ENGG2013Unit 3 RREF and

Applications of Linear EquationsJan, 2011.

A TRAFFIC FLOW PROBLEMA motivating example

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A Traffic flow problem

• A round-about connecting 5 roads• The traffic within the circle is counter-clockwise.• Question: model the traffic in each section of the round-about

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Modeling a round-about

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100

150

50

150 10050

300

400

200100

x1

x2

x4

x3

x5

Can we find x1, x2, x3, x4 and x5?

The unit is numberof vehicles perhour.

In-flow = out-flow

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100

150

50

150 10050

300

400

200100

x1

x2

x4

x3

x5

We need to solve …

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x1 x2 x3 x4 x5

Representationusing augmented matrix

Last time: Gaussian elimination

• Step 1: Try to transform the matrix into upper triangular form

• Step 2: Backward substitution

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Row operations

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(5) (5)+(1)

(5) (5)+(2)

(5) (5)+(3)

(5) (5)+(4) Equation 5 is redundant

Choose a free variable• In this example, we can pick

any variable as the “free variable”.

• Let’s pick x5 as the free variable for example.

• Expressed in terms of x5, we get:

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x4 = x5 – 100

x3 = (x5 – 100) + 50 = x5 – 50

x2 = (x5 – 50) – 100 = x5 – 150

x1 = (x5 – 150) + 50 = x5 – 100

But x1 to x5 are trafficflow in the round-aboutand cannot be negative.This restricts the value ofx5 to be at least 150. 0

0

0

0

General solution

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where f is a real numberlarger than or equal to 150

Solution:

System of linear equations:

This is called the general solutionbecause all possible solutionscan be put in this form

Discussions• The system of linear equations is

underdetermined.– A car endlessly going around the circle without

exiting is undetectable in this model.– We cannot determine the traffic flow uniquely– There are infinitely many solutions.– How to remove redundant equalities in general?

• The variables in this example are subject to non-negativity constraint.– In many applications, the variables cannot be

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REDUCED ROW ECHELON FORM

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Pivot

• Example from last time

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pivot

(2) (2) – (1)

(3) (3) + (2)/2pivots

Sometimes we cannot find pivot

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(2) (2) – 2(1)

(3) (3) – 3(1)

Cannot find a pivotin the second column

Row Echelon Form (REF)• A leading entry in a row means the first nonzero entry

from the left.• A rectangular matrix is in row echelon form if

– All nonzero rows are above any all-zero row.– All entries below a leading entry are zeros.– In any pair of adjacent nonzero row, say row i and row i+1,

the leading entry in row i is to the left of the leading entry of row i+1.

• Examples (triangles indicate the leading entries)

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Non-examples of REF

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– All nonzero rows are above any all-zero row.– All entries below a leading entry are zeros.– In any pair of adjacent nonzero row, say row i and row i+1, the leading

entry in row i is to the left of the leading entry of row i+1.

Reduced Row Echelon Form (RREF)

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– All nonzero rows are above any all-zero row.– All entries above and below a leading entry are zeros.– In any pair of adjacent nonzero row, say row i and row i+1, the leading

entry in row i is to the left of the leading entry of row i+1.– All leading entries are equal to 1.

Examples

The concept of RREF appliesto all matrices in general, not just augmented matrix.

Non-examples of RREF

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Theorem

• By applying the three types of elementary row operations, we can reduce any rectangular matrix to a matrix in reduced row echelon form (RREF).(In other words, any matrix is row equivalent to a matrix in RREF)

• Furthermore, the RREF of a matrix is unique.

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A row reduction algorithm

1. Scan the columns from left to right.2. Start from the 1st column.3. If this column contains a pivot (a nonzero

entry), move the pivot to the top by exchanging rows

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Algorithm (cont’d)

4. Make all entries below and above the pivot equal to 0.

5. Move to the next column and try to locate a nonzero entry which is not in any row already containing a pivot.

6. Repeat step 5 until you can find such column.

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Algorithm (cont’d)

• Repeat step 3 to step 6 until we reach the right-most column.

• Finally, normalize all leading entries to 1.

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The RREF of is

SOLVING LINEAR EQUATIONS USING RREF

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Parametric representation

• How to represent a solution set?• Example: How to plot points on a straight line?

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y

x

2x+3y=2

We can solve for yin terms of xy = (2 – 2x) / 3

x (2 – 2x) / 3

0 2/3

1 0

2 -2/3

3 -4/3

x is a parameter whose value can be freely chosen.

Parametric representation of circle

• How about a circle x2+y2=1?

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y

x

We can also solve for yin terms of xy = (1 – x2)^(1/2)

x (1 – x2)^0.5

0 1

0.2 0.9798

0.4 0.9165

0.6 0.8

0.8 0.6

1 0

x is a parameter whose value can be freely chosen.

There are many choices for parameters

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y

x

2x+3y=2

x = (2 –3y)/2y = free

We can pick y as the parameter

y

x

We can pick as the parameter

x = cos y = sin

between 0 and 2

Parametric representations of a plane

• 2x + 3y – z = 5

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-2-1

01

2

-2

-1

0

1

2-15

-10

-5

0

5

xy

z

x = freey = freez = 2x+3y–5

If x and y are the parameters,the representation is

x = (5+z–3y)/2 y = freez = free

If y and z are the parameters,the representation is

x = freey = (5+z–2x)/3 z = free

If x and z are the parameters,the representation is

Parametric representation ofthe solutions to a linear system

1. First row reduce the system of linear equations to a reduced row echelon form.

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Solve

Transform to RREF

Pick the free variable(s)

2. Pick the variable(s) which is/are not associated with a column with pivot as “free variable”.

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Pick z as a free variable

x y z

Solve for the non-free variables

• Express the “non-free” variables in terms of the “free” variables.

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General solution

• The general solution to

can be represented parametrically as

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General solution means1) All solutions can bewritten in this form2) Every (x,y,z) in thisform is a solution.

The solutions in set notation

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stands for the set of all triples with real numbers as components

means “belong to”

How to plot the solutions?

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z x = 2+z y= – 1 – 2z

-2 0 3

-1 1 1

0 2 –1

1 3 –3

2 4 –5

3 5 –7

We get (0,3,-2), (1,1,-1), (2,-1,0) etc, as solutions to

z is the parameter

Solution Set

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-5

0

5 -5

0

5

-10

-5

0

5

10

y

x

z

The solutions form a straightline in the 3-D space.

A PRODUCTION MODEL IN ECONOMICS

Application 1

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A production model in economics• Consider a close economy

– one steel plant– one coal mine

• To produce 1 ton of steel, 0.5 ton of coal is consumed by the steel plant.• To produce 1 ton of coal, 0.1 ton of steel is used.

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Steel Plant

50 tons of coal

100 tons of steel

10 tons of steel

Coal mine

100 tons of coal

Question• We want to produce 400 tons of steel and 300 tons of coal.

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30

300

400

200To produce 1 ton of steel, we need 0.5 ton of coal.To produce 1 ton of coal, we need 0.1 ton of steel.

Does not work!

Formulation as a linear system• Suppose that the total output of steel plant is xS and the total output of coal mine

is xC.

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400

0.1 xCTotal output ofthe steel plant

0.1 xC goes tothe coal mine

400 tonsare exported

300

0.5 xS

Total output ofthe coal mine

0.5 xS goes tothe steel plant

300 tonsare exported

Solve a system of two equations

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Solution:xS = 452.63 xC = 526.32

In augmented matrix form

Internal consumption

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Steel Plant

300 tons of coal

Coal mine

400 tons of steel

226.32 tonsof coal

52.63 tonsof steel

xS = 452.63

xC = 526.32

The red links indicate internal consumption

Leontief’s input-output model

• Proposed by Prof. Wassily Leontief (1905~1999) from Harvard.

• He modeled the economy of USA using linear algebra.

• From wikipedia: “Around 1949, Leontief used the primitive computer systems available at the time at Harvard to model data provided by the U.S. Bureau of Labor Statistics to divide the U.S. economy into 500 sectors. Leontief modeled each sector with a linear equation based on the data and used the computer, the Harvard Mark II, to solve the system, one of the first significant uses of computers for mathematical modeling”.

• Nobel prize in economics (1973)kshum ENGG2013 41

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EQUILIBRIUM PRICE IN AN ECONOMY

Application 2

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An example of three industries

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Steel

Coal

Electric Power

0.5

0.4

0.10.6

0.3

0.1

0.2

0.1

0.7

0.1 of the total output fromcoal industry goes to thesteel industry, and 0.7 ofthe total output goes to the electric power industry.

0.4 of the total output fromthe electric power industrygoes to the coal mines, and0.5 of the total output goesto the steel plants

0.6 of the total output from the steel industry goes to the electric powerindustry, 0.3 of the output goes to the coal industry.

Distribution of outputs

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Steel

Coal

Electric Power

0.5

0.4

0.10.6

0.3

0.1

0.2

0.1

0.7

Output FromPurchased bySteel Coal Power

0.1 0.1 0.5 Steel

0.3 0.2 0.4 Coal

0.6 0.7 0.1 Power

Question

• Can we find the prices of steel, coal, power, such that the cost to each industry is balanced with the income – can we find an equilibrium price ?

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Balancing income and expenditure

• Let the price, or value, of steel, coal and electric power be Ps, Pc and Pe respectively.

• From “Expenditure = Income”, we get three equations

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Output FromPurchased bySteel Coal Power

0.1 0.1 0.5 Steel

0.3 0.2 0.4 Coal

0.6 0.7 0.1 Power

Solve the system of equations

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Short-hand notationusing augmented matrix

Find the solutions from reduced row echelon form

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Transform to RREF

Ps Pc Pe

Choose Pe as the free variable

Ps = 44/69 Pe

Pc = 17/23 Pe

Pe = any positive real number

Pivots

free

The equilibrium price• There are infinitely many solutions.

– Any constant multiple of them is also a solution.• For example, if Pe =1, we have

– Ps = 44/69 = 0.64– Pc = 17/23 = 0.74– Pe = 1 (electric power is the most valuable in this example)

• If Pe =100, the equilibrium prices are– Ps = 44/69 = 64– Pc = 17/23 = 74– Pe = 100

• There is no unique solution. It depends on the currency, RMB, Yen, USD etc.

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Summary

• In many problems, the solutions are not unique.

• Reduced row echelon form is a useful in solving for the general solution, especially when the system of linear equations is under-determined.

• Linear algebra has applications in economics, for example in finding equilibrium.

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