Homogenous Linear Equations

8/7/2019 Homogenous Linear Equations

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This presentation includes the following

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Introduction to linear equationsHomogeneous linear equationsSolutions of

homogeneous linear equationsTrivial solution

Non trivial solution

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8Examples

ApplicationsHow to solvehomogeneous linear equations


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Introduction to linear equationsA LINEAR EQUATION is an algebraic equation inwhich each term is either a constant or product of aconstant and a single variable of power 1 . It can bewritten as

a 1x1 + a 2x2 + . . . . .+ a nxn = b;A SYSTEM OF LINEAR EQUATIONS (or linear system) is a collection of linear equations involvingthe same set of variables.

mnmn2 m2 1m1

2 n2n2 22 121

1n1n2 12 111

b x a.... x a x a

b x a.... x a x a

b x a.... x a x a

!

!

!

/


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Homogeneous linear equations A linear equation will be homogeneous if the constant

term is zero.a 1x1 + a 2x2 + . . . . .+ a nxn = 0

Sy stem of homogeneous linear equat ions is acollection of homogeneous linear equations having thesame set of variables. They can be represented as

The above homogeneous system can be written in thematrix form as

0

10

0

2211

2222121

1212111

!

!

!

nmnmm

nn

nn

xa xa xa

xa xa xa

xa xa xa

....

)....

....

/

..

AX=0


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Where A is the matr ix of c o-eff ici ents i.e

X is the c o lumn ve c tor of var iab les i.e .,

And 0 i s the nu ll matr ix.

-

!

mnmm

n

n

aaa

aaa

aaa

-

///

-

-

11

22121

11211

A

-

!

n x

x

x

/

2

1

X

-

!

0

0

0

/o


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SOLUTIONS OFHOMOGENEOUS LINEAR EQUATIONS

The solution obtained by solving the homogeneouslinear equations can be unique or there can beinfinite number of solutions.

These are classified into:

Trivial solutionNon trivial solution

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Trivial solution As, homogeneous linear equations can be written inmatrix form as

AX=0 . (1)It is easy to notice that the solution to (1) is

X=0or x1=0, x 2=0, x =0,.., x n=0

and certainly this is the case. Although it is easy toget this solution, usually this is not the desiredsolution, so it is called the "trivial solution. So wereach a very simple theorem (on the next slide).

Moreover if det A 0

then it is the unique solution.


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Theorem 1H omogeneous system of linear equations is always

consistent as such a system in n variables alwayshas the solution

x 1 = 0, x 2 = 0,. , x n = 0.


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Non trivial solution Along with the trivial solution, a homogeneoussystem can have an infinite number of solutionswhich are called non-trivial solutions. As, homogeneous linear equations can be written

in matrix form asAX=0 . (1)

R eferring to eq (1), whendet A = 0

there is an infinite number of solutions which arecalled non-trivial solutions.


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Theorem 2An n x n matr ix that represents a homogeneous

s ystem of linear equat ions has a nontr ivia l so lut ion if and on ly i f its determ inant = 0PROOF:(1) Assume that the matrix has a nontrivial solution .

(2) Assume that its determ inant 0(3) By Cramer's R ule, if determ inant 0 , then the

matrix has a unique solution. But if matrix has aunique solution, then it does not have a nontrivial

solution (As, unique solution will be the trivialsolution).

(4) Therefore we have a contradiction and we rejectour assumption in step # 2 and conclude that if

determ inant =0 , we can get a non-trivial solution.


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THEOREM 3

STATEMENT: A homogeneous system of m linear equations in nun-knowns will have a nontrivial solution if m < n.PROOF:

Consider a matrix B which is in the reduced rowechelon form and is row equivalent to the coefficientmatrix of the system. Let m < n. Let r be the number of nonzero rows in the matrix B. Then

r m < n and hencen r > 0

and so the number n r of arbitrary unknowns is in

fact positive and we are forced to use the free


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to get the solution. Taking one of these unknowns tobe t, an arbitrary constant, gives us a nontrivialsolution.


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Solving H omogeneous L inear equations As we know that we can get two types of solutionsfrom homogeneous linear equations, namely trivialand non-trivial solutions. Trivial solutions can befound easily by taking all variables equal to zero. Onthe other hand non trivial solution is not as easy to

find as trivial solution is. So, non trivial solution canbe found by different ways. Some of them areElementary OperationsGauss Elimination Method

Guass Jordan EliminationBefore going into the detail of these methods, we

should be familiar with some of the basic conceptsstated below.


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SOME BASIC CONCEPTSBefore discussing these methods in detail, we shouldget familiar with some of the basic concepts.Row equ iva len c eTwo matrices are said to be row equivalent if one can

be changed to the other by a sequence of elementaryrow operations.Free Var iab lesIt can be difficult to describe the non trivial solution.Typically, some of the variables are introduced whichare allowed to take any value, and the remainingvariables are dependent on the values of the freevariables. These variables are called free variables.


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Augmented Matr ixThe augmented matrix represents all the importantinformation in the system of equations. The namesof the variables are ignored, and the onlyconnection with the variables is the location of their coefficients in the matrix.

For example, the augmented matrix of thesystem in eq.1 is

-

0

0

0

11

22121

11211

/

/

/

/

/

-

///

-

-

mnmm

n

n

aaa

aaa

aaa


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Elementary OperationsH omogeneous linear equations can be solved bythe following simple operations, also calledelementary operations.Interchange the two equationsMultiply a non-zero constant throughout anequationAdd a multiple of one row to another row.

These operations can be performed on thesimple equations or by converting them into thematrix form. The other methods also employ theseoperations to solve the equations.


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EXAMPLE 1

Solve the following homogeneous system

SOLUTIONFirst of all, well check whether the above system hasa non trivial solution or not. For this, we should have:

det A =0

(where A is the coefficient matrix of the above system)Taking L. H .S.

2x+z=0x-y-z=0

x-y=0

0 13

11110 2

det !


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0 2 2 1(2)0 1)2(

3)11( 3)0(0 1)2(0

!

So, the system has non-trivial solution.R eferring to the system

Subtracting1/2 E 1 from E 2 and 3/2 E 1 from E 3

E1: 2x+z=0E 2: x -y-z=0E : x -y=0

E1: 2x+z=0E2: -y-3/ 2 z=0E3: -y-3/ 2 z=0


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Multiplying E 1 by E1: x+ 1 /2 z=0

Multiplying E 2 by -1E2: y+ 3/2 z=0

Add E 2 on E 3E3: 0=0

Let z = t

or z= 2 tPutting this value in the above equations we have

E1: x+1 /2 (2z) = 0

x= -z


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E2: y+ 3/2 (2 t) = 0y = - 3t

So, the non trivial solution isx = -z

y = -3tz = 2 t


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Gaussian EliminationGaussian elimination is a three step method for solving systems of linear equations:1) Write the system as an augmented matrix (usezeroes for a particular variable of the system if it

doesn't appear in one of the equations)2) Use elementary row operations to reduce thematrix to row echelon form.3) Use back substitution to finally solve the system. In

other words, write the system of linear equationscorresponding to the reduced echelon form.Then, take a free variable and assign it different

values to get the non-trivial solution.


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Gauss-Jordan Elimination

It involves the following stepsLocate the leftmost column that does not consist

entirely of zero.Interchange the top row with another row, if necessary, to bring a nonzero entry to the top fromstep 1.If the entry that is now at the top is a constant, divideentire row by it. Add multiples to top row to the rows below such thatall entries have 1 as leading term.Cover top row and begin with step 1 applied to

submatrix.

Through th is method , the augmented matr ix i s redu c ed into a form s imp le enough su c h that s ystem of equat ion are so lved b y i nspe c tion .

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APPLICATIONSH

omogeneous linear equations find their applications inBalancing a chemical equationElectrical Network Analysis

Kirchhoffs Laws

Conditions of equilibrium


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Application to ChemistryConsider the following combustion reaction in which achemical equation relates how propane molecules(C 3H 8 ) combine with oxygen atoms (O 2 ) to formcarbon dioxide (CO 2) and water (H 2O). This reactionis given as:

x1 (C 3H 8 )+ x2 (O 2 ) x3 (CO 2)+ x 4 (H 2O)When a chemist wants to "balance this equation,"

whole numbers x 1, x2, x 3 , x 4 must be found so thatthe number of atoms of carbon (C), hydrogen (H ) andoxygen (O) on the left match their respective number on the right.

To balance the equation, write three equations

which keep track of the number of carbon, hydrogen


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and oxygen atoms, respectively.Carbon : 3x1+ 0 x 2 = x 3 + 0x 4H ydrogen : 8 x1+ 0 x 2 = 0x 3 + 2x 4Oxygen : 0x 1+ 2 x 2 = 2x 3 + x 4

The next example shows how to solve this system.


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Ex ample 2

Q. B a lan c e the propane-o xy gen equat ion b yso lvi ng the homogeneous linear s ystemCarbon : 3x1+ 0 x 2 = x 3 + 0x 4Hydrogen : 8x 1+ 0 x 2 = 0x 3 + 2x 4Oxy gen : 0x 1+ 2 x 2 = 2x 3 + x 4

SOLUTION:

We have the following equations3x1- x3 = 08 x1-2x 4 = 0

2 x 2- x3-x4 = 0


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We rite t em i ma tri f r m as

AXWhe r e

As , e have

A

-

!

-

!

-

!4

3

2

1

x

x

x

x

X

0

0

0

B

12 2 0

2 0 0 8

0 10 3

A,,

-

!

-

-

0

0

0

x

x

x

x

12 2 0

2 0 0 8

0 10 3

4

3

2

1


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Form the augmented matrix M = [ A, B]

Find the reduced row echelon form of the augmentedmatrix M = [ A, B] by Gaussian Elimination method asfollows. We have the following system

-

0

0

0

1220

008

0103

2

-

1220

2008

0103


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3238

3

8

1220

200

0103

R R By

-

2138

43

8

3

1220

200

003

R R B y

-

32310

38

43

3

4

1220

00

003

R R By

-

23

23

310

38

43

4

3

200

00

003

R R B y

-


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By multiplyingR

1 by 1/ 3,R

2 by3/8

andR

3 by-1/2, weshall get the required matrix in row reduced echelonform that will be

This linear system is equivalent to:

-

43

45

41

1

1

1

0 x4

5 - x

0 x4

1 - x

42

41

!

!


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There is one free variable required which we chooseto be

x4 = 4t.The above equation will be used in computing x 1, x2and x 3 .Solving the previous equations for x 1, x2 and x 3 .

3

3 !

5t(4t)4

5 x

4

5 x

t(4t)41

x41

x

42

41

!!!

!!!


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By putting these values in the chemical equations,we shall get the required balanced equation

t (C 3H8)+ 5 t (O 2 ) 3t (CO 2)+ 4 t (H2O)This is the required balanced equation

t3( t)43

x4

3 x 43 !!!


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Electrical Network Analysis

K ir c hoffs LawH omogeneous linear systems can be used insolving electrical circuits in the form of Kirchoffsfirst law, which can be written mathematically as

I1+I2+I3+...+I n=0


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Conditions of equilibriumBoth the conditions of equilibrium yield homogeneousequations that can be written mathematically as

F=0or F 1+F 2+F 3+...+F n=0

and

=0or 1+ 2+ 3+...+ n=0

A body will be said to be in equilibrium if and only if itsatisfies both the above mentioned equations.


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Homogenous Linear Equations

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