His willing this mathematics learning material can bepresented. Shalawat and peacefull always address to my belove ProhetMuhammad SAW.

This learning material consists of:1. Competency standard and basic competency as in Curriculum of Education

Unit Level, Kurikulum Tingkat Satuan Pendidikan (KTSP).2. Benefits and competencies that hoped can be reached by student after learn

each chapter.3. Material explanation, involves: 1) Theory that generally begin with the reality

or presenting the relevan illustration, 2) examples and 3) drill and test

competency try out.4. In some parts serves the computer program application to facilities the

accounting or contributes the related mathematical jobs. Its aimed to give thewider perspectives for student in managing mathematical jobs, especially formathematics application.

5. Some enrichments are also given to strenghten and axpand the studentsknowledge.

This learning material especially is served for student in XI Grade Scienceprogram and also can be used as a reference for teachers. For anyone that wantto get this learning material can download it from the E-Learning of msynacom in

website: msyna.com/site.

The author thank to Mr. Drs. H. Busra as the head master of SMA Negeri 1Martapura, the mathematics teachers of SMA Negeri 1 Martapura, my colleaguesin Mathematics Teacher Conference (MGMP Matematika), and also to all mystudents of SMA Negeri 1 Matapura for their loyalty in following my mathematicsinstruction. Those all give me stimulus to write this learning material. My thankalso to all people that have contributes me, especially my family.

I hope this simple learning material can be benefits for all of us. I also hope thesuggestions for improving it.

Martapura, Januari, 2011Author,

Syaiful YazanPembinaNIP 19691209 199412 1 003


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Contents

Preface .............................................................................................. ii

Introduction ....................................................................................... iii

Contents ........................................................................................... iv

Chapter 1 Integral .............................................................................. 6

Chapter 2 Linear Program .............................................................. 20

Chapter 3 Matrix .............................................................................. 30

Chapter 4 Vector ............................................................................. 55

Chapter 5 Transformation ................................................................ 72

Chapter 6 Sequence and Series ..................................................... 94

Chapter 7 Exponent and Logarithm ............................................... 113

References .................................................................................... 114


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INTEGRALCompetency Standard1. Use the concept of integral in solving

problems.

Basic Competency

1.1 Understanding the concept of the certain anduncertain integral.

1.2 Calculating the certain and uncertain integral of simple algebra andtrigonometry functions.

1.3 Using integral to calculate the area of a region under a curve andthe volumn of twisted objects.

Lesson Benefit

Students would have more understanding and methods ofmeasuring the area and volumn of geometry object by applying theintegral.

Will be Reached Competencies

Students are capable in:

1. Explaining the concept of the certain and uncertain integral.2. Calculating the certain and uncertain integral of simple algebra and

trigonometry functions.3. Using integral to calculate the area of a region under a curve and

the volumn of twisted objects.

Material

A. The Concept of IntegralIntegral is related hard to derivative. It is like someone at the first

time directly watch the result of the derivative of a function, then heasks what function is that has the derivative like it?

Illustration:If 2 3 then thederivative of to is

2 2

What function is that hasderivative to , 2 2? Thisproblem is stated as the integralof

2 2toward

, and

symbolized as

2 2 . The

answer is 2

Chapter 1


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Here, the answer state as 2 not directly as 2 3,becauce for any number will cause the derivative of 2 equalto 2 2. Because the value of is uncertained so the integral like2 2

is called as uncertained integral.

The relation between derivative and integral is shown in the Picture 1.

Here:If then:

its derivative,

its integral toward , Drill:Evaluate the following ones:1.

2

2. 3 5 3. 3 4 5 4. 12 6 5 5. 6. 4 1 2 7.

8.

2

9. 2 10. 1 11. 12. sin 13. 14. 2 sin 15. 2 s in c os

Picture 1.

2

2 2

2 2


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B. Integral Involving Chain rule

According to the chain rule,

1 . . then,

. 1 1 Example:

1. 2 22.2 2233 2. 22 34. 4 3 22355 3. .

Drill:1. Evaluate the following ones:

a. 2 4 2 b. 3 2 c.

2

3

3

6

6

d. . 2. Fill the blanks bellow in order the statements are true.a. 3 4 b. 3 3 c. 4 4 d. .

C. Certain Integral

1) Definition of Riemann Integral

Georg Friedrich

Bernhard Riemann(1826-1866)


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By intuitive way, it tries to count the area of the region under thecurve , above the axis and the vertical borders and by using some rectangles such shown in the above picture,where the width of all the rectangles are same,

equal to

.

Of course the area aproximetly equals to1 2 3 4 5 6 7 .

. The result is more acurate if the number of rectangles are big ortheir width, closed to zero, and its stated as

lim

1 .

In this case, also should be closed to .

2) Basic Theorem of Calculus

Proof:If . Then make thepartitions base on this asumtion, such:

Meanwhile, by using trick minus and add, then Base on the Average Value Theorem that applicated on at theinterval

, ,

For one choice of in the open interval . So,

If is continue at , and any integral of, then

lim

1 . Riemann has given the definition, that the certainintegral of from to as


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In the left side there is a constanta, while in the right side is the

form on Riemann integral for in interval , . If the limit for 0, then lim

3) Calculate the certain integralManagement of calculating a certain integral. If is theuncertain integral of then

|

Example:

a) 4 4 | 4 . 3 c 4.2c 1 2 8 4 Because 0. This calculation can be simplipied as

4 4| 4 . 3 4 . 2 1 2 8 4 Also,

b) 6 3| 3 . 4 3 . 1 4 8 3 4 5

Average Value Theorem

If continue in the interval , and can be found the derivativeat any point in , , then at least there is one in , whereis

or


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Drill:

1. 2.

3. 3 2 4 4. 4 5 5. 4 5

6. 7.

8. cos 9. cos 10. sin 3cos3

D. The Application of Integral1. The area of a plane

As has been mentioned previously in discussion of the certain

integral, one of the application of integral is calculating the areaof a plane such as:Case 1:

The shadow area is

Case 2:

The shadow area is

|| or

0

0


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Case 3:

The shadow area is

Case 4:

The shadow area is

Here is recomended:1. Sketch the picture adequately.2. Since the calculation of area by integral uses the

approacment of calculating the area of rectangles, in abovecases with the height or and the width , thenwhen or is negative, it must be taken the absolutevalue.

0

0


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Example:

Find the area of the shadow part below:

Answer:Find the interval of , the intersection of= 9 and =2 9;

9=

2 9

2 0

0 ; 2Then,

2 9 2 9 2 2

2 2 0 4

0

= 9=2 9

0


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Drill:Find the area of shadow parts below:

1. 4.

2. 5.

3. 6.

= 8

= 2

sin

0

cos

0 2

cossin

0 454

= 2

2 4

= 2


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2. The volumn in a spaceWhat is the volumn? A volumn indicates the measurement of thecontent of an space object, geometrically, if it is a prism then its

formulated as the product of the base area and the height of theobject.

The area, . Illustration:

Before

1.

After twisting

In Integral, finding the volumn of twisted plane

around the axis above by approachment ofcalculating the volumn of a cylinder (the basearea multiple by the height, . , here is theradius and is the height).So the volumn in the interval , is

0 0


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2.

Analog, but in this case the plane is twisted around the

axiz, then the volumn in the interval , is Example:Find the volumn that made of the twisted plane with borders are , axis, line 4 that is twisted around the axis, asbelow:

Answer:

12 4 12

412 012 163

00

4

0

12


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Drill:Find the volumn that made of the twisted plane with borders are:

1. , axis, line 4that is twisted around the

axis, as below:

3. s in and axis, thatis twisted around the

axis,

as below:

2. = 2, axiz, line 6 that is twisted around the axis, as below:4. and 6

that is twisted around the axis, as below:

s i n

0

4

=

2

0

0

6

0

2

6


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Competency Test Try Out

For number 1 until 6, find the area of shadow parts below:

1. 4.

2. 5.

3. 6.

=

12 4

sin

0

sin

0

cos

sin

0 2

= 10

1

= 2

2


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7. , axis, line 3that is twisted around the axis, as below:

9. 2 sin and axis, thatis twisted around the axis,as below:

8. = 2, axiz, line 6 that is twisted around the axis, as below: 10. and 8that is twisted around the axis, as below:

2 sin

0

12 3

0

= 6

0 0

2

8


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LINEAR PROGRAMMING

Competency Standard2. Solving the linear program problem.

Basic Competency

2.1 Solving inequality system of two variables2.2 Designing the mathematical model of linear program problem2.3 Solving the mathematical model of linear program problem and

the interpretation

Lesson Benefit

Student would be familiar to identify the factors of a problem, thequalifications and the destination functions, then able to solve theproblem.



1) Solving inequality of two variables by analyzing and geometricalapproachment.

2) Making the mathematical model of program linear problem.

3) Solving the mathematical model of linear program problem and theinterpretation

Material

A. What is the Linear Programming?Among the problems in real life such as to get the maximum orminimum value, need mathematical help in solving the suitableconditions to get the best result, and when the used mathematical

model involves the system of linear equation, usually in form oflinear inequality, this case is called as linear programming

In Wikipedia explains:Linear programming (LP) is a mathematical method for determining a way toachieve the best outcome (such as maximum profit or lowest cost) in a givenmathematical model for some list of requirements represented as linearequations.

Chapter 2


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B. Linear Function, Equation, Inequality and the Position of aPoint, a Region and the Graph of Function in the CartesianPlane

1. A linear Equation with two variable

Take 2 4 (the function or equation f) , if then 2 4. (the function or equation y) Futhermore it can bepresented as:2 x y 4 0 2 4 02 x y 4 2 x y 4The last two forms are acknowledged as an equation with twovariables, that is

and

.

How to draw the graph of 2 4 or 2 x y 4 in the CartesiansField?Answer:

There are many ways to draw it, but someone need to makeanalyzing it, one of them such as:

Analyzing,

When

0, it means the graph cuts

axis, then

2 . 0 4 4. The coordinat point is 0, 4. When 0, it means the graph cuts axis, then 0 2 4; 2. The coordinat point is 2,0.Using table:

If

0

0

2 4 2.0 4 0 2 4then 4 2The

Coordinat0,4 2,0

Intercept axis axis0, 4

2, 0


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2. Inequality with two variables

When 2x y 4, maybe:2x

y

4; 2x

y

4

;2x

y

4

; 2 x y 4, then these

forms are called as inequality with two variables.Find the region that fulfils:

a. 2x y 4b. 2x y 4c. 2x y 4d. 2x y 4

Answer:

The procedures:

First, make the graph of 2x y 4 on the Cartesians FieldThen, choose arbitary one point(recommended the easy one), forexample 0, 0 then subtitutes to 2x y,that is 2.0 0 0 4, then all pointswhich located at the same posisition as(0,0) when viewed from the line2x

y

4 are satisfied 2x

y

4,

meanwhile all points which located atthe opposite posisiton of (0,0) whenviewed from the line line 2x y 4 aresatisfied 2x y 4. For more detailcan be presented as bellow:

The shadow partindicates

2x y 4Sample point 2, 3;2.2 3 7 4It uses the dashed line.

0 2

4

2x y 4

0 2

4

2x y 4

,


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2x y 4Sample point 2, 3;2.2 3 7 4It uses the continues line.


2x y 4Sample point 0,0;2

.0

0

0

4

It uses the dashed line.


2x y 4Sample point 0,0;2.0 0 0 4It uses the continues line.

0 2

4

2x y 4

,

0 2

4

2x y 4

0 2

4

2x y 4


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The skill to state the region that satisfies an inequality as shownabove is so infortant in solving a linear programming problem.

C. Linear Programming Problem

There are two main parts, the first is the linear equation system asthe qualification that must be satisfied and the second is the targetfunction or the destination function, usually to find the maximum orminimum value of the function.

Example:The maximum value of 2 3 if must be satisfied 2 5; 2 4 ; 0 ; 0 is ...Answer:The linear inequality system as the qualifications is:2 5; 2 4 ; 0 ; 0Analyzing,a. Line 2 5 from 2 5 the line 2 5;

If

0

0

2 5 2 . 0 5 0 2 5then 5 52The Coordinat 0, 5 52 , 0

Intercept axis axisb. Line

2from

2 4 the line 2 4;

If 0 0 12 2 12 . 0 2 0 12 2then 2 4

The Coordinat 0, 2 4, 0Intercept axis axis


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c. The linear equation system: 2 5; 2, then

2 5 2

4 1 0 4 4 4 1 03 10 4; 3 6; 2; 2.2 5 1So the intersection of lines 2 5 and 2 is 2,1

d. Drawing the graph

e. The critical points are 0, 2, 2, 1, 2.5,0f. Check the destination function,, 2 3 , 0, 2 2, 1 2.5,0, 2 3 2.03.2 2.23.1 2. 2.5 3.06 7 5

The critical point 2, 1 gives the maximum result, that is 7.If the linear programming problem is in a problem solving type thenhere is recommended to use the procedures:1. Make the mathematical model of the problem, It should consists

of the linear inequality or equation system as the qualificationand the destination function inequality or equation system,decide the satisfication region, usually by using the shadow part.

2. Identify the critical points, some of them are the intersectionpoints of several lines.

3. Subtitutes the critical points to the destination function, it canuses table. Analyze and state the suitable point that gives thebest result, gives the maximum or the minimum value.

2 5

1

2 2

2,125

2.50 4


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Example:A fitness equipment dealer wants to buy 2 types of fitnessequipment. The cost for each regular type of the fitness equipmentis Rp. 1.600.000, 00 and for the lux type costs Rp. 2.000.000, 00.

The fitness equipment purchased at most 25 pieces and availablecapital Rp.48.000.000, 00. Income earned for each regular fitnessequipment is Rp. 600.000, - and the lux one is Rp.700.000, - . Howmany of each equipment to be purchased in order to obtain themaximum profit

Answer:

Analyzing:

If the number of:Regular fitness equipment = Lux fitness equipment = Make the table like,

EquipmentType

Reguler Lux Model

Purchasing

Number

2 5

Cost (Rp) 1.600.000 2.000.000 1.600.0002.000.00048.000.000 Profit 600.000 700.000 600.000700.000So the mathematical model is:

Description Model

A. Qualifications

Purchasing

2 5

Capital 1.6000002.00000048.000000Regular and luxpurchasing number

0 ; 0B. Destination

Profit , 600.000700.000


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Making the graphs:

25 25

If 0

0 2 5 0 2 5 0 2 5

then 2 5 2 5The Coordinat 0,25 25,0

Intercept axis axis

1.600.0002.000.00048.000000

1.600.000 2.000.000 48.000000Simplipied:16204804 5 1 2 0If 0 0

4 5 1 2 0

4 . 0 5 1 2 0

4 5 . 0 1 2 0

then 2 4 3 0The Coordinat 0,24 30,0Intercept axis axis

The intersection between 2 5 and4 5 1 2 0

From 2 5, then 2 5 , subtitute in 4 5 1 2 0, then4 5 2 5 1 2 04 1 2 5 5 1 2 05 ; 5 and 20


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From the graph there are found critical points 25,0, 0,24, 5,20,subtitute them to the destination function, profit function, 600.000700.000Use the table:

, , 600.000700.000 Result25,0 25,0 600.000. 25 700.000.0 15.000.0000,24 0,24 600.000.0700.000.24 16.800.0005,20 5,20 600.000. 5 700.000. 20 17.000.000The critical point 5,20 gives the maximum result, that is17.000.000

. So to get maximum profit, the dealer must buy

5

reguler type and

20lux type of the fitness equipments.

Competency Test Try Out1. The minimum result for 3 6 that satisfies 0 , 0 , 12 and 2 1 6 is ....2. The sum of , and that satisfies the linear equation system

2 3 4 3 , 2 3 2 ,and

3 2 2 1is ....

3. The maximum of , 6 8 with the handicaps 0 , 0 , 2 8 0 , 6 0 is ....

25

25

24

300

5,20


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4. At the graph below, the solution set of inequality system 2 8,6 7 4 2 , and 2 8 is ....

5. The shadow part below indicates the all set of , that satisfy ....

6. A passengger plane has 200 seats. Every main class passenggermay bring 60 kg, meanwhile the economic class may bring 40 kg.The plane can brings only 8000 kg. The cost for ticket for mainclass is Rp1.000.0000,00 and the economic class is Rp600.000,00.

The maximum profit would be gotten if the number of main classseat is ....

7. A parking place has area 1000 m2 only can be occupied by 80 busand cars. Every car need 8 m2 and bus 32 m2. The parking cost fora car is Rp1.000,00 and for the bus is Rp2.000,00. If the parkingplace is full, the parking maximum cost is ....

8. The purchase price of flour A type is Rp4.000 and sold for 5000 perkg, meanwhile for B type is Rp5.000 and sold for Rp.6.500 per kg.A seller has money Rp.500.000,00 and if his shop can contains 200

kg, then he can get maximum profit if buy ....

8

4

4

8

7

6

0

IIIIIIIV V

010

20

10 25


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MATRIXCompetency Standard

3. Using the concept of matrix, vector, andtransformation in solving problem

Basic Competency

3.1 Using the natures and the operation of matrix to show that asquare matrix is the inverse of other square matrix.

3.2 Determining the determinant and inverse of matrix 2 x 2.

3.3 Using determinant and inverse in solving linear equation systemof two variables.

3.4 Using the natures and algebra operation of vector in solvingproblems.

Lesson Benefit

This is a big expantion of numbers. Previously numbers are viewed asindividual that interacts with another numbers, but now, numbers aregrouped in matrix form and it interacts with numbers or other matrixes.

Students would be aware with the useful of matrix in solving problem,especially in solving the linear system equation.



1) Identifying the properties of a matrix.2) Doing operation of a matrix.3) Determining the determinant and inverse of matrix 2 x 2.

4) Using determinant and inverse in solving linear equation system oftwo variables.

5) Using the natures and algebra operation of vector in solvingproblems.

Chapter 3


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Material

A. Definition of MatrixIllustration:

Given some rectangles with their celss like:

The rectangles Fill with any numbers In matrix form

So a matrix can be supposed as a set of numbers that arrange in arectangle form, if it has number of row and number of column,then it is called as a matrix with ordo A range of numbers in a row could be supposed as a row vector,

meanwhile a range of numbers in a column could be supposed as acolumn vector

So a matrix can also defined as a set of some row and columnvectors.

Generally a matrix can be presented as

2x2

2x3

3x3

4x4

1 20 3

1 0 -23 2 4

2 3 12 0 44 -2 1

1 0 1 53 4 2 15 2 4 10 1 4 3

1 2

0 3

1 0 2

3 2 4

2 3 1

2 0 4

4 2 1

1 0 1 5

3 4 2 1

5 2 4 1

0 1 4 3


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The ordo of matrix A is , where is the number of row and is the number of column.

It shows that is the element of matrix at the 2nd

row and 3rt

column.

B. Types of Matrix1. Square matrix is a martix that has same number of row and

column. Example:

1 2

0 3

2 2

,

1 2 1

3 0 4

2 5 3

3 3

2. Symmetric matrix is a matrix that the element in the row and column is same with the element in row and column , here is . Example:

1

1

9

9

1

2

5

3

2

2

4

4

3

3

7

7

3. Diagonal matrix is a symmetric matrix that all elements exceptthe main diagonal are zero. Example,

1 0 0 0

0 2 0 0

0 0 5 0

0 0 0 3

4. Scalar matrix is a diagonal matrix that all elements in the maindiagonal are same. Example,

3 0 0

0 3 0

0 0 3


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5. Identity matrix is a scalar matrix that all elements of the maindiagonal are one and symbolized as capital letter . Example,

1 0

0 1

,

1 0 0

0 1 0

0 0 1

,

1 0 0 0

0 1 0 0

0 0 1 0

0 0 0 1

6. Sign matrix is a diagonal matrix that the elements of the maindiagonal are 1 or -1. Example,

1 0 0

0 1 0

0 0 1

7. Symmetric and oblique matrix is a symmetric matrix than the

element in the row and column is same with thenegative of element in row and column , here is

. Example:

3

3

1

1

4

9

2

2 7

9

0

0

4

0

7 0

8. Triangle matrix is a square matrix, but not a diagonal matrix,that all elements above or under main diagonal are zero.

Example,

1 2 4

0 4 5

0 0 3

,

2 0 0

2 3 0

7 5 4


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9. Vertical matrix is a matrix that the number of row is greaterthan the number of its column. Example,

2 4

4 2

6 3

10. One-Nol matrix is a matrix that the elements are 1 or 0.Example,

1 0 1

1 0 0

0 1 1

11. Nol matrix is a matrix that all elements are 0. Example,

0 0

0 0

0 0

12. J matrix is a matrix that all elements are 1. Example,1 1 1

1 1 1

1 1 1

C. Operation of MatrixLike numbers that can be operated such as addition, substraction,multiplication, matrixes can also be operated.

1. AdditionExample:2 31 4 4 32 4 2 4 3 31 2 4 4 6 63 8

3 3 1 2 3 1 3 2 4 5

1 2 7 4 1 7 2 4 8 6

0 4 3 5 0 3 4 5 3 9

+ + + = + + = + +


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2. Substraction

Example:

6 35 4 4 22 1 6 4 3 25 2 4 1 2 13 3

3 3 1 2 3 1 3 2 2 1

1 2 7 4 1 7 2 4 6 2

8 9 3 5 8 3 9 5 5 4

= =

The addition and substraction can be applied among thematrixes that has same ordo

and

Then

3. MultiplicationIllustration:Given a linear system equation2 3 83 2 7Some approachments are made to convert this case to a matrixmodelFirst model:

2 33 2 . 87

It hopes that:


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2 33 2 . 2. 33. 2 87But actually it more fits for

2 33 2 . 2. 33. 2 87as in addition, but what can be

hoped to get 87, its confused.Then lets try another model

Second model:2 33 2 . 87Briefly, state the rules for above expression as

2 33 2 . 2 3 3 2 87

Its more comfortable and simple rather than previous model,So Generally it definesi) Two matrix can be multipled if the number of column of the

first matrix is same with the number or row of the secondmatrix. and , then .

ii) Here if

and

Then,

.

. . . . . . . . . . . . . . . . . . . . . . . . . . .

.


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iii) The result element in the row and column is thecombination operation from the row of the first matrix andthe column of the second matrix. So, . . . .

Example:

1. 2 43 5 . 23 2.24.33.25.3 11212. 1 45 2 . 2 31 4 1.2 4.1 1.3 4.45.2 2.1 5.3 2.4 6 1912 233.

2 11 32 4 . 2 3 43 2 1

2.2 1.3 2.3 1.2 2.4 1.11.2 3.3 1.3 3.2 1.4 3.12.2 4.3 2.3 4.2 2.4 4.1

7 8 911 9 714 14 12

Drill:

1. If 6 4 54 5 3 , 2 4 21 3 5 and 1 2 03 4 2, find:a. b. c. d.

2 3

2. Find the result of the following:

a. 1 22 1 . 2 3 45 4 3b. 1 12 23 4 . 1 2 23 2 1c. 1 1 22 3 34 4 1 .

1 22 33 4D. The transpose of a matrix

Sometime its imfortant to change a row into a column andviceversa.Illustration:Given the data in arrange:1 2 2 5 4 6 74 5 3 2 6 7 4


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Then its need to be rearranged, so the rows become the columnsand viceversa. The result is

1 42 52 35 24 66 77 4

It also works in a matrix, called the transpose of a matrix.Example:

1 22 33 4

, the transpose of is written as 1 2 32 3 4

Notice a symmetric matrix like

1

1

9

9

0

0

0

0

2

2

4 7

4

7

3

3

B

=

, the transpose of is written as So a matrix is a symmetric matrix if and only if .

Drill:

1. The transpose of 4 3 21 3 2 is ....2. If 2 4 13 4 2 and 6 3 93 4 2, find the transpose of , 3. If

2 34 5

and

2 3 13 4 2

, find the transpose of

,

In Ms Excel it can be handled:

1. Select the cells that want to be transposed, copy the

selection cells.2. Click mouse in new location (the position that would serve

the result of the trasposing.

3. Click paste menu, select transpose.


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E. Determinant and Inverse of a Matrix1. Determinant of a matrix

Given a square matrix with ordo 3 x 3 Generally

Would be taken for each row one element, but there is noelement from the same column, or would be taken for eachcolumn one element, but there is no element from the same row,so:a. For each taking would be 3 elements, example , , .

Negative and positive pair of two elements

From

, , , would be arranged two element as pairs,

and it would be = 3 pairs, they are:, , , , , It defines that a pair , is:

Negative pair if and or and Here is ,

Positive pair if and or and Here are

, , ,

So from the arangement , , , there are exist:1 negative pair and 2 positive pairs. If the number or negativepair for an arrangement is stated as , then for , , its 1

b. The number possibilities of the arrangements are 3! That isif the arangement is taken continuously from 1st, 2nd, and 3rdrow:

In 1st

row would be 3 possibilitiesIn 2nd row would be 2 possibilitiesIn 3rd row would be 1 posibilitySo all the possibilities are 3.2.1 = 3! = 6, they are:, , 0, , 1, , 1, , 2, ,

3

, , 2


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Now, it would state the main idea of determinant byconcepted a function such:1. . . 1. . . 1. 12. 21. 331

. 12. 23, 31 1

. 13. 22, 31 1

. 13. 21, 32

. . . . . . . , . , . , . . . . . . . . . . . , . . . . . . . . . . . , It would give a saclar real number as long as the elements are real, andit is called as the determinant of the matrix, written as det or ||d e t || . . . . . . . . . . . ,

Visually, to demostrate the way finding the determinant is

Or


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Drill:1. Find the determinant of the following matrix:

a.

1 42 3

b.

1 22 3

c.

5 46 3

2. Find the determinant of the following matrix:a. 1 2 33 2 12 4 6 b. 2 2 21 2 13 3 3 c. 2 2 23 3 14 5 8

2. Minor and CofactorIf one row and one column of a square matrix are delated, for

example in a matrix

, the 2nd row and 3rd

column are deleted, then the determinant of the remain matrix = . . and it is called as the minor ofelement , written as .Then it makes the sign minor such as 1. . . it is called as the cofactor of element ,written as .Now find the result of

. . . !Answer: 1. . . . . 1. . . 1. . . . . Then,

. . . . . . . . . . . . . . . . . . . , . . . .

. . . . . . . . . . . , d e t


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So, the determinant of a matrix can also be found by added allproduct of every element in one row or coulumn with itscofactor.Here for the matrix

,det

By row:. . . or. . . or. . . det and 1 , 2 , , By column:

. . . or

. . . or. . . det and 1 , 2 , , Drill:

1) Find the determinant of . 1 42 6

by using cofactor based onsecond row!

2) Find the determinant of . 1 2 13 2 12 3 5 by using cofactor based onfirst row!

3) Find the determinant of . 1 2 13 2 12 3 5 by using cofactor based onthird column!

3. The natures of determinant matrix

a. The determinant of the transpose of a matrix is same with thedeterminant of the matrix.

,

|| ||


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b. If two rows or columns of a matrix A are reversibled to get anew matrix B, then .

, || ||c. Determinant of a matrix that has two rows or columns are

same, equal to zero.

|| 0d. The linear combination of a row (column) of a matrix with the

cofactors of adjusted another row (column) equal to zero.

. . . 0 ,

. . . 0 , 4. Inverse of a matrixIf is a number not zero, then is its inverse, since . 1,where 1 is the identity of multiplication. How about a matrix?Here are the explanation.

Take a matrix

, then

d e t

,

1 , 2 , , , and

also

0 , 1 , 2 , , and

1 , 2 , , ,but

. In

another word, if all cofactors are arranged to become amatrix , then ||. || . ||

Then

||is called as the inverse of

.

is called the adjoint

matrix


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So the inverse of a matrix , written as is the adjoint ofmatrix , written , divided by the determinant of matrix.

Drill:1. Find the adjoint and the inverse of the following matrixes:

a. 1 42 6 b. 3 25 6 c. . 3 2 42. Find the adjoint and the inverse of the following matrixes:

a. 1 2 13 2 12 3 5 b. 2 1 11 2 12 3 4 c. . 2 1 10 2 12 4 33. Find the inverse of matrix:

1 2 0 1

2 3 1 1

1 2 1 1

2 4 0 1

F. Solving Linear Equation SystemGenerally,Given a linear equation system ,In matrix model:

If , , , then the model is . , and is the inverse of, then. . . . .

.


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To solve this formula, do analyzing:

1. Find the adjoint matrixMinor

Sign Minor (Cofactor)

Cofactor Matrix

Adjoin Matrix

Take: The minors: ; ; ; The sign minor (Cofactor): 1 ; 1 ; 1

; 1

The cofactor matrix: The adjoint matrix: 2. Find the determinant of the matrix

Since the determinant of a matrix =d e t . . . . , then .

. . Example:Given a linear equation system with two variables:

2 83 4 1 8In matrix model:1 23 4 818So

4 23 1 1.42.3 818 12 4 23 1 818


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2 132

12

818 2.81.1832

. 8 12

. 18 23 2 ; 3Another way to solve a linear equation system1 2 3 11 2 3 21 2 3 3..

.1 2 3 in which the matrix model like

...

...

To get subtitute the 1st column of A with the result matrix,

...

,

then:


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Also with, , ,

Next,

Example:

a. 1 23 4 818

8 218 41 23 4

8 218 41 23 4

8.42.181.42.3

3 2 3 64 6

42 2


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1 83 181 23 4

1 83 181 2

3 4 1.188.31.42.3 1 8 2 44 6 62 3

So 2 ; 3 .b. 12 2 3

The matrix model,

1 1 12 0 10 1 1 526

To find the determinant of a matrix 3x3, may use

or

So,

5 1 12 0 16 1 1 1 1 12 0 10 1 1

5.0.1 1.1.61.2.1 1.0.6 1.2.15.1.11.0.1 1.1.01.2.1

1.0.0 1.2.11.1.1

8 72 3 155 3


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5 1 12 0 16 1 1

1 1 12 0 10 1 1 105 2

5 1 12 0 16 1 1 1 1 12 0 1

0 1 1

205 4

So 3 ; 2 ; 4 .Using ms excel program application

To find the determinant of the matrix 1 1 12 0 10 1 1 by using msaxcel is demostrated by:

and after pressing enter it would give the result 5.

Drill:1. Find the solution of the following model:

a. 2 14 1 610b. 3 12 1 14c.

4 23 4 1418


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2. Find the solution of the following model:

2 1 11 2 13 1 2

2711

3. Solve the following linear equation system:

a. 2 83 4 1 8b. 2 83 4 182 2 1 8

G. The transformation MatrixesHere is recommended to the students to learn Chapter 5 abouttransformation first to understand better about the transformationmatrixes. They are:1. Reflection of point, to

a. Line axis, the result is ,The matrix model:

1 00 1

b. Line axis, the result is ,The matrix model:1 00 1

c. Line , the result is2,The matrix model:

2 1 00 1 2 d. Line , the result is,2

The matrix model:

1 00 2 1 2


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e. Line , the result is , .The matrix model:

0 11 0

f. Line , the result is , .The matrix model: 0 11 0

g. Point 0, 0, the result is ,.The matrix model:

1 00 1

h. Point , , the result is 2,2.The matrix model:

2 1 00 2 1 2 2

i. Line , the result is2 21 , 2 21 .The matrix model:

2 2 1 1 00 2 2 1 1

2 2 1 2

2

1

2. Translation of a point, by , the result , .The matrix model:

1

0

0 1


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3. Dilatation of a point, with:a. scale factor and center 0,0, the result ,.

The matrix model:

b. scale factor and center ,, the result 1 , 1 .The matrix model:

1

0

0 1 1 1

4. Rotation of a point, with rotation angle and center0, 0, the result cos . sin . , sin . cos . .The matrix model:cos sin sin cos cos . sin . sin . cos .

Competency Test Try Out

1. The value of that satisfy the equation matrix 1 22 2 3 10 2 5 27 32. The transpose of the matrix is. If 5 24 7, then .3. 2 13 15 45 94 3 40 5094 60, then .UMPTN914. Given 252

165 7212 1, then . UMPTN99

5. If 4 02 3 2

8 02 7

, then . UMPTN2000


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6. 1 2 , 1 0 and 1 10 1. If ,then 2 . UMPTN 2001

7. If , and are real numbers that satisfy the equation 101 2 110 011 121, then . SPMB 20048. If 1 11 1 and 0 11 0, then is the matrix .... SPMB 20049. Lets the matrix equation

12 1 2 33 11 101 . The value of

2 3 . UN 2009/2010 IPA A-P1910. The value of that satisfy the equation matrix 1 22 3 3 2 8 416 9 62 5 is ....

UN 2009/2010 IPA P19 B-P52

11. If

and

1

3 , and

, in which

is the transpose of . The value of 2 . UN 2006/2007.12. If 1 1 1 and 1 1 4 then .13. The value of 2 that satisfies the equation matrix2 2 11 3 6 24 1 2 11 1 0 12 414. If 3 24 1 and 1 50 3, then the determinant of the matrix is ....15. If 1 23 4, then .16. If the matrix 1

4 1, 6 4

1 1 and 1 8

1 5. If

, in which

is the inverse of matrix

, then

.


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17. Given 4 23 4 and 5 32 1 . If and is theinverse of , then the determinant of is ....

18. The inverse of the matrix 2 1 11 0 12 3 2 is ....19. The value of that satisfies 8 43 4 1418 is ....20. The value of that satisfies 4 2 22 4 2

3 1 2

414

11

is ....Analyzation:

Every year, the national examination includes one item of the matrixproblem

Tips:

Understand clearly about the matrix equation, the linear equationsystem, the multiples of matrixes, the transpose, determinant andinverse of matrixes.


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VECTOR

Competency Standard

3. Using the concept of matrix, vector, andtransformation in solving problem

Basic Competency

3.4 Using the natures and algebra operation of vector in solvingproblems.

Lesson Benefit

Students would know a vector as a directed quantity, something that iscomposed of the quantity and the direction. It prepares students to thebase knowledge such as navigation, force, dynamics, especially inphysics lesson.


Students are capable in:1) Defining a scalar and a vector.2) Analyzing the unit vector.3) Do operation of vectors.4) Using the comparation formula of two vectors.

Material

A. Definition of a Scalar and a VectorIn linear algebra, real numbers are called scalars and relate tovectors in a vector space through the operation of scalarmultiplication, in which a vector can be multiplied by a number toproduce another vector (http://en.wikipedia.org/wiki/Scalar).

A vector is a quantity that has both magnitude and direction such asforce.

Chapter 4


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Illustration:The box is moved strightly from to position such below. Anarrow

is parallel to

and its lenght is equal to

, here

is a

vector of the boxs movement.

Then given,

In Cartesian Field, in 2 dimension (2D) or 3 dimension (3D), avector is indicated by its components such as for , and axis. Ifany arbitary point in space is related to origin 0, 0, then called as a position vector. In another word, a vector that the startpoint in origin

0, 0is a posistion vector. Any vector can be

presented as a position vector by translated it so the isin the origin 0, 0.

Horizontal distance

Vertical distance

2

vectorscalar


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Here for any position vector:1. Two dimension

A vector

,

2. Three dimension

A vector , ,

=3x63,63, 6

3,4,5

3,4,5

3

4

5


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B. The Magnitute or lenght of a vectorIllustration:

The magnitute of 3,4,5 (written as ) is foundby using the Phytagorean Theorem, so 3 4 5 9 1 6 2 5 5 2

Generally, if a vector

, ,

, then

| | Drill

1. Find the lenght of the position vector

, when

is the point:

a. 3, 4 b. 6, 8 c. 12,13d. 1,3,4 e. 2,3,2f. 4,9,122. If1,2,3 and 4,5,7, then the lenght of vector is ....

C. Unit VectorA unit vector is a vector that has the magnitute or the lenght one

unit. Example

1,0; , ; , , Here a vector , , is a unit vector if and only if| | 1.Standard basis vectorsThe vectors 1,0,0,0,1,0and 0,0,1 arecalled the standard basis

vectors. Of course , and areunit vectors

11

1


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Any vector in analytic space is equal to a combination of multiplesof the standard basis vectors. For example , , can bepresented as . So, 2,3,4 234 .For mor simple it is written as

2 3 4 .

D. Operation of Vectors1. Addition

Given two different vectors such as,

Then, add both of them by translation. It may be like:

a. b.

c.

Example 2 3 4 and 46 8 , then 2 4 3 6 4 8 2912 2,9,12


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Or

If 2

34 and 4

68, then 24

3 64 8 2

912

Generally if and , then Or

If

and

, then

By using the cosinus rule,

the lenght of || || || 2|||| cos180 || || || 2|||| cosDrill:

1) If 2,4,5 and 6,7,8, then find:a. b. ||

2) If 4,1,3 and 3,5,2, then find:a. b. ||


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2. SubstractionBe back to,

Then, the substraction , can be proccessedsuch as,a. b.

Example 2 15 9 and 46 8 , then 2 4 1 5 6 9 8 69

6,9,1OrIf 2159 and

468 , then 241 5 69 8

691Generally if

and

, then

OrIf and

, then


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By using the cosinus rule,

the lenght of || || || 2|||| cosDrill:

1) If 4,6,8 and 2,3,4, then find:a. b. ||2) If 4,1,3 and 3,5,2, then find:

a. b. ||3. Multiplication

a. Scalar Multiplication

If is a scalar (any real number) and

123, then the

scalar multiplication of and is

.. 123

123

Example:

The scalar multiplication of 3 and 234 is

3.3.234 3.23.33.4 6912


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Parallel vectorsIf and are two vectorssuch that 0 and 0,and there is a real number

such that , then and are parallel vectors, and bewritten as .Drill:

1) If 4,6,8 and 2,3,4, and 4, thena. b.

2) If 4,1,3 and 3,5,2, and 4, thena. c. b. || d. ||b. Two Vectors Multiplication

If

1

2

3and

1

2

3, then their multiplication is

defined as

.

123

.

123

1. 1 2. 2 3. 3 . Two vectors multiplication or product is also called as dotproduct or inner product.

Example:

If 2

34and 3

21, then the multiplication

. 2 . 3 3 . 2 4 . 1 6 6 4 1 6


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Properties of the Dot ProductLet , and are vectors and is a scalar, then1.

. | |

2.

..

3. . .. 4. . . . .. .5. .0,0,0,0Drill:

1. Find the multiplication of 2,0,4 and 2,3,7

.

2. Find the multiplication of 325 and342 3. Find if the inner product of 3,2,5 and ,3 ,7 is 27.4. A shop sells three product , and for 2, 3 and 4

liras respectively. In one day the shop sells 7 ofproduct , 5 of product and 8 of product . Use thedot product of two vectors to calculate the amountearned by the shop.

3) Distance of a point to a lineIllustrationFind the distance of point , to the line 0.Solution:

Briefly, the

axis is moved or translated until cuts the

intersection point of line 0 and axis. It isindicated by As the consequencies:The line 0 becomes 0The point , becomes , The intersection point between the line 0 and

axis ,

0, becomes

0,0.

They are presented as bellow:


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. .

Because a distance always positive, then,

| | 4) Proof of . | || |

Given a simple one of

1, 2and

1, 2and the

angle formed is ,lenghBy definition . 1. 1 2. 2 ... (1)

0 0 on , on

On ,0, on 0,0

, is a real number


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The equation of line is or 0The distance of1, 2 to vector or line 0 is | | Meanwhile,

is

| | cos(it would be discussed more detail in

the next part).

Notice the right triangle , | | cos || ,

| |

cos

2

| | ||cos 2 2 2 1. 1 2. 2Here,

| | ||cos . .

0

1, 21, 2


67/115


| ||| c o s . . Because . 1. 1 2. 2, then

. | || | 5) The projection of a vector

The lenght of the projection of to is ||called as the scalar orthogonal vector projection of to .

| | | |

Since. || cos, then| | .

.

| |.

If the angle between the vectors and is:a. zero, 0 then . ||. b. Perpendicular (orthogonal, ), 90, then . 0Example:For what value of

, the vectors

4,6,and

4,,2 are orthogonal?


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Solution: , then . 04 . 4 6 2 0

1 6 8 08 1 6 2.The vector projection of on its self is and written asproj . That is the multiplication of || and the unit vector of

proj ||.

|| cos. . . .

.

Example:

Find the vector projection of 3,5,7 on 1,1,0.Solution:

proj . 2 .

3.15. 1 7.012 12 . 1,1,0 22 . 1,1,0 1,1,0.Drill:

1) Find the vector projection of 2,5,8 on 1,1,2.2) Find the vector projection of

2,3,4on

2,0,3.


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70/115


, , 2,2,3then

32 4.53 4 2; 3.2 4. 53 4 2; 3.4 4. 33 4 0;So

2,2,0.

Drill:

1) Let and are position vectors, 2,3,4 and4,4,2. Point lies on line, if:1:2, thenthe vector position is ....

2) Let and are position vectors, 4,8,2 and1,0,5

. Point

lies on line

, if

: 3 : 2, then

the vector position is ....Competency Test Try Out

1. The following picture shows that = .... UMPTN 97

2. If 433 , 223 ,

425 and 2 3, then thelenght of vector is ....

3. Given a square and the lenght of each sides is 2, then . .

= .... UMPTN 99


71/115


4. If 0, 0, 6, 2 and 4,8, then the kind of the triangle is ....UMPTN 98.

5. Given the rectangle

,

1 2 , 5 .If

and

,

then . = .... UMPTN 99.6. Given the vector 453 and the point 2,1,3. If thelenght of is equal to the lenght of , but the direction of is inopposite of , then the coordinate of is .... UMPTN 99.

7. 1,5,4; 2,1,2; 3,,. If the points , and are inone line (colinner), then the value of and respetively are ....UMPTN 97.

8. Given 3,2,1,2,1,0 and 1,2,3. If and then the projection of vector on is .... UN .10 B-P 52.9. Given 23 , 2 5 and 3 . If

the vector is perpendicular to , then 2 .10. Given 628 , 4 810 and 23

5. If the vector

is perpendicular to

, then

.

11. Given the triangle , 0,0,0, 2,2,0 and 0,2,2. Theorthogonal projection of on is ....12. Given the vector 112 ,

222 and 02. If the

lenght of the projection vector on is 1 and perpendicular to ,then the value of

.

13. If || 12, . 112 and . 48, then .14. Given 24 , 4 2 2 and 4 2 6 . If is perpendicular to , the orthogonal projection of on is ....15. Given || 14, 14 and 42. If is the angle

between

and

, then

t a n .


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TRANSFORMATION

Competency Standard3. Using the concept of matrix, vector, and

transformation in solving problem

Basic Competency

3.5 Using the natures and the operation of matrix to show that asquare matrix is the inverse of other square matrix.

3.6 Determining the determinant and inverse of matrix 2 x 2.3.7 Using determinant and inverse in solving linear equation system

of two variables.3.8 Using the natures and algebra operation of vector in solvingproblems.

Lesson Benefit

This is a big expantion of numbers. Previously numbers are viewed asindividual that interacts with another numbers, but now, numbers aregrouped in matrix form and it interacts with numbers or other matrixes.

Students would be aware with the useful of matrix in solving problem,

especially in solving the linear system equation.Will be Reached Competencies


5) Identipying the natures of a matrix.6) Doing operation of a matrix.7) Determining the determinant and inverse of matrix 2 x 2.8) Using determinant and inverse in solving linear equation system of

two variables.

9) Using the natures and algebra operation of vector in solvingproblems.

Chapter 5


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Material

A. The transformation Matrixes

It begins with the explanation of some types oftransformations, then would be followed by discussion of itsmatrix.

First it would show the transformations in blank area. In thiscase the use of geometry tools is so important. Then thediscussion is brought to the Cartesian Field with two axis, and

axis or using 2D (two dimentions)

1. Reflection

The base of the reflection of an object is the mirror or thesymmetry axis of the reflection

The properties of reflections are:

a. The object and its reflection result are congruent.b. The object and its reflection result are in opposite

direction, symmetry.

m, The symmetry axis

A

A

T


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c. The distance of the object to the reflector (symmetry axis) issame with the distance of its reflection result to the reflector,.

d. The line that relates the object to its reflection resullt is

perpendicular to the reflector line, .

Reflection in Cartesian Field


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a. Reflector axis,

Generally, when, is reflected to the result is,The matrix model that suitable for it,

1 00 1 Drill:

1. Find the reflection results of the following points to and show by using the matrix.a.

2,5c.

2,8

b.

2,5d.

3,6

2. If2,3 is reflected to the result is 4,6,then ....

3. If , is reflected to the result is 4,36,then ....

4. If

4,8is reflected to

the result give the result

,

then the distance of 2, 3 to the point is ....

,

,

43

3


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b. Reflector axis

Generally, when

,is reflected to

the result is

,The matrix model that suitable for it,1 00 1 Drill:

1. Find the reflection results of the following points to

and show by using the matrix.a. 3,5 c. 6,8b. 2,4 d. 8,6

2. If2,3 is reflected to the result is 4,7,then ....

3. If , is reflected to the result is 4,64,then ....

4. If4,8 is reflected to the result give the result,then the distance of

3, 4to the point

is ....

,

,


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c. Reflector

Generally, when, is reflected to the result is2 , or2,In above case2,3 is reflected to 6, then2 .62,3 10,3The matrix model that suitable for it,

2 1 00 1 2 Drill:

1. Find the reflection results of the following points to 4and show by using the matrix.a. 4,5 c. 4,6b. 2,3 d. 3,5

2. If , is reflected to 3 the result is 8,7, then ....3. If 2,5 is reflected to the result is 8,5, then

....

4. If 4,5 is refected first to 6, then the result is reflectedto 3, the final result is ....

,

,

6


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d. Reflector

Generally, when, is reflected to the result is,2 or,2In above case4,3 is reflected to 6, then4 ,2 .63 4, 9The matrix model that suitable for it,

1 00 2 1 2 Drill:

1. Find the reflection results of the following points to 6and show by using the matrix.a.

4,5c.

4,6

b.

2,3d.

3,5

2. If , is reflected to 3 the result is 8,8, then ....3. If 2,5 is reflected to the result is 2,9, then ....4. If 4,5 is refected first to 6, then the result is reflected

to 3, the final result is ....

,

,

4

9

3

6


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e. Reflector

Generally, when, is reflected to the result is,.In above case

6,2is reflected to

, then

2, 6The matrix model that suitable for it is0 11 0 Drill:


and show by using the matrix.

a. 3,5 c. 6,8b. 2,4 d. 8,62. If3,24 is reflected to the result is4,12, then ....3. If ,3, is reflected to the result is

9,64, then

....

,

,


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f. Reflector

Generally, when, is reflected to the result is,.In above case

1,4is reflected to

, then

4,1The matrix model that suitable for it, 0 11 0 Drill:1. Find the reflection results of the following points to

and show by using the matrix.

a.

3,5c.

6,8

b. 2,4 d. 8,62. If3,24 is reflected to the result is4,12, then ....3. If ,3, is reflected to the result is 9,64,

then ....

,

,


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g. Reflector the original point , It seems to reflect a point to axis then the result bereflected to

axis, or to

axis then the result be

reflected to

axis.

Generally, when

,is reflected to

0,0the result is

,.In above case2,3 is reflected to 0,0, then2,3The matrix model that suitable for it,

1 00 1

Drill:1. Find the reflection results of the following points to 0,0and show by using the matrix.a. 2,7 c. 6,9b. 3,4 d. 9,6

2. If3,24 is reflected to 0,0 the result is12,4, then ....3. If ,3, is reflected to 0,0 the result is81,12, then 2 ....

,

,


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h. Reflector the point , Actually it same with the reflection, first to thenfollowed by the reflection to to , or first to thenfollowed by the reflection to to

.

Generally, when

,is reflected to

,the result is

2 , 2 or2,2.In above case2,1 is reflected to 6,3, then2 .62,2 .31 10,5.The matrix model that suitable for it,

2 1 0

0 2 1

2 2

Drill:1. Find the reflection results of the following points to 4,6

and show by using the matrix.a. 2,3 c. 7,9b. 3,2 d. 8,6

2. If ,4 is reflected to 2,5 the result is 5,2,then ....

3. If

4,9is reflected to

2,4and the result is

then the

lenght of is ....

, ,

6

3,

,


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i. Reflector

Given a point

, be reflected to the line

. It

seems more difficult , but there is always the way to find thereflection result. One of the way is given such as:1) Find the line through, that perpendicular to the line . If the line is , then the gradient is and the

equation of is 1

1

1 2) Find the intersection point of line and 1

Here, 1

, ,

6

5,

,

2 1 7


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1

1

. 1 1

Golden bridge:

Then, 1 It seems to reflect to the point

21 ,

21

and the result is

2 21 , 2 21 .At the case2,3 and the reflector 2 1 7,then21 , 21

..

22

1 , 2 ..

22

1 17 6, 5

It seem to reflect 2,3 to the point 6, 5 and the result is2 26 2, 3 25 3 10,7The matrix model that suitable for it,

2 2 1 1 00 2 2 1 1

2 2 1 2 2 1


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Drill:


2 3

a. 4,5 c. 4,8b. 3,1 d. 5,62. If3, is reflected to 2 4 the result is 4,1,then ....

3. If 2,4 is reflected to 2 3, and the result isreflected to 2 3 gives the result . Find thearea of the triangle

2. Translation

The base of the translation of an object includes:

The distance in horizontal direction

The distance in vertical direction

The natures of translation:

a. The object and its translation result are congruent.b. The objests and its translation result have the same

direction.

Horizontal distance

Vertical distance


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In Cartesian Field,

a translation of a point , that moved unithorizontally then the result is moved to unit verticallyis indicated by, is translated by , and theresult is

, .

In the above case 2, 2 is translated by 34, and theresult is2 3 , 2 4 5, 6.The matrix model that suitable for it is

1

0

0 1

Drill:

1. Find the translation results if the point 3, 5 is translatedby

a. 34 c. 45 b.

32 d.

3

,

,

3 4


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2. If A 2,3 is translated by gives the result10 ,10, then the equation of a line through and,

is ....

3. If 4,6 is translated by 2 3 gives the result4,7, then ....3. Dilatation (Scaling)

The base of the dilatation of an object includes:

The center of dilatation

The scale factor

Center of dilatation

Scale factor


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Generally, when, is dilatated with scale factor andcenter 0,0 the result is,.In the above case, 4,2 is dilatated with scale factor 3 andcenter 0,0 the result is3.4,3.2 12,6.The matrix model that suitable for it is

But for the center not 0,0, such as:

,

,

,

,

,


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Generally, when, is dilatated with scale factor andcenter , the result is , 1 , 1 .In the above case, 4, 8 is dilatated with scale factor 3 andcenter 1,11 the result is3 . 4 11 3,3 .8111 2 10,2.The matrix model that suitable for it is

1 00 1 1 1

Drill:

1. Complete the following table:

Part PointScaleFactor

DilatationCenter

DilatationResult

a. 2, 3 2 0,0 ...b. . 4 0, 0 20,24c.

4, 7

2

1, 2 ...

d. 4 2, 3 2, 3e. 2, 3 2, 2 2, 7f. 3, 2 3 5, 4

2. If2, 2and 2,0 are dilatated with the center 0, 0and scale factor 2, gives the result respectically

and

,

then the area of trapesium is ....


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4. Rotation

The base of the rotation of an object includes:

The center of rotation

The angle of rotation

Center of dilatation

Angle of rotation


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Generally, when , is rotated by with center 0, 0, then , cos cos sin sin , sin cos cos sin cos . sin . , sin . cos . cos . sin . , sin . cos . .The matrix model that suitable for it is

cos sin sin cos cos . sin . sin . cos . Example:2, 4 is rotated by 60 with center 0, 0, thenis found by,

cos60

sin60

sin60 cos60 24

12 3232 12

24 12 . 2 32 . 432 . 2 12 . 4

1 233 2

,

,


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Drill:Complete the following table:

Part PointRotation

Angle

Rotation

Center

Rotation

Resulta. 2, 3 30 0,0 ...b. 4, 7 45 0, 0 ...c. 5, 6 60 0, 0 ...d.

4, 3

120

0, 0...

e. 6, 8 315 0, 0 ...f. 90 0, 0 2, 3g. 4, 5 . 0, 0 4,5

Competency Test Try Out1. If4,3 is reflected to the result is 4,6, then2 ....2. If2, 4 is reflected to the point , and the result is 8,10, then .3. If 4,5 is refected first to 7, then the result is reflected to

3, the final result is ....

4. If 4,5 is reflected to the result is 4,9, then ....5. The line 3 is refelected to axis and be dilatated base on0,0 with scale factor 2 is .... UN 10 A-P 19.6. If3,24 is reflected to the result is 4,12, then ....7. If

,3,is reflected to

0,0the result is

64,12, then

2 ....


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8. Given a circle with center 3, 4 and 4 3 1 0 is one of itstangent line. If the circle is reflected to 2, the the result cuts axis in point and . The lenght of is ....

9. A paraball 3 is rotated on the center 1,2 and theangle 90 . The result is ....10. The point 2,5 is dilatated by 1,2, 5. 1,2 is the center

of dilatation and 5 is the scale factor. The result is ....

11. A point , is transformated by matrix 0 11 0. The result istransformated by 1 0

0 1 and the result is 5, 7, .

12. The vector is rotated with the center 0,0 and theangle 90. The result is refelcted to the origin 0,0 and gets theresult . If , then .

13. The matrix that shows the rotation with angle and the rotation

center 0,0 then followed by the reflection to line is ....14. The transformation

is maped

, into

,and the

transformation is maped , into ,. If thetransformation is the transformation that followed by thetransformation , then the matrix of is ....

15. A circle with center 2, 3 and radius 3 is rotated by center 0,0and angle

, and followed by dilatation 1,2, 2, the circleequation of the result is ....

16. If

3, 5is reflected to the line

2 4gives the result

, then

the lenght of is ....17. The line ; 2 2 is reflected to line 1 gives the result. The cute angle that made by and is ....18. The points 2, 4, 4, 7, and lies on , such that : 2: 3, then the equation of line throught and perpendicular to

is ....


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SEQUENCES AND SERIES

Competency Standard

4. Using the concept of sequences and seriesin problems solving

Basic Competency

4.1 Determining the formula of the th term of arithmetic andgeometry sequences and also their series formulas.

4.2 Using the sigma notation in series and mathematics induction inproofing.

4.3 Constructing the mathematical model of problems related toseries.

4.4 Solving the mathematical model of problems related to seriesand giving the interpretation.

Lesson Benefit

Students would observe more detail about the style of numberssequences and their series, finding the formulas of them and use themto solve problems related to sequences and series. Its one kind offinding the functions that can be applied in certain sequences andseries.



1. Determining the formula of the nth term of arithmetic and geometry

sequences and also their series formulas.2. Using the sigma notation in series and mathematics induction inproofing.

3. Constructing the mathematical model of problems related to series.4. Solving the mathematical model of problems related to series and

giving the interpretation.

Chapter 6


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Material

A. The Sigma NotationThe symbol

sig

mameans sum and its defined as

1 1 2 1 Note: is natural number, consequently , 1 , 2 , , 2 , 1 , are also natural number, and So, here is discussed about the value of the function withnatural number (1, 2, 3, ...) as its domain.

Example:

12 12 12 12 14 18 116 4 2 116 716Drill:

1. Expanda. 2 c. 3 2

e. 2 b. 24

d. 12

f. 2 2

2. Write in sigma notationa. 2 . 2 2 . 3 2 . 4 2 . 8b. 4 4 4 4c. 4 2 5 2 6 2 12 2d. . . . .e.

s insin2 sin3 sin9

f.

log

4.2log

4.3log

4 . 4 l og

4 . 8


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B. Properties of Sigma1. If is a constant, then

Proof:

1 2 1 2

Example:a.

3

3 32 3 4 5

3 . 14 42b.

12

12 12 1 2 3

12 . 1 4 72. If

and

are constants, then

1 Proof:

1 2


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1 2 1 time 1

Example:

3 2 First method:

3 2

3 .12 3 . 2 2 3 . 3 2 3 . 4 2 5 8 1 1 1 4 3 8

Second method:

3 2 3 4 1 12 31 2 3 4

8 3 83. If

,

,

, , are functions, then


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Example:

2 3 5 2

3

5

2. 2 2 . 3 2 . 4 2 . 5 3.23.33.43.5+

5 2 15

8 1 8 3 2 5 0 6 9 1 2 1 5 20 1 0 8 4 2 2 0 8 64. If 1, then

Example:

3 3 3 3 31 2 3 4 5 6 7 69 Drill:1. Calculate

a.

5

c.

2 3

e.

22 4 3

b. 14 d. 12 1

f. 12 12

2. If 12 , determine the value of 4 .3. If

18 ,determine the value of

.

4. If 30 , determine the value of 2 4 .5. If 13 , 40 and 10 ,

then determine the value of 2 4 4


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C. Arithmetic SequenceIllustration:

2 2 2

0 2 4 6A frog jumps 2 meters for each jumping, the distance of thefrogs movement is indicates bellow:

The frequncy ofjumping

The distance, 0

2 . 0 0

1 time 2 . 1 22 times 2 2 2 . 2 43 times 2 2 2 2 . 3 64 times 2 2 2 2 2 . 4 8... 1 times 2 2 2 2 2 . 1

times

2 2 2 2 2 2 .

If the distances are elaborated as 0,2,4,6,8, then it is calledas one of arithmetic sequence, and the th term is symbolized as, here 2 1 or 2 2.From 0,2,4,6,8,, the common differnce is 2 (from 2 0 or 4 2or 6 4 or 8 6 , its fixed common difference)Another examples of arithmetic sequences:

a. 3,5,7,9,11, the common difference is 2b. 2,5,8,11,14, the common difference is 3c. 7,12,17,22,27, the common difference is 5In general, if and the common difference is , then

2

2 3 ...


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2 standard method

smart method

Take again:a. 3,5,7,9,11, the common difference is 2b. 2,5,8,11,14, the common difference is 3c. 7,12,17,22,27, the common difference is 5Then their th term or are given such as:Part.

Standard Method

Smart Method

a.

3 12 3 2 2 2 1 2 3 2 2 1b.

2 13 2 3 3 3 1 3 2 3 3 1

c.

7 15

7 5 5 5 2 5 7 5 3 2By smart method , here9, 13, 17, 21, 25,

4 5

calculate

9 4 5

write 44the common difference


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Drill:1. Complete the table bellow:

The sequence Formula of

Find the term

a. 1,3,5,7,9,11, ... b. 4 6 c. 3,10,17,24,31,38, ... d. 5 8

2. Find the formula of the

term of the sequence

100,90, 80 ,

3. Find the

2000th term of the sequence

1, 1 , 1 , 1 ,2 ,

4. Determine the number of terms of the sequence12,15,18,,20105. If is the 30 th term and is the 100 th term of the sequence24,28,32,36,, then 2 .

D. Arithmetic SeriesHistory:Carl Friedrich Gauss

(30 April 1777 23 February 1855)German mathematician who issometimes called the "prince ofmathematics." He was a prodigiouschild, at the age of three informing hisfather of an arithmetical error in acomplicated payroll calculation andstating the correct answer. In school,when his teacher gave the problem ofsumming the integers from 1 to 100(an arithmetic series Eric Weisstein'sWorld of Math) to his students to keepthem busy, Gauss immediately wrotedown the correct answer 5050 on hisslate.

Resource: http://scienceworld.wolfram.com/biography/Gauss.html

Here, explains the way of Gauss to count the sum of integer from 1to 100.

If the sum is , then


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1 2 3 9 8 9 9 1 0 0, then it is also true for 1 0 0 9 9 9 8 3 2 1,: 1 2 3 9 8 9 9 1 0 0

1 0 0 9 9 9 8 3 2 12 101 101 101 101 101 101100 2100.10150.101 5050This method also works in any arithmetic series, for example:

3 5 7 9 11 2 1 2 1 2 1 2 1 5 32 2 4 2 4 2 4 2 4 2 2 4 2 4

In general, if 2 2 1then, 2 1 1 2 2 2 1 2 1 2 1 2 1

2 2 1

2 1 or or 1 or 2 2

+

+

+

Standard method

Smart method


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Example:Find the series of the first terms of the following ones:a. 2 5 8 11 14 the common difference is 3b.

7 12 17 22 27 the common difference is

5

Solution:

Part.standard method 2 2 1

smart method 2 2 a.

2 2.2 132 4 3 3

2 3 12

3 2 . 2 32 3 1

b.

2 2.7 152 1 4 5 52 5 92 5 2 . 7 52 5 9

By smart method

, here

9 13 17 21 25

4 14 calculate 2 . 9 4 1 4

write 4

4 the common difference


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Drill:1. Complete the table bellow:

The sequence Formula of

Find the series

e. 1 3 5 ... f. 4 6 g. 3 1 0 1 7 ... h. 5 8

2. Find the formula of the series for the first

term of the series

100 90 80

3. Find the series for the first 2000 th term of the series 1 1 1 1 2 4. Determine the series of the following one12 15 18 20105. If is the series of the first 10 th term and is the series of thefirst 20 th term of the following one2 4 2 8 3 2 3 6 , then 2 .

E. Expand Sequence1.Second Degree

In forming of the following angles:a.

Number of angle, 1 ()b.

Number of angle,

3(

,

,

)

O

A

B

O

A

C

B

Number of

line

Number of angle

1 0 1. 2 1 2. 3 3

3.

4 6 4. 5 10 5. . .

. .

. .

n . 12


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c.

Number of angle, 6 (,,,,,)Etc.

U1=0 U2=1 U3=3 U4=6 U5=10

1 2 3 4

1 1 1

After 2 times the proccess of subtraction, it gets fixed result. Itmeans the formula for the th term is Since it is the second degree, supposed the th terms are: , they are:

. 1 . 1

. 2 . 2 4 2 . 3 . 3 9 3 . 4 . 4 16 4 . 5 . 5 25 5 The analyzing of common difference:

U1=a+b+c U2=4a+2b+c U3= 9a+3b+c U4=16a+4b+c U5=25a+5b+c

3a +b 5a+b 7a+b 9a+b

2a 2a 2a

If this case is applied to the counting of angles in the previousproblem, then:

U1=0 U2=1 U3=3 U4=6 U5=10

1 2 3 4

1 1 1

O

A

C

B

D


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03 12 1Solved:

,

,

0

So, Un = 0 12.Third Degree

Since it is the third degree, supposed the th terms are:

,they are:

. 1 . 1 . 1 . 2 . 2 . 2 8 4 2 . 3 . 3 . 3 2 7 9 3 . 4 . 4 . 4 6 4 1 6 4 . 5 . 5 . 5 1 2 5 2 5 5 The analyzing of common difference:

8 4 2 2 7 9 3 6 4 1 6 4 1 2 5 2 5 5 7 3 1 9 5 37 7 61 9 122 182 242

6 6Example:

1 3 9 31 812 6 22 504 16 28

12 12


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18 4 2 32 7 9 3 96 1 2 ; 2Solved:2 1 1 ...(1)16 4 2 3 4 2 13 ...(2)5 4 9 3 9 9 3 4 5 ...(3)From(1) and (2): 14 2 1 33 1 2 ...(4)From (2) and (3):

4 2 1 39 3 4 55 3 2 ...(5)From (4) and (5)3 1 25 3 22 20;10 ...(6)From (4): 3 12 310 1 2 1 8From (1):

1 1 0 1 8 1 9Then: 2 10 1 8 9Recheck: 2 . 1 10.1 1 8 . 1 9 2 1 0 1 8 9 1 2 . 2 10.2 1 8 . 2 9 1 6 4 0 3 6 9 3 2 . 3 10.3 1 8 . 3 9 5 4 9 0 5 4 9 9 2 . 4 10.4 1 8 . 4 9 1 2 8 1 6 0 7 2 9 3 1

2 . 5 10.5 1 8 . 5 9 2 5 0 2 5 0 9 0 9 8 1

Drill:1. Find the next two terms of the following sequence:

a. 2,4,7,11,, c. 5, 6, 10, 17, ..., ...b. 1 , 4 , 8 , 1 3 , , d. 2, 4, 10, 20, ..., ...

2. Find the formula for the th term of the following second degreesequences:a.

4,7,11,16,,

b. 4 , 8 , 1 3 , 1 9 , , c. 6, 10, 17, 27, ..., ...


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3. Find the the next two terms and also the formula for the th term ofthe following third degree sequences:a. 1, 3, 6, 11, 19, ..., ...b. 2, 4, 8, 15, 25, ..., ...

c. 3, 4, 6, 10, 17, ..., ...

F. Geometric SequenceIllustration:The number of a bactery increases two times of theprevious number in every one hour. If the origin number ofthe bactery is 3, then it is indicated as:

The time until The number of the bactery,

0 hour 31 hour 2 . 3 3 . 2 62 hours 2 . 2 . 3 2. 3 3 . 2

3. Math for XII Science Trial Editionx

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