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Unit 3 RREF and
Applications of Linear Equations
Jan, 2011.
A motivating example
A TRAFFIC FLOW PROBLEM
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http://www.beyond.ca/how-to-navigate-a-traffic-circle-roundabout/210.html
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
100
200
150
400
x1
100
x5
300
x2
x4
50
The unit is number
of vehicles per
hour.
x3
50
150
100
Can we find x1, x2, x3, x4 and x5?
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In-flow = out-flow
100
200
150
400
x1
100
x5
x2
300
x4
50
x3
50
150
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100
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We need to solve …
x1
x2
x3
x4
x5
Representation
using augmented matrix
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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
(5)  (5)+(1)
(5)  (5)+(2)
(5)  (5)+(3)
(5)  (5)+(4)
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Equation 5 is redundant
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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
x = x – 100
0
get:
4
5
x3 = (x5 – 100) + 50 = x5 – 50  0
But x1 to x5 are traffic
flow in the round-about
and cannot be negative.
This restricts the value of
x5 to be at least 150.
x2 = (x5 – 50) – 100 = x5 – 150  0
x1 = (x5 – 150) + 50 = x5 – 100  0
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General solution
System of linear equations:
Solution:
This is called the general solution
because all possible solutions
can be put in this form
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where f is a real number
larger than or equal to 150
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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
negative.
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REDUCED ROW ECHELON FORM
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Pivot
• Example from last time
pivot
(2)  (2) – (1)
(3)  (3) + (2)/2
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pivots
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Sometimes we cannot find pivot
(2)  (2) – 2(1)
(3)  (3) – 3(1)
Cannot find a pivot
in the second column
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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
– 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.
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Reduced Row Echelon Form (RREF)
– 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.
The concept of RREF applies
to all matrices in general,
not just augmented matrix.
Examples
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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.
The RREF of
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is
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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?
We can solve for y
in terms of x
y = (2 – 2x) / 3
y
2x+3y=2
x
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.
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Parametric representation of circle
• How about a circle x2+y2=1?
We can also solve for y
in terms of x
y = (1 – x2)^(1/2)
y
x
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.
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There are many choices for
parameters
y
y
2x+3y=2

x
x
We can pick y as the parameter
We can pick  as the parameter
x = (2 –3y)/2
y = free
x = cos 
y = sin 
 between 0 and 2
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Parametric representations of a
plane
• 2x + 3y – z = 5
5
If x and y are the parameters,
the representation is
z
0
-5
-10
x = free
y = free
z = 2x+3y–5
-15
2
1
2
1
0
0
-1
If y and z are the parameters,
the representation is
x = (5+z–3y)/2
y = free
z = free
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y
-1
-2
-2
If x and z are the parameters,
the representation is
x = free
y = (5+z–2x)/3
z = free
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Parametric representation of
the solutions to a linear system
1. First row reduce the system of linear
equations to a reduced row echelon form.
Solve
Transform to RREF
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Pick the free variable(s)
2. Pick the variable(s) which is/are not
associated with a column with pivot as “free
variable”. x
y
z
Pick z as a
free variable
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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
General solution means
1) All solutions can be
written in this form
2) Every (x,y,z) in this
form is a solution.
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The solutions in set notation
stands for the set of all triples with real numbers as components
means “belong to”
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How to plot the solutions?
z is the parameter
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
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Solution Set
The solutions form a straight
line in the 3-D space.
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z
5
0
-5
5
-10
-5
0
0
5
-5
y
x
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Application 1
A PRODUCTION MODEL IN
ECONOMICS
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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.
Steel Plant
50 tons
of coal
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Coal mine
100 tons
of steel
10 tons
of steel
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100 tons
of coal
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Question
• We want to produce 400 tons of steel and 300 tons of coal.
400
200
Does not work!
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To 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.
300
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•
Formulation as a linear system
Suppose that the total output of steel plant is xS and the total output of coal mine
is xC.
400
0.1 xC
Total output of
the steel plant
0.1 xC goes to
the coal mine
400 tons
are exported
0.5 xS
Total output of
the coal mine
0.5 xS goes to
the steel plant
300 tons
are exported
300
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Solve a system of two equations
In augmented matrix form
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Solution:
xS = 452.63
xC = 526.32
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Internal consumption
xS = 452.63
Steel Plant
400 tons
of steel
226.32 tons
of coal
52.63 tons
of steel
Coal mine
300 tons
of coal
xC = 526.32
The red links indicate internal consumption
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• 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)
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http://en.wikipedia.org/wiki/Wassily_Leontief
Leontief’s input-output model
Application 2
EQUILIBRIUM PRICE IN AN
ECONOMY
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An example of three industries
Steel
0.1
0.6
0.1
0.1 of the total output from
coal industry goes to the
steel industry, and 0.7 of
the total output goes to the
electric power industry.
0.5
0.1
0.3
0.4
0.7
Coal
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Electric Power
0.4 of the total output from
the electric power industry
goes to the coal mines, and
0.5 of the total output goes
to the steel plants
0.6 of the total output from the steel
industry goes to the electric power
industry, 0.3 of the output goes to
the coal industry.
0.2
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Distribution of outputs
Steel
0.1
0.1
0.6
Electric Power
0.5
0.1
0.3
0.4
Output From
Steel
Coal
Power
Purchased by
0.1
0.1
0.5
Steel
0.3
0.2
0.4
Coal
0.6
0.7
0.1
Power
0.7
Coal
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0.2
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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
Output From
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Steel
Coal
Power
Purchased by
0.1
0.1
0.5
Steel
0.3
0.2
0.4
Coal
0.6
0.7
0.1
Power
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Solve the system of equations
Short-hand notation
using augmented matrix
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Find the solutions from reduced
row echelon form
Choose Pe as the free variable
Ps = 44/69 Pe
Pc = 17/23 Pe
Transform
to RREF
Pe = any positive real number
Pivots
free
Ps
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Pc
Pe
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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 underdetermined.
• Linear algebra has applications in economics,
for example in finding equilibrium.
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