7 linear algebraic equations.pdf

8/11/2019 7 Linear Algebraic Equations.pdf

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Linear Algebraic Equations

Cheng-Liang ChenPSELABORATORY

Department of Chemical EngineeringNational TAIWAN University


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Chen CL 1

Linear Algebraic Equations and Gauss Elimination

x+y+ 2z = 2 (1)3x y+z = 6 (2)

x+ 3y+ 4z = 4 (3)

(3)(1) 2y+ 2z = 2 (4)(2)+3(1) 2y+ 7z = 12 (5)

(5)

(4) 5z = 10 (6) ( z= 2)z=2 in (4) y = 1 (7)

z=2,y=1 in (1)

x = 1 (8)


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Chen CL 2

Linear Algebraic Equations and Gauss Elimination

1. Equation (1) is thepivotequation

2. Multiply it by

1 and add the result to (3)

to obtain2y+ 2z = 2 (equivalent toy+ z = 1)

x+y+ 2z = 2 (1)3x

y+z = 6 (2)

x+ 3y+ 4z = 4 (3)

3. Next multiply (1) by3 and add the result to (2)to obtain 2y+ 7z= 12

Thus we have a new set of two equations in twounknowns (y, z):

y+z = 1 (4)

2y+ 7z = 12 (5)

4. Equation (4) is the newpivotequation.

5. Multiply (4) by2 and add the result to (5)to obtain 5z = 10 (orz = 2)

6. Substitutez = 2 into (4) to obtain y+ 2 = 1 (ory = 1)

7. Then substitutey= 1 andz = 2 into (1)to obtainx 1 + 4 = 2 (orx= 1)


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Chen CL 3

Test Your Understanding

T6.1-1

Solve the following equations using Gauss elimination:

6x 3y+ 4z = 4112x+ 5y

7z =

26

5x+ 2y+ 6z = 14

(Answer:x= 2, y= 3, z = 5.)


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Chen CL 4

Singular and Ill-Conditioned Problems

3x 4y = 56x 10y = 2

unique solution, x= 7, y= 4

3x 4y = 56x 8y = 10

singular, infinite # solution

3x

4y = 5

6x 8y = 3 singular, no solution

C C


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Chen CL 5

Ch CL 6


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Chen CL 6

function P363b

x = [6 : 0.25 : 8];

y_1 = (3*x-5)/4;

y_2 = (6*x-2)/10;

subplot(3,1,1)plot(x,y_1,x,y_2),...

xlabel(x),ylabel(y),...

title(Unique solution),...

legend(Line 1,Line 2)

y_3 = (6*x-10)/8;

subplot(3,1,2)plot(x,y_1,x,y_3,o),...


title(Singular, No unique soln),...


y_4 = (6*x-3)/8;

subplot(3,1,3)

plot(x,y_1,x,y_4),...


title(Singular, No solution),...


Ch CL 7


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Chen CL 7

Homogeneous Equations (zero RHSs)

6x+ay = 0 (1)

2x+ 4y = 0 (2)

(1)3(2)

(a 12)y = 0

Case 1: y = 0 only ifa = 12y=0 in (1)

x = 0

Case 2: 0y = 0 ifa= 12

x = 2y infinite # of solutions

Ch CL 8


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Chen CL 8

Ill-Conditioned EquationsAnIll-Conditionedset of equations is a set that is close to being

singular

3x 4y = 56x 8.002y = 3

y = 3x 54

x= 4668y =

3x 1.54.001

y= 3500

But if we had carried only two significant figures We would have rounded the denominator of the latter

expression to 4.0

Two expressions for y have the same slope (parallel)

No solution ill-conditioned status (soln dep.s on accuracy)

Ch CL 9


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Chen CL 9


T6.1-2Show that the following set has no solution.

4x+ 5y = 10

12x 15y = 8

T6.1-3

For what value ofb will the following set have a solution in which

both x and y are nonzero? Find the relation between x and y.

4x by = 03x+ 6y = 0

(Answer: Ifb= 8, x= 2. Ifb = 8, x=y= 0.)

Chen CL 10


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Chen CL 10

Matrix Methods for Linear Equations

2x1+ 9x2 = 5

3x1

4x2 = 7

2 9

3 4

A

x1

x2

x

=

5

7

b

Ax = b

a11x1+a12x2+ +a1nxn = b1a21x1+a22x2+ +a2nxn = b2

am1x1+am2x2+ +amnxn = bm

a11 a12 a1na21 a22 a2n

am1 am2 amn

A

x1

x2...

xn

x

=

b1

b2...

bm

b

Ax = b

Chen CL 11


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Chen CL 11

Determinant and Singular Problems

A=

3 4 16 10 2

9 7 8

A = [3,-4,1; 6,10,2; 9,-7,8];

det(A)

ans =

8


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Chen CL 13


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Chen CL 13

Left-division Method With Three Unknowns

Example: Use the left-division method to solve the following set:

3x+ 2y 9z = 659x 5y+ 2z = 16

6x+ 7y+ 3z = 5

Solution:

A =

3 2

9

9 5 26 7 3

We can use MATLAB to check the determinant ofA to see

whether the problem is singular.

Chen CL 14


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Chen CL 14

A = [3, 2, -9; -9, -5, 2; 6, 7, 3];

b = [-65; 16; 5];det_A = det(A),...

soln = A\b ,... % Ax=b

A*soln

det_A =

288

soln =

2.0000-4.0000

7.0000

ans =

-65.0000

16.0000

5.0000

Chen CL 15


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Chen CL 15

Example: An Electrical-resistance Network

The circuit shown in the following Figure has five resistances and applied voltages.Assuming that the positive directions of current flow are in the directions shown inthe figure, Kirchhoffs voltage law applied to each loop in the circuit gives

v1+R1i1+R4i4 = 0R4+R2i2+R5i5 = 0R5i5+R3i3+v2 = 0

Conservation of charge applied at each node in the circuit gives

i1 = i2+i4

i2 = i3+i5

You can use these two equations to eliminate i4 andi5 from the first three

Chen CL 16


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Chen CL 16

equations. The results is:

(R1+R4)i1 R4i2 = v1

R4i1+ (R2+R4+R5)i2

R5i3 = 0

R5i2 (R3+R5)i3 = v2

Thus we have three equations in three unknowns: i1, i2, and i3.

Write a MATLAB script file that uses given values of the applied voltages v1andv2 and given values of the five resistances to solve for the currents i1, i2, and

i3. Use the program to find the currents for the case R1= 5, R2= 100,

R3= 200, R4= 1504, R5= 250 k, v1= 100, andv2= 50 volts. (Note that 1

k = 1000)

Chen CL 17


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Chen CL 17

Solution:

% File resist.m % Solvers for the currents i_1, i_2, i_3

R = [5, 100, 200, 150, 250]*1000;v1 = 100; v2 = 50;

A1 = [R(1) + R(4), -R(4), 0];

A2 = [-R(4), R(2) + R(4) + R(5), -R(5)];

A3 = [0, R(5), -(R(3) + R(5))];

A = [A1; A2; A3];

b = [v1; 0; v2];current = A\b;

disp(The currents are:)

disp(current)

>> resist % run resist.m

The currents are:

1.0e-003*

0.9544

0.3195

0.0664 % i_1,i_2,i_3 = 0.9544, 0.3195, 0.0664 mA

Chen CL 18


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Chen CL 18

Example: Ethanol Production

Engineers in the food and chemical industries use fermentation in many processes.

The following equation describes Bakers yeast fermentation.

a(C6H12O6) +b(O2) +c(N H3)

C6H10N O3+d(H2O) +e(CO2) +f(C2H6O)

The variables a, b, . . . , f represent the masses of the products involved in theequation. In this formula C6H12O6 representsglucose, C6H10N O3 representsyeast, andC2H6O representsethanol. This reaction produces ethanol, in additionto water and carbon dioxide. We want to determine the amount of ethanol fproduced. The number ofC, O, N, and Hatoms on the left must balance thoseon the right side of the equation. This gives four equations:

Cbalance 6a = 6 +e+ 2f

O balance 6a+ 2b = 3 +d+ 2e+f

N balance c = 1

Hbalance 12a+ 3c = 10 + 2d+ 6f

Chen CL 19


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Chen CL 19

The fermentor is equipped with anoxygensensor and acarbon dioxidesensor.These enable us to compute the respiratory quotient R:

R=CO2

O2=

e

b

Thus the fifth equation is Rb e= 0. The yeast yield Y (grams of yeastproduced per gram of glucose consumed) is related to a as follows .

Y =

144

180a

Where 144 is the molecular weight of yeast and 180 is the molecular weight of

glucose. By measuring the yeast yield Ywe can compute a as follows:

a= 144/180Y. This is the sixth equation.

Write a user-defined function that computes f, the amount of ethanol produced,

withR andYas the functions arguments. Test your function for two cases where

Y is measure to be 0.5: (a) R= 1.1 and (b) R= 1.05.

Chen CL 20


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Chen CL 20

Solution: let x1 b, x2 d, x3 e, x4 f

x3

2x4 = 6

6(144/180Y)

2x1 x2 2x3 x4 = 3 6(144/180Y)2x2 6x4 = 7 12(144/180Y)

Rx1 x3 = 0

In matrix form:

0 0 1 2

2 1 2 10 2 0 6

R 0 1 0

x1

x2x3

x4

= 6 6(144/180Y)

3 6(144/180Y)7 12(144/180Y)

0

The function file and the the session:

Chen CL 21


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funtion E = ethanol(R,Y)

% Computes ethanol produced

% from yeast reaction.

A = [0, 0,-1,-2; 2,-1,-2,-1; ...0,-2, 0,-6; R, 0,-1, 0];

b = [6 - 6*(144./(180*Y)); ...

3 - 6*(144./(180*Y)); ...

7 - 12*(144./(180*Y)); 0];x = A\b;

E = x(4);

E_1 = ethanol(1.1, 0.5)

E_2 = ethanol(1.05,0.5)

E_1 =

0.0654E_2 =

-0.0717

Chen CL 22


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Matrix Inverse

Ax = b

A1A = AA1 = I

A1Ax = A1b

x = A1b

Chen CL 23


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Example: Calculation Of Cable Tension

A mass m is suspended by three cables attached at the three points B, C, and D,

as shown in the following figure. Let T1, T2, and T3 be the tensions in the threecables AB, AC, and AD, respectively. If the mass m is stationary, the sum of thetension components in the x, in the y, and in the z directions must each be zero.This requirement gives the following three equations:

T135

3T234

+ T3

42= 0

3T135

4T342

= 0

5T135

+ 5T2

34+ 5T

342

mg = 0

Use MATLAB to findT1, T2, andT3 in terms of an unspecified value of theweight mg.

Solution: set mg= 1

Chen CL 24


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A =

135

334

142

335

0 442

535

534

542

x =

T1

T2

T3

b =

0

0

1

% File cable.m

% Computes the tensions in three cables.

A1 = [1/sqrt(35), -3/sqrt(34), 1/sqrt(42)];

A2 = [3/sqrt(35), 0, -4/sqrt(42)];

A3 = [5/sqrt(35), 5/sqrt(34), 5/sqrt(42)];

A = [A1; A2; A3];

b = [ 0; 0; 1];

x = A\b;

disp(The tension T_1 is:)

disp(x(1))disp(The tension T_2 is:)

disp(x(2))

disp(The tension T_3 is:)

disp(x(3))

The tension T_1 is:

0.5071 % 0.5071 mg

The tension T_2 is:

0.2915 % 0.2915 mg

The tension T_3 is:0.4166 % 0.4166 mg

Chen CL 25


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Example: The Matrix Inverse Method

Solve the following equations using the matrix inverse:

2x+ 9y = 5

3x 4y = 7

Solution:

A = 2 9

3 4

Its determinant is|A| = 2(4) 9(3) = 35, and its inverse is

A

1

=

1

35 4 93 2 = 135 4 93 2 The solution is

x = A1b = 1

35 4 9

3 2 5

7 = 1

35 83

1

Chen CL 26


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The Matrix Inverse inMATLAB

A = [2, 9; 3, -4];

b = [5; 7]x = inv(A)*b

x =

2.37140.0286

Chen CL 27


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T6.2-1

Use the matrix inverse method to solve the following set by handby using MATLAB: (Answer:x= 7, y= 4.)

3x 4y = 5

6x 10y = 2

T6.2-2

Use the matrix inverse method to solve the following set by hand

and by using MATLAB:

3x 4y = 56x 8y = 2

(Answer: no solution.)

Chen CL 28


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Cramers Method

a11x+a12y = b1

a21x+a22y = b2 a22(a11x+a12y) = a22b1a12(a21x+a22y) = a12b2

x = b1a22

b2a12

a22a11 a12a21 =

b1 a12

b2 a22

a11 a12

a21 a22

D1

D

y = b2a11 b1a21a22a11 a12a21

=

a11 b1

a21 b2

a11 a12a21 a22

D2

D

Note: IfD= 0 but D1 = 0 then x is undefinedIfD= 0 andD1= 0 then x has infinitely many soln.s

Chen CL 29


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T6.3-1Use Cramers method to solve for x and y in terms of the

parameter b. For what value ofb is the set singular ?

4x

by = 5

3x+ 6y = 3

(Answer: x= (10 +b)/(8 b), y = 9/(8 b) unless b= 8.)

Chen CL 30


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T6.2-2

UseCramers method to solve for y.

UseMATLAB to evaluate the determinants.

2x+y+ 2z = 17

3y+z = 6

2x 3y+ 4z = 19

(Answer: y= 1.)


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Chen CL 32


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2x 4y+ 5z = 44x 2y+ 3z = 4

2x+ 6y 8z = 0A = [2,-4,5; -4,-2,3; 2,6,-8];

b = [-4, 4, 0];

soln_left_div = A\b

soln_pseudoin = pinv(A)*b

Warning: Matrix is singular

to working precision.

soln_left_div =

Inf

Inf

Inf

soln_pseudoin =

-1.2148

0.2074

-0.1481

Chen CL 33


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Matrix Rank

Anm

n matrix A has a rank r

1 if and only if

|A

|contains a

nonzeror r determinant and every square sub-determinant withr+ 1 or more rows is zero

Example: The rank ofA=

3 4 16 10 2

9 7 3

is 2 because|A| = 0

whereas A contains at least one nonzero 2 2 subdeterminant.

eg,|A| =

10 2

7 3

= 44

A = [3,-4,1; 6,10,2; 9,-7,3];

rank(A)

ans =

2

Chen CL 34


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Existence and Uniqueness of Solutions

Existence and Uniqueness of SolutionsThe set Ax= b with m equations and n unknowns has solutions

if and only ifrank(A) = rank([A b])

Letr= rank(A).

1. If previous condition is satisfied and ifr=n,

then the solution is unique

2. If previous condition is satisfied and but r < n,

an infinite number of solution exists andrunknown variables can

be expressed as linear combination of the other n r unknownvariables, whose values are arbitrary.

Chen CL 35


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Homogeneous Case

1. For the homogeneous set Ax = 0, rank(A) = rank([A b])

always, and thus the set always has the trivial solution x= 0.

2. A nonzero solution (at least one unknown is nonzero) exists ifand only ifrank(A) < n

3. Ifm < n, the homogeneous set always has a nonzero solution

Chen CL 36


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Example: A Set Having A Unique Solution

Determine whether the following set has a unique solution, and ifso,find it:

3x 2y+ 8z = 486x+ 5y+z = 129x+ 4y+ 2z = 24

Solution:

A =

3 2 8

6 5 1

9 4 2

b =

48

12

24

x =

x

y

z

Chen CL 37


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A = [3, -2, 8; -6, 5, 1; 9, 4, 2];

b = [48; -12; 24];

rank(A)

ans =3

rank([A b])

ans =

3

x=A\b

x =

2

-1

5

Chen CL 38

Y U d di


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T6.4-1

UseMATLAB to show that the following set has a unique solutionand then find the solution:

3x+ 12y 7z = 55x 6y 5z = 8

2x+ 7y+ 9z = 5

(Answer: The unique solution is

x= 1.0204, y= 0.5940, z= 0.1332.)

Chen CL 39

Th Mi i E lid N S l i


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The Minimum Euclidean Norm Solution

Thepinvcommand can obtain a solution of an

underdetermined set

v = [x y z]

N =

vTv =

[x y z]T

xy

z

=

x2 +y2 +z2

Chen CL 40

E l A U d d i d S


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Example: An Under-determined Set

Show that the following set does not have a unique solution. How many of the

unknowns will be undetermined ? Interpret the results given by the left-divisionmethod.

2x 4y+ 5z = 44x 2y+ 3z = 4

2x+ 6y 8z = 0

Solution:A = [2, -4, 5; -4, -2, 3;...

2, 6, -8];

b = [-4; 4; 0];r1 = rank(A)

r2 = rank([A b])

r1 =

2

r2 =

2

infinite no. of solutions

soln = pinv(A)*b

soln =

-1.2148

0.2074

-0.1481

Chen CL 41

E A S i ll I d i P bl


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Ex: A Statically-Indeterminate Problem

Determine the forces in the three equally spaced

supports that hold up a light fixture. The supports

are5 feet apart. The fixture weights 400 pounds,

and its mass center is 4 feet from the right end.

(a) Solve the problem by hands. (b) obtain the

solution using the MATLABleft-division method

and the pseudoinverse method.

Solution:

vertical force must cancel; total moments about right endpoint are zero

Chen CL 42


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T1+T2+T3 400 = 0400(4)

10T1

5T2 = 0

or T1+T2+T3 = 400

10T1+ 5T2+ 0T3 = 1600

1 1 110 5 0

A

T1

T2

T3

x

= 400

1600

b

[A b] = 1 1 1 40010 5 0 1600 T2 =

1600 10T15

= 320 2T1T1 = T3 80T2 = 320

2T1= 320

2(T3

80) = 480

2T3

Chen CL 43

l l ft A\b


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A = [1, 1, 1; 10, 5, 0];

b = [400; 1600];

rank(A)

ans =

2

rank([A b])

ans =

2

soln_left = A\b

soln_pseu = pinv(A)*b

soln_left =

160.0000

0240.0000

soln_pseu =

93.3333 % min norm soln

133.3333

173.3333

norm_left = sqrt(sum(soln_left.^2))norm_pseu = sqrt(sum(soln_pseu.^2))

norm_left =

288.4441

norm_pseu =

237.7674


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Chen CL 45

Th R d d R E h l F


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The Reduced Row Echelon Form

T1

= T3

80

T2 = 480 2T3 T1 T3 = 80

T2+ 2T3 = 480

1 0 10 1 2

T1

T2

T3

= 80480

1 0 1 800 1 2 480

rref([A b])command provides a procedure to reduce an underdetermined set tosuch a reduced row echelon form

Its output is the augmented matrix [C d] that corresponds to the equation set

Cx= d

Chen CL 46

E l A Si l S t


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Example: A Singular Set

The following under-determined equation set was analyzed in

previous Example. There it was shown that an infinite number of

solutions exists. Use the pinv and the rrefcommands to obtain

solutions.

2x 4y+ 5z = 4

4x 2y+ 3z = 42x+ 6y 8z = 0

Solution:

Chen CL 47

A = [2 4 5; 4 2 3;


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A = [2, -4, 5; -4, -2, 3;...

2, 6, -8];

b = [-4; 4; 0];

x = pinv(A)*b

x =-1.2148

0.2074

-0.1481

rref([A b])

ans =

1 0 -0.1 -1.2000

0 1 -1.3 0.40000 0 0 0

The answer corresponds to the augmented matrix [C d],

[C d] =

1 0 0.1 1.20 1 1.3 0.40 0 0 0

The matrix corresponds to the matrix equation Cx= d, or

x+ 0y 0.1z = 1.20x+y 1.3z = 0.40x+ 0y+ 0z = 0.0

x = 0.1z 1.2y = 1.3z+ 0.4

z = arbitrary value

Chen CL 48

E a le P od ctio Pla i


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Example: Production Planning

The following table shows how many hours reactors A and B need to produce 1ton each of the chemical products 1, 2, and3. The two reactors are available for

40 hours and 30 hours per week, respectively. Determine how many tons of eachproduct can be produced each week.

Hours Product 1 Product 2 Product 3

Reactor A 5 3 3

Reactor B 3 3 4

Solution:

5x+ 3y+ 3z = 40

3x+ 3y+ 4z = 30

A =

5 3 3

3 3 4

b =

40

30

x =

x

y

z

Chen CL 49

Note: rank(A) = rank([A b]) = 2 which is less than the number of unknowns


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Note: rank(A) = rank([A b]) = 2, which is less than the number of unknowns(3). Thus an infinite number of solution exists, and we can determine two of thevariables in terms of the third.Usingrref([A b]), whereA=[5,3,3; 3,3,4]andb=[40;30], we obtain the following

reduced echelon augmented matrix,

1 0 0.5 50 1 1.8333 5

x 0.5z = 5y+ 1.8333z = 5

x = 5 + 0.5zy = 5 1.8333z

Suppose we make a profit of$400, $600, $100 per ton for products 1, 2 and3,

Chen CL 50

respectively


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respectively.

P = 400x+ 600y+ 100z

= 400(5 + 0.5z) + 600(5

1.8333z) + 100z

= 5000 800z

To maximize profit P, choose z = 0 x= y = 5 tons.

Chen CL 51

Example: Traffic Engineering


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Example: Traffic Engineering

A traffic engineer wants to know whether measurements of traffic flow entering

and leaving a road network are sufficient to predict the traffic flow on each street

in the network. For example, consider the network of one-way streets shown in the

following Figure. The numbers in the figure give the measured traffic flows in

vehicles per hour. Assume that no vehicles park anywhere within the network. If

possible, calculate the traffic f1, f2, f3, andf4. If this is not possible, suggest

how to obtain the necessary information.

Chen CL 52

Solution:


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Solution:

100 + 200 = f1+f4

f1+f2 = 300 + 200

600 + 400 = f2+f3

f3+f4 = 300 + 500

A =

1 0 0 1

1 1 0 00 1 1 0

0 0 1 1

b = 300

5001000

800

x = f1

f2f3

f4

rref([A b])

1 0 0 1 3000 1 0 1 2000 0 1 1 800

0 0 0 0 0

f1 = 300 f4

f2 = 200 +f4

f3 = 800 f4

Chen CL 53



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T6.4-3

Use the rref and pinv commands and the left-division method to solve thefollowing set:

3x+ 5y+ 6z = 6

8x

y+ 2z = 1

5x 6y 4z = 5

(Answer: The set has an infinite number of solutions. The result obtainedwith the rref commands is x = 0.2558 0.3721z, y = 1.0465 0.9767z, z isarbitrary. The pinv commands gives x= 0.0571, y = 0.5249, z = 0.5340. The

left-division method generates an error message.)

Chen CL 54

T6.4-4


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T6.4 4Use the rref and pinv commands and the left-division method to solve thefollowing set:

3x+ 5y+ 6z = 4x 2y 3z = 10

(Answer: The set has an infinite number of solutions. The result obtained withtherrefcommands isx = 0.2727z + 5.2727 0.3721z,y = 1.3636z 2.3636,z is arbitrary. The pinv commands gives x= 4.8000, y = 0, z = 1.7333. Thepseudoinverse method gives x= 4.8396, y= 0.1972, z= 1.5887.)

Chen CL 55

Overdetermined Systems: The Least


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Overdetermined Systems: The LeastSquares Method

Suppose we have the following three data points, and we want to find the straightline y=mx+b that best fits the data in some sense.

x y0 2

5 6

10 11

(a) Find the coefficients m andb by using the least squares criterion. (b) Find the

coefficients by using MATLAB to solve the three equations (one for each data

point) for the two unknowns m andb. Compare the answers from (a) and (b).

Solution:

Chen CL 56


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J=

i=3i=1

(mxi+b yi)2

J= (0m+b 2)2

+ (5m+b 6)2

+ (10m+b 11)2

Chen CL 57

J2(5 b 6)(5) 2(10 b 11)(10)


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m = 2(5m+b 6)(5) + 2(10m+b 11)(10)

= 250m+ 30b 280 = 0J

b

= 2(b

2) + 2(5m+b

6) + 2(10m+b

11)

= 30m+ 6b 38 = 0

250m+ 30b = 280

30m+ 6b = 38 m= 0.9, b= 11

6 (y = 0.9x+ 11/6)

Evaluate y=mx+b at each data point:

0m+b = 2

5m+b = 6

10m+b = 11


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Chen CL 59

Example: An Over-determined Set


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Example: An Over-determined Set

(a) Solve the following equations by hand and (b) solve them using MATLAB.

Discuss the solution for two cases: c= 9 andc= 10.

x+y = 1

x+ 2y = 3

x+ 5y = c

Solution:

A =

1 1

1 2

1 5

[A b] = 1 1 1

1 2 3

1 5 c

c= 9 rank(A) = rank([A b]) = 2 A\bgives unique solution x= 1, y= 2


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Chen CL 61

Fitting Models by Least Squares


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Fitting Models by Least Squares

inputs = xi1, xi2,

, xip output=yi, i= 1, . . . , n

(y1 ; x11, x12, , x1p) (1st observation data)(y2 ; x21, x22, , x2p) (2nd observation data)

... ...

(yn ; xn1, xn2, , xnp) (nth observation data)y = 1x1+2x2+ +pxp (linear model)

y1 = 1x11+2x12+ +px1p+1y2 = 1x21+2x22+ +px2p+2

... ...

yn

obs= 1xn1+2xn2+ +pxnp model output

+ n

error


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Chen CL 64



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T6.5-1

Use MATLAB to solve the following set:

x 3y = 23x+ 5y = 7

70x

28y = 153

(Answer: The unique solution, x = 2.2143, y = 0.0714, is given by theleft-division method.)

T6.5-2

Use MATLAB to solve the following set:

x 3y = 23x+ 5y = 7

5x

2y =

4

(Answer: no exact solution.)

Chen CL 65

AMATLABProgram to Solve Linear Equations


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g t S q t

Chen CL 66

% Script file lineq.m


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p q

% Solve the set Ax=b, given A and b

% Check the ranks of A and [A b]

if rank(A) == rank([A b]) % A, [A b]: equal ranks

size_A = size(A);

if rank(A) == size_A(2) % rank of A = no unknowns

disp(There is a unique solution: )

x = A\b % solve using left division

else % rank of A # no unknownsdisp(There is an infinite no of solutions.)

disp(The augmented matrix of reduced system: )

rref([A b]) % compute augmented matrix

endelse % A, [A b]: not equal ranks

disp(There are no solutions )

end

Chen CL 67

A = [1, -1; 1, 1]; b = [3; 5];


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lineq

There is a unique solution:

x =

41

A = [1, -1; 2, -2]; b = [3; 6];

lineq

There is an infinite no of solutions.

The augmented matrix of reduced system:

ans =

1 -1 3

0 0 0

A = [1, -1; 2, -2]; b = [3; 5];

lineq

There are no solutions

Chen CL 68


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Thank You for Your AttentionQuestions Are Welcome

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