Matematik merupakan satu bidang ilmu yang pengetahuan yang sangat penting. Tokoh

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Algebra ialah salah satu cabang matematik yang membincangkan tentang prinsip operasi dan

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Comments (31)

http://ndu2009algebra.blogspot.com/

Linear algebra and electricity of a car

The diagram below shows some of a car's electrical network. The battery is on the left, drawn

as stacked line segments. The wires are drawn as lines, shown straight and with sharp right

angles for neatness. Each light is a circle enclosing a loop.

The designer of such a network needs to answer questions like: How much electricity flows

when both the hi-beam headlights and the brake lights are on? Below, we will use linear

systems to analyze simpler versions of electrical networks.

For the analysis we need two facts about electricity and two facts about electrical networks.

The first fact about electricity is that a battery is like a pump: it provides a force impelling the

electricity to flow through the circuits connecting the battery's ends, if there are any such

circuits. We say that the battery provides a potential to flow. Of course, this network

accomplishes its function when, as the electricity flows through a circuit, it goes through a

light. For instance, when the driver steps on the brake then the switch makes contact and a

circuit is formed on the left side of the diagram, and the electrical current flowing through

that circuit will make the brake lights go on, warning drivers behind.

The second electrical fact is that in some kinds of network components the amount of flow is

proportional to the force provided by the battery. That is, for each such component there is a

number, it's resistance, such that the potential is equal to the flow times the resistance. The

units of measurement are: potential is described in volts, the rate of flow is in amperes, and

resistance to the flow is in ohms. These units are defined so that .

Components with this property, that the voltage-amperage response curve is a line through

the origin, are called resistors. (Light bulbs such as the ones shown above are not this kind of

component, because their ohmage changes as they heat up.) For example, if a resistor

measures 2 ohms then wiring it to a 12 volt battery results in a flow of 6 amperes.

Conversely, if we have flow of electrical current of 2 amperes through it then there must be a

4 volt potential difference between it's ends. This is the voltage drop across the resistor. One

way to think of a electrical circuits like the one above is that the battery provides a voltage

rise while the other components are voltage drops.

The two facts that we need about networks are Kirchhoff's Laws.

Current Law. For any point in a network, the flow in equals the flow out.

Voltage Law. Around any circuit the total drop equals the total rise.

In the above network there is only one voltage rise, at the battery, but some networks have

more than one.

For a start we can consider the network below. It has a battery that provides the potential to

flow and three resistors (resistors are drawn as zig-zags). When components are wired one

after another, as here, they are said to be in series.

By Kirchhoff's Voltage Law, because the voltage rise is 20 volts, the total voltage drop must

also be 20 volts. Since the resistance from start to finish is 10 ohms (the resistance of the

wires is negligible), we get that the current is (20 / 10) = 2 amperes. Now, by Kirchhoff's

Current Law, there are 2 amperes through each resistor. (And therefore the voltage drops are:

4 volts across the 2 oh m resistor, 10 volts across the 5 ohm resistor, and 6 volts across the 3

ohm resistor.)

The prior network is so simple that we didn't use a linear system, but the next network is

more complicated. In this one, the resistors are in parallel. This network is more like the car

lighting diagram shown earlier.

We begin by labeling the branches, shown below. Let the current through the left branch of

the parallel portion be i1 and that through the right branch be i2, and also let the current

through the battery be i0. (We are following Kirchoff's Current Law; for instance, all points

in the right branch have the same current, which we call i2. Note that we don't need to know

the actual direction of flow— if current flows in the direction opposite to our arrow then we

will simply get a negative number in the solution.)

The Current Law, applied to the point in the upper right where the flow i0 meets i1 and i2,

gives that i0 = i1 + i2. Applied to the lower right it gives i1 + i2 = i0. In the circuit that loops

out of the top of the battery, down the left branch of the parallel portion, and back into the

bottom of the battery, the voltage rise is 20 while the voltage drop is , so the Voltage Law

gives that 12i1 = 20. Similarly, the circuit from the battery to the right branch and back to the

battery gives that 8i2 = 20. And, in the circuit that simply loops around in the left and right

branches of the parallel portion (arbitrarily taken clockwise), there is a voltage rise of 0 and a

voltage drop of 8i2 − 12i1 so the Voltage Law gives that 8i2 − 12i1 = 0.

The solution is i0 = 25 / 6, i1 = 5 / 3, and i2 = 5 / 2, all in amperes. (Incidentally, this

illustrates that redundant equations do indeed arise in practice.)

Kirchhoff's laws can be used to establish the electrical properties of networks of great

complexity. The next diagram shows five resistors, wired in a series-parallel way.

This network is a Wheatstone bridge (see Problem 4). To analyze it, we can place the arrows

in this way.

Kirchoff's Current Law, applied to the top node, the left node, the right node, and the bottom

node gives these.

Kirchhoff's Voltage Law, applied to the inside loop (the i0 to i1 to i3 to i0 loop), the outside

loop, and the upper loop not involving the battery, gives these.

Those suffice to determine the solution i0 = 7 / 3, i1 = 2 / 3, i2 = 5 / 3, i3 = 2 / 3, i4 = 5 / 3,

and i5 = 0.

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THE $25,000,000,000 EIGENVECTOR THE LINEAR ALGEBRA BEHIND

GOOGLE

Google’s success derives in large part from its PageRank algorithm, which ranks the importance of webpages

according to an eigenvector of a weighted link matrix. Analysis of the PageRank formula provides a wonderful

applied topic for a linear algebra course.

PageRank algorithm, which quantitatively rates the importance of each page on the web.

Google calculates web page rankings using standard linear algebra application. It does three basic things: 1. Crawl the web and locate all web pages with public access. 2. Index the data from step 1, so that it can be searched efficiently for relevant keywords or phrases.

3. Rate the importance of each page the subset of pages in the database been found, the most important pages

can be presented first.

The importance score for any web page will always be a non-negative real number.

An arrow from page A to page B indicates alink from page A to page B.

xk denotes the importance score of page k in the web

x1 = 2, x2 = 1, x3 = 3, and x4 = 2, so that page 3 would be the most important,

pages 1 and 4 tie for second, and page 2 is least important. A link to page k becomes a vote for

page k’s importance.

This approach ignores an important feature ranking algorithm that a link to page k from an important page

should boost page k’s importance score more than a link from an unimportant page.

we will boost page k’s score by xj/nj , rather than by xj .

xk = ∑ xj/ nk j ϵ L

k ( a link from a page to itself will not be counted.)

Let’s apply this approach :

x1 = x3/1 +

x4/2, since pages 3 and 4 are backlinks for page 1 and page 3 contains only one link, while page 4 contains two

links (splitting its vote in half). Similarly, x2 = x1/3, x3 = x1/3 + x2/2 + x4/2, andx4 = x1/3 + x2/2. These linear

equations can be written Ax = x, where x = [x1 x2 x3 x4]T

This transforms the web ranking problem into the “standard” problem of finding an eigenvector for a square

matrix! (Recall : eigenvalues _ and eigenvectors x of a matrix A satisfy the equation Ax = λx, x != 0 by

definition.)

x1 ~=0.387, x2~ = 0.129,x3 ~= 0.290, and x4 ~=.194.

Note that this ranking differs from that generated by simply counting backlinks. It might seem surprising that page 3, linked to by all other pages, is not the most important.

To understand this, note that page 3 links only to page 1 and so casts its entire vote for page 1. This, with the

vote of page 2, results in page 1 getting the highest importance score.

http://aix1.uottawa.ca/~jkhoury/leonteif.htm

Application to Leontief input-output model

Introduction In order to understand and be able to manipulate the economy of a country or a

region, one needs to come up with a certain model based on the various sectors of this

economy. The Leontief model is an attempt in this direction. Based on the assumption that

each industry in the economy has two types of demands: external demand (from outside the

system) and internal demand (demand placed on one industry by another in the same system),

the Leontief model represents the economy as a system of linear equations. The Leontief

model was invented in the 30’s by Professor Wassily Leontief (picture above) who developed

an economic model of the United States economy by dividing it into 500 economic sectors.

On October 18, 1973, Professor Leontief was awarded the Nobel Prize in economy for his

effort.

1) The Leontief closed Model Consider an economy consisting of n interdependent

industries (or sectors) S1,…,Sn. That means that each industry consumes some of the

goods produced by the other industries, including itself (for example, a power-

generating plant uses some of its own power for production). We say that such an

economy is closed if it satisfies its own needs; that is, no goods leave or enter the

system. Let mij be the number of units produced by industry Si and necessary to

produce one unit of industry Sj. If pk is the production level of industry Sk, then mij pj

represents the number of units produced by industry Si and consumed by industry Sj .

Then the total number of units produced by industry Si is given by:

p1mi1+p2mi2+…+pnmin.

In order to have a balanced economy, the total production of each industry must be equal to

its total consumption. This gives the linear system:

If

then the above system can be written as AP=P, where

.

A is called the input-output matrix.

We are then looking for a vector P satisfying AP=P and with nonnegative components, at

least one of which is positive.

Example Suppose that the economy of a certain region depends on three industries: service,

electricity and oil production. Monitoring the operations of these three industries over a

period of one year, we were able to come up with the following observations:

1. To produce 1 unit worth of service, the service industry must consume 0.3 units of its

own production, 0.3 units of electricity and 0.3 units of oil to run its operations.

2. To produce 1 unit of electricity, the power-generating plant must buy 0.4 units of

service, 0.1 units of its own production, and 0.5 units of oil.

3. Finally, the oil production company requires 0.3 units of service, 0.6 units of

electricity and 0.2 units of its own production to produce 1 unit of oil.

Find the production level of each of these industries in order to satisfy the external and the

internal demands assuming that the above model is closed, that is, no goods leave or enter the

system.

Solution Consider the following variables:

1. p1= production level for the service industry

2. p2= production level for the power-generating plant (electricity)

3. p3= production level for the oil production company

Since the model is closed, the total consumption of each industry must equal its total

production. This gives the following linear system:

The input-output matrix is

and the above system can be written as (A-I)P=0. Note that this homogeneous system has

infinitely many solutions (and consequently a nontrivial solution) since each column in the

coefficient matrix sums to 1. The augmented matrix of this homogeneous system is

which can be reduced to

.

To solve the system, we let p3=t (a parameter), then the general solution is

and as we mentioned above, the values of the variables in this system must be nonnegative in

order for the model to make sense; in other words, t≥0. Taking t=100 for example would

give the solution

2) The Leontief open Model The first Leontief model treats the case where no goods

leave or enter the economy, but in reality this does not happen very often. Usually, a

certain economy has to satisfy an outside demand, for example, from bodies like the

government agencies. In this case, let di be the demand from the ith

outside industry,

pi, and mij be as in the closed model above, then

for each i. This gives the following linear system (written in a matrix form):

where P and A are as above and

is the demand vector.

One way to solve this linear system is

Of course, we require here that the matrix I-A be invertible, which might not be always

the case. If, in addition, (I-A)-1

has nonnegative entries, then the components of the vector

P are nonnegative and therefore they are acceptable as solutions for this model. We say in

this case that the matrix A is productive.

Example Consider an open economy with three industries: coal-mining operation, electricity-

generating plant and an auto-manufacturing plant. To produce $1 of coal, the mining

operation must purchase $0.1 of its own production, $0.30 of electricity and $0.1 worth of

automobile for its transportation. To produce $1 of electricity, it takes $0.25 of coal, $0.4 of

electricity and $0.15 of automobile. Finally, to produce $1 worth of automobile, the auto-

manufacturing plant must purchase $0.2 of coal, $0.5 of electricity and consume $0.1 of

automobile. Assume also that during a period of one week, the economy has an exterior

demand of $50,000 worth of coal, $75,000 worth of electricity, and $125,000 worth of autos.

Find the production level of each of the three industries in that period of one week in order to

exactly satisfy both the internal and the external demands.

Solution The input-output matrix of this economy is

and the demand vector is

By equation (*) above,

where

Using the Gaussian-elimination technique (or the formula B-1

=(1/det(B))adj(B)), we find that

which gives

So, the total output of the coal-mining operation must be $229921.59, the total output for the

electricity-generating plant is $437795.27 and the total output for the auto-manufacturing

plant is $237401.57.

If you like to know more about the subject and the life and achievements of W. Leontielf,

check the following links:

Wassily Leontief (life and achievements of W. Leontielf)

http://online.redwoods.cc.ca.us/instruct/darnold/laproj/Fall2001/Iris/lap

http://www.facstaff.bucknell.edu/ap030/Math345LAApplications/Flow.html

Flow: From Traffic to Pipes -

An Application of Linear

Algebra

The most practical applications of mathematics are

those that apply to the most people. When it comes

to linear algebra, one wide-reaching application is to

the problem of flow. Whether it is data through a

network, water through pipes, or cars through traffic,

arranging an efficient flow is a crucial element of

design. Just these three examples already show that

solving flow problems is a wide topic that has

relevance in many aspects of everyday life.

Many of the problems that arise with networks and

traffic involve optimizing the flow through the

system. This can be done in many ways and depends

a lot on the particular situation. However, by looking

at systems of linear equations, we can simplify flow

problems. Linear algebra allows us to study these

systems using matrices and determine which

variables have forced values and which are free to be

chosen. After this distinction has been made using

linear algebra, we can use other techniques to choose

the free variables.

In order to begin understanding these types of

problems, it is best to look at a simple example.

Consider a system of streets with traffic flow as

described in the diagram below, where each number

represents the number of cars passing through in one

hour. In order to optimize traffic flow, the city

planners need to determine how to set the traffic

lights and how many cars to let through streets A, B,

C, D, and E each hour.

First, the picture yields the following four equations:

B + E = 20 + A

C + 30 = B + 40

60 = C + D

D = E + 15

Next, the equations can be rearranged:

-A + B + E = 20

-B + C = 10

C + D = 60

D - E = 15

Now they can be placed into a matrix:

After row-reducing, the matrix becomes:

Finally, this gives four new equations:

A = 15

B + E = 35

C + E = 45

D - E = 15

From these equations, the city planners can see that

A must be 15 cars per hour but they have some

choice for the other streets. If they set E at 10 cars

per hour, for example, B must be 25, C must be 35,

and D must be 25. As another example, consider

shutting down E street completely. Then it follows

that B = 35, C = 45, and D = 15. And if the city

reverses the direction of traffic flow on E and then

allows 20 cars per hour the other streets are affected

as follows: B = 55, C = 65, and D also changes

direction and must admit 5 cars per hour.

When looking at the other applications mentioned

earlier, such as data through a network or water

through pipes, the system can be set up similarly.

The reason that linear algebra is so key in solving

these types of problems is that it allows for solving

many linear equations simultaneously. Matrices use

the symmetry of the equations to solve for the

variables.

Now that it is clear how useful linear algebra can be

in this application, it is important to look at the

process in much more detail.

The Process

The first step is to come up with equations after

looking at the picture, or actual situation. For a more

general map, there needs to be one equation for each

node on the graph. A node is a point where two or

more lines intersect. At each node, the incoming

traffic or flow needs to equal the outgoing traffic or

flow.

The second step is to rearrange these equations in

order to get the constant on one side and all of the

variables on the other side. This is a very simple

process and can be done with general algebra

knowledge.

The third step is to create a matrix. This step is a

little bit more involved. Matrices are measured by

how many rows and columns they have. In general,

we call a matrix with m rows and n columns an m-

by-n matrix . The number of equations from steps 1

and 2 will be the number of rows the matrix has and

the number of columms will be equal to the number

of variables plus 1. Next, the matrix needs to be

filled in. Each column of the matrix corresponds to

one variable, so the coefficients of that variable in

each respective equation should be filled in that

column. The final column will contain the constants

from each equation.

The fourth step is to row-reduce the matrix. There

are four rules that have to be followed:

The Four Rules:

1. All nonzero rows must have a 1 as the furthest left

nonzero entry (called a leading 1).

2. All columns containing leading 1s must have only

zeros everywhere else.

3. All leading 1st must be further to the right than

leading 1s in the above rows.

4. All rows containing only zeros must be at the

bottom.

This step can be accomplished by using the three

elementary row operations below:

The Three Operations:

A. Any two rows can be switched.

B. A row can be multiplied by a nonzero constant, if

each element of the row is multiplied by the same

constant.

C. A row can be replaced by the sum of itself plus

another row.

The fifth step is to convert this matrix back into

linear equations by using the reverse process of step

3 and using the entries in the matrix to determine the

coefficients of the variables and the constants in the

equations.

Now these equations can be used to see which

variables have forced values and which are free to be

chosen. Depending on the goal of the exercise,

different methods can be used to choose the free

variables. Free variables exist when there are more

variables than there are equations. In the above

example, we had five variables, but only four

equations (because there were four nodes).

In problems involving data flowing through

networks, a key issue that arises is efficiency and

speed of travel, but also having a large amount of

data traveling at once. When it comes to water

flowing through pipes, the capacity of the pipes

needs to be taken into consideration, and also any

valves that might restrict flow to only one direction.

For street traffic, the needs of the intersections might

be dependent on time, capacity, direction, or a

number of other factors. In all three situations, cost

can also play a large role as companies and cities

always need to reduce costs. The strength of linear

algebra is that it is broad enough to be used in any

and all of these possible situations and can be applied

to help solve all of these issues.

Linear algebra can be used to solve increasingly

more complex systems of equations, with hundreds

or even thousands of variables. The reason linear

algebra is so useful for these types of problems is

that it can be thought of as an algorithm and the steps

can be used in a computer program to facilitate the

process. Many programmers use matrices to store

and manipulate data due to these reasons. The

concepts explained here can be applied to a variety

of situations in the real world.

A Last Example

Now examine the following pipe flow problem:

One optimization problem could be to maximize the

quantity of water flowing through pipe B. Using the

process described above, the picture can be

transformed into four equations, which can in turn be

turned into a matrix. After row reduction, the

following four equations will be apparent:

A + E = 60

B - F = 80

C - D = -10

D + E + F = 0

Because the goal in this particular situation is

maximizing B, the focus should be on the equation

involving B. Because B and F are output pipes, F

cannot have negative flow. Therefore the minimum

flow F can have is zero, making the maximum value

for B equal to 80. Interestingly, the flow through B

does not actually depend on A or D even though

those pipes interact with the same node as B.

In the real world, it frequently occurs where many

variables are not significant to the final solution of

the system. It all depends on what the network is

designed to do and what parameters are relevant.

There are many directions in which to take this

application.

Sami Prehn

Bucknell University

Math 345, Fall 2009

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