1. ENGG2013 Unit 3 RREF and Applications of Linear Equations
Jan, 2011.

2. A motivating example
A TRAFFIC FLOW PROBLEM
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3. A Traffic flow problem A round-about connecting 5 roads
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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4. Modeling a round-about
100
200
150
400 x1 x2 x4 x3 x5
100
300
The unit is number of vehicles per hour. 50 50
150
100
Can we find x1, x2, x3, x4 and x5?
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5. In-flow = out-flow 100 200 150 400 x1 x2 x4 x3 x5 100 300 50 50 150
\begin{align*}
x_1 - x_2 &= = 50\\
x_2 - x_3 &= = -100\\
x_3 - x_4 &= = 50\\
x_4 - x_5 &= = -100\\
x_5 - x_1 &= = 100
\end{align*} 50
150
100
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6. We need to solve … x1 x2 x3 x4 x5
Representation using augmented matrix
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7. Last time: Gaussian elimination
Step 1: Try to transform the matrix into upper triangular form
Step 2: Backward substitution
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8. Row operations (5)  (5)+(1) (5)  (5)+(2) (5)  (5)+(3) (5)  (5)+(4)
Equation 5 is redundant
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9. 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:
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.
 0
x4 = x5 – 100
x3 = (x5 – 100) + 50 = x5 – 50
x2 = (x5 – 50) – 100 = x5 – 150
x1 = (x5 – 150) + 50 = x5 – 100
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10. General solution System of linear equations: Solution:
where f is a real number larger than or equal to 150
This is called the general solution because all possible solutions can be put in this form
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11. 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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12. REDUCED ROW ECHELON FORM
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13. Pivot Example from last time pivot pivots (2)  (2) – (1)
(3)  (3) + (2)/2
pivots
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14. 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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15. 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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16. 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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17. 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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18. Non-examples of RREF
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19. 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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20. A row reduction algorithm
Scan the columns from left to right.
Start from the 1st column.
If this column contains a pivot (a nonzero entry), move the pivot to the top by exchanging rows
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21. Algorithm (cont’d)
Make all entries below and above the pivot equal to 0.
Move to the next column and try to locate a nonzero entry which is not in any row already containing a pivot.
Repeat step 5 until you can find such column.
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22. 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 is
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23. SOLVING LINEAR EQUATIONS USING RREF
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24. 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 x
(2 – 2x) / 3
2/3 1 2
-2/3 3
-4/3
2x+3y=2 x
x is a parameter whose value can be freely chosen.
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25. 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
 (1 – x2)^0.5 1
0.2
0.9798
0.4
0.9165
0.6
0.8
0.8
0.6 1 x
x is a parameter whose value can be freely chosen.
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26. There are many choices for parameters
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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27. Parametric representations of a plane
2x + 3y – z = 5
If x and y are the parameters, the representation is
x = free
y = free
z = 2x+3y–5
If y and z are the parameters, the representation is
If x and z are the parameters, the representation is
x = (5+z–3y)/2
y = free
z = free
x = free
y = (5+z–2x)/3
z = free
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28. Parametric representation of the solutions to a linear system
First row reduce the system of linear equations to a reduced row echelon form.
Solve
Transform to RREF
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29. Pick the free variable(s)
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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30. Solve for the non-free variables
Express the “non-free” variables in terms of the “free” variables.
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31. 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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32. The solutions in set notation
stands for the set of all triples with real numbers as components
means “belong to”
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33. How to plot the solutions?
z is the parameter z
x = 2+z
y= – 1 – 2z -2 3 -1 1 2 –1 –3 4 –5 5 –7
We get (0,3,-2), (1,1,-1), (2,-1,0) etc, as solutions to
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34. Solution Set The solutions form a straight line in the 3-D space.
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35. A PRODUCTION MODEL IN ECONOMICS
Application 1
A PRODUCTION MODEL IN ECONOMICS
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36. 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
Coal mine
50 tons of coal
100 tons of steel
10 tons of steel
100 tons of coal
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37. Question We want to produce 400 tons of steel and 300 tons of coal.
200
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.
Does not work! 30
300
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38. 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
Total output of the steel plant
0.1 xC goes to the coal mine
400 tons are exported
0.1 xC
300
0.5 xS
Total output of the coal mine
0.5 xS goes to the steel plant
300 tons are exported
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39. Solve a system of two equations
In augmented matrix form
Solution: xS =
xC =
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40. Internal consumption xS = 452.63 Steel Plant 400 tons of steel
tons of coal
52.63 tons
of steel
Coal mine
300 tons of coal
xC =
The red links indicate internal consumption
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41. 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)
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42. EQUILIBRIUM PRICE IN AN ECONOMY
Application 2
EQUILIBRIUM PRICE IN AN ECONOMY
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43. An example of three industries
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.
Steel
0.1
0.6
0.1
0.5
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
0.1
0.3
0.4
Electric Power
0.7
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
Coal
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44. Distribution of outputs
Steel
0.1
0.1
0.6
Electric Power
0.5
0.1
0.3
0.4
Output From
Purchased by
Steel
Coal
Power
0.1
0.5
0.3
0.2
0.4
0.6
0.7
0.7
0.2
Coal
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45. 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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46. 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
Purchased by
Steel
Coal
Power
0.1
0.5
0.3
0.2
0.4
0.6
0.7
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47. Solve the system of equations
Short-hand notation using augmented matrix
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48. Find the solutions from reduced row echelon form
Choose Pe as the free variable
Ps = 44/69 Pe
Pc = 17/23 Pe
Pe = any positive real number
Transform to RREF
Pivots
free Ps Pc Pe
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49. 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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50. 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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