mth term paper

8/2/2019 Mth Term Paper

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TOPIC: - BIOPOLYMERS.

SUBMITTED TO: SUBMITTED BY:

JAGANJOT KAUR

(SUBJECT TEACHER) GURMUKH SINGH

SECTION: E4001

ROLL NO.: B42

REGD.NO:-11006705


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ACKNOWLEDGEMENT

Thanks giving is a sacred Indian culture. So, first of all, I would liketo thank our subject teacher, MISS JAGANJOT KAUR for his

humble support and encouragement which enhanced me through the

project. His likeness towards my topic uplifted my spirit which in

turn helped me throughout.

Secondly, I express my gratitude to my Parents for being a

continuous source of encouragement and the financial aid given tome.

Thirdly, I would love to thank my beloved friends who

helped me in their little ways. They were of great help and support as

they helped me a lot with their innovative ideas. They helped me a lot

in completing my work in time.

Finally, I would like to thank God for showering his

blessings upon me.

GURMUKH SINGH


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CONTENTS

1. INTRODUCTION 12. HISTORY 23. ALGORITHM OVERVIEW 34. EXAMPLES 3-55. APPLICATIONS 6-96. SYSTEM OF LINEAR EQUATIONS 10-157. GAUSS-JORDAN METHOD 15-168. EXAMPLES 17-189. CONCLUSION 19

REFERENCES


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INTRODUCTION

In linear algebra, Gaussian elimination is an algorithm for solving systems of

linear equations, finding the rank of a matrix, and calculating the inverse of an

invertible square matrix. Gaussian elimination is named after German

mathematician and scientist Carl Friedrich Gauss, which makes it an example of

Stigler's law.

Elementary row operations are used to reduce a matrix to row echelon form.

GaussJordan elimination, an extension of this algorithm, reduces the matrix

further to reduced row echelon form. Gaussian elimination alone is sufficient for

many applications, and is cheaper than the -Jordan version.

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History

The method of Gaussian elimination appears in Chapter Eight, Rectangular

Arrays, of the important Chinese mathematical text Jiuzhang suanshu or The

Nine Chapters on the Mathematical Art. Its use is illustrated in eighteen

problems, with two to five equations. The first reference to the book by this title

is dated to 179 CE, but parts of it were written as early as approximately 150

BCE.[1]

It was commented on by Liu Hui in the 3rd century.

The method in Europe stems from the notes of Isaac Newton.[2]In 1670, he wrote

that all the algebra books known to him lacked a lesson for solving simultaneous

equations, which Newton then supplied. Cambridge University eventually

published the notes as Arithmetica Universalis in 1707 long after Newton left

academic life. The notes were widely imitated, which made (what is now called)Gaussian elimination a standard lesson in algebra textbooks by the end of the

18th century. Carl Friedrich Gauss in 1810 devised a notation for symmetric

elimination that was adopted in the 19th century by professional hand computers

to solve the normal equations of least-squares problems. The algorithm that is

taught in high school was named for Gauss only in the 1950s as a result of

confusion over the history of the subject.

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Algorithm overview

The process of Gaussian elimination has two parts. The first part (Forward

Elimination) reduces a given system to either triangular or echelon form, or

results in a degenerate equation with no solution, indicating the system has nosolution. This is accomplished through the use ofelementary row operations. The

second step uses back substitution to find the solution of the system above.

Stated equivalently for matrices, the first part reduces a matrix to row echelon

form using elementary row operations while the second reduces it to reduced

row echelon form, or row canonical form.

Another point of view, which turns out to be very useful to analyze the

algorithm, is that Gaussian elimination computes matrix decomposition. The

three elementary row operations used in the Gaussian elimination (multiplying

rows, switching rows, and adding multiples of rows to other rows) amount to

multiplying the original matrix with invertible matrices from the left. The first

part of the algorithm computes LU decomposition, while the second part writes

the original matrix as the product of a uniquely determined invertible matrix

and a uniquely determined reduced row-echelon matrix.

Example:

Suppose the goal is to find and describe the solution(s), if any, of the following

system of linear equations:

The algorithm is as follows: eliminate x from all equations below L1, and then

eliminate y from all equations below L2. This will put the system into triangular

form. Then, using back-substitution, each unknown can be solved for.

In the example, x is eliminated from L2 by adding to L2. X is then eliminated

from L3 by adding L1 to L3. Formally:

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The result is:

Now y is eliminated from L3 by adding 4L2 to L3:

The result is:

This result is a system of linear equations in triangular form, and so the first part

of the algorithm is complete.

The last part, back-substitution, consists of solving for the known in reverse

order. It can thus be seen that

Then, z can be substituted into L2, which can then be solved to obtain

Next, z and y can be substituted into L1, which can be solved to obtain

The system is solved.

Some systems cannot be reduced to triangular form, yet still have

at least one valid solution: for example, if y had not occurred in L2 and L3 after

the first step above, the algorithm would have been unable to reduce the system

to triangular form.

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However, it would still have reduced the system to echelon form. In this case, the

system does not have a unique solution, as it contains at least one free variable.

The solution set can then be expressed parametrically (that is, in terms of the

free variables, so that if values for the free variables are chosen, a solution will be

generated).

In practice, one does not usually deal with the systems in terms of equations but

instead makes use of the augmented matrix (which is also suitable for computer

manipulations). For example:

Therefore, the Gaussian Elimination algorithm applied to the augmented matrix

begins with:

Which, at the end of the first part (Gaussian elimination, zeros only under the

leading 1) of the algorithm, looks like this:

That is, it is in row echelon form.

At the end of the algorithm, if the GaussJordan elimination (zeros under and

above the leading 1) is applied:

That is, it is in reduced row echelon form, or row canonical form.

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APPLICATIONS

Finding the inverse of a matrix

Suppose A is a matrix and you need to calculate its inverse. The

identity matrix is augmented to the right of A, forming a matrix

(the block matrix B = [A, I]). Through application of elementary row operations

and the Gaussian elimination algorithm, the left block of B can be reduced to the

identity matrix I, which leaves A 1

in the right block of B.

If the algorithm is unable to reduce A to triangular form, then A is not

invertible.

General algorithm to compute ranks and bases

The Gaussian elimination algorithm can be applied to any matrix A. If

we get "stuck" in a given column, we move to the next column. In this way, for

example, some matrices can be transformed to a matrix that has a

reduced row echelon form like

(The *'s are arbitrary entries). This echelon matrix T contains a wealth of

information about A: the rank of A is 5 since there are 5 non-zero rows in T; the

vector space spanned by the columns of A has a basis consisting of the first,

third, fourth, seventh and ninth column of A (the columns of the ones in T), and

the *'s tell you how the other columns of A can be written as linear combinations

of the basis columns.

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Analysis

Gaussian elimination to solve a system of n equations for n unknowns

requires n (n+1) / 2 divisions, (2n3

+ 3n2 5n)/6 multiplications, and (2n

3+ 3n

2

5n)/6 subtractions, for a total of approximately 2n3 / 3 operations. So it has a

complexity of .

This algorithm can be used on a computer for systems with thousands of

equations and unknowns. However, the cost becomes prohibitive for systems

with millions of equations. These large systems are generally solved using

iterative methods. Specific methods exist for systems whose coefficients follow a

regular pattern (see system of linear equations).

The Gaussian elimination can be performed over any field.

Gaussian elimination is numerically stable for diagonally dominant or positive-

definite matrices. For general matrices, Gaussian elimination is usually

considered to be stable in practice if you use partial pivoting as described below,

even though there are examples for which it is unstable.

Higher order tensors

Gaussian elimination does not generalize in any simple way to higher order

tensors (matrices are order 2 tensors); even computing the rank of a tensor of

order greater than 2 is a difficult problem.

Gaussian elimination is a method for solving matrix equations of

the form

(1)

To perform Gaussian elimination starting with the system of equations

(2)

compose the "augmented matrix equation"

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(3)

Here, the column vector in the variables is carried along for labeling the matrix

rows. Now, perform elementary row operations to put the augmented matrix

into the upper triangular form

(4)

Solve the equation of the th row for , then substitute back into the equation of

the st row to obtain a solution for , etc., according to the formula

(5)

In Mathematical, Row Reduce performs a version of Gaussian elimination, with

the equation being solved by

Gaussian Elimination [m_?MatrixQ, v_?VectorQ] :=

Last /@ Row Reduce [Flatten /@ Transpose [{m, v}]]

LU decomposition of a matrix is frequently used as part of a Gaussian

elimination process for solving a matrix equation.

A matrix that has undergone Gaussian elimination is said to be in echelon form.

For example, consider the matrix equation

(6)

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In augmented form, this becomes

(7)

Switching the first and third rows (without switching the elements in the right-

hand column vector) gives

(8)

Subtracting 9 times the first row from the third row gives

(9)

Subtracting 4 times the first row from the second row gives

(10)

Finally, adding times the second row to the third row gives

(11)

Restoring the transformed matrix equation gives

(12)

Which can be solved immediately to give , back-substituting to obtain

(which actually follows trivially in this example), and then again back-

substituting to find

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Systems of Linear Equations: Gaussian Elimination

It is quite hard to solve non-linear systems of equations, while linear systems are

quite easy to study. There are numerical techniques which help to approximatenonlinear systems with linear ones in the hope that the solutions of the linear

systems are close enough to the solutions of the nonlinear systems. We will not

discuss this here. Instead, we will focus our attention on linear systems.

For the sake of simplicity, we will restrict ourselves to three, at most four,

unknowns. The reader interested in the case of more unknowns may easily

extend the following ideas.

Definition. The equation

a x + b y + c z + d w = h

Where a, b, c, d, and h are known numbers, while x, y, z, and w are unknown

numbers, is called a linear equation. If h =0, the linear equation is said to be

homogeneous. A linear system is a set of linear equations and a homogeneous

linear system is a set of homogeneous linear equations.

For example,

And

Are linear systems, while

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is a nonlinear system (because of y2

). The system

Is a homogeneous linear system.

Matrix Representation of a Linear System

Matrices are helpful in rewriting a linear system in a very simple form. The

algebraic properties of matrices may then be used to solve systems. First,

consider the linear system

Set the matrices

Using matrix multiplications, we can rewrite the linear system above as the

matrix equation

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As you can see this is far nicer than the equations. But sometimes it is worth to

solve the system directly without going through the matrix form. The matrix A is

called the matrix coefficient of the linear system. The matrix C is called the

nonhomogeneous term. When , the linear system is homogeneous. The

matrix X is the unknown matrix. Its entries are the unknowns of the linear

system. The augmented matrix associated with the system is the matrix [A|C],

where

In general if the linear system has n equations with m unknowns, then the matrix

coefficient will be a nxm matrix and the augmented matrix an nx (m+1) matrix.

Now we turn our attention to the solutions of a system.

Definition. Two linear systems with n unknowns are said to be equivalent if and

only if they have the same set of solutions.

This definition is important since the idea behind solving a system is to find an

equivalent system which is easy to solve. You may wonder how we will come up

with such system. Easy, we do that through elementary operations. Indeed, it is

clear that if we interchange two equations, the new system is still equivalent to

the old one. If we multiply an equation with a nonzero number, we obtain a new

system still equivalent to old one. And finally replacing one equation with the

sum of two equations, we again obtain an equivalent system. These operationsare called elementary operations on systems. Let us see how it works in a

particular case.

Example. Consider the linear system

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The idea is to keep the first equation and work on the last two. In doing that, we

will try to kill one of the unknowns and solve for the other two. For example, if

we keep the first and second equation, and subtract the first one from the last

one, we get the equivalent system

Next we keep the first and the last equation, and we subtract the first from the

second. We get the equivalent system

Now we focus on the second and the third equation. We repeat the same

procedure. Try to kill one of the two unknowns (y or z). Indeed, we keep the first

and second equation, and we add the second to the third after multiplying it by 3.We get

This obviously implies z = -2. From the second equation, we get y = -2, andfinally from the first equation we get x = 4. Therefore the linear system has one

solution

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Going from the last equation to the first while solving for the unknowns is called

backsolving.

Keep in mind that linear systems for which the matrix coefficient is upper-triangular are easy to solve. This is particularly true, if the matrix is in echelon

form. So the trick is to perform elementary operations to transform the initiallinear system into another one for which the coefficient matrix is in echelon

form.

Using our knowledge about matrices, is there any way we can rewrite what we

did above in matrix form which will make our notation (or representation)

easier? Indeed, consider the augmented matrix

Let us perform some elementary row operations on this matrix. Indeed, if we

keep the first and second row, and subtract the first one from the last one we get

Next we keep the first and the last rows, and we subtract the first from the

second. We get

Then we keep the first and second row, and we add the second to the third after

multiplying it by 3 to get

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This is a triangular matrix which is not in echelon form. The linear system for

which this matrix is an augmented one is

As you can see we obtained the same system as before. In fact, we followed the

same elementary operations performed above. In every step the new matrix was

exactly the augmented matrix associated to the new system. This shows that

instead of writing the systems over and over again, it is easy to play around with

the elementary row operations and once we obtain a triangular matrix, write the

associated linear system and then solve it. This is known as Gaussian

Elimination.

GAUSS-JORDAN METHOD:

The Gauss-Jordan method is a version of Gaussian elimination in solving

systems of linear equations. The variables' coefficients, instead of merely being

reduced to a triangular shape, are reduced to a diagonal. This eliminates the

need for successive substitution, allowing one to just read off the solutions.

Gauss-Jordan Elimination

Gauss-Jordan elimination goes the extra step of using such operations to

eliminate variables above the diagonal as well.

As a result, one can just read off the solution, for example, that x1 = -1, x2

= 2, and so on. The need for back-substitution to solve for each variable,

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As in Gaussian substitution, is therefore eliminated.

Difference from Gaussian Elimination

The additional operations Gauss-Jordan performs to put the

variables into a diagonal form triples the number of computations

required, even with Gaussian elimination's back-substitution operations.

The gain, however, is in being able to read answers off immediately.

Disadvantages

1. The additional operations of Gauss-Jordan add to rounding error andcomputer time. A disadvantage of both Gaussian and Gauss-Jordan

elimination is that they require the right vector, for example, (4,1,-3,4)

above, to be known. If these numbers are to be learned later, a method

called matrix factorization can prepare a triangular shape for easy

calculation when the vector is known. If the vector changes, the effort in

factorizing has saved time as well.

Elimination Method for Solving Systems of LinearEquations

A technique for solving systems of linear equations of any size is the Gauss-

Jordan Elimination Method. This method uses a sequence of operations on a

system of linear equations to obtain an equivalent system at each stage. An

equivalent system is a system having the same solution as the original system.

The operations of the Gauss-Jordan method are:1. Interchange any two equations.

2. Replace an equation by a nonzero constant multiple of itself.

3. Replace an equation by the sum of that equation and a constant multiple of

any other equation.

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EXAMPLES:

1. Let's solve the system:-2x

1- 3x

2= -6

5x1 + 4x2 = 31

We form the matrix (A|b) and reduce it:

This is the matrix (A|b).

Row 1:= 1/2 x row 1

Row 2:= row 2 - 5 x row1

Row 2:= 2/23 x row 2

Row 1:= row 1 + 3/2 row 2

The reduced matrix represents the system

1x1 + 0x2 = 3

0x1 + 1x2 = 4

Which has the unique solution (x1,x2) = (3,4). This is also the unique

solution to the original system.

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2. Let's solve the system

x1 + x2 + x3 = 5

x1 + 2x2 + 3x3 = 8

We form the matrix (A|b) and reduce it:

This is the matrix (A|b)

Row 2:= row2 - row1 the matrix is in echelon form

row1:= row1- row2 the matrix is reduced

The reduced matrix represents the system

1x1 + 0x2 - x3 = 2

0x1 + 1x2 + 2x3 = 3

Which has infinitely many solutions:

x1 = 2 + x3, x2 = 3 - 2x3, x3 = any number

We wrote the solution to the exercise on the before page as two equations:

X1 = 2 + x3, x2 = 3 - 2x3, (x3 R)

Notice that these two equations have the perfect same meaning as one

vector equation:

When we write the equation this way, it looks clear that the solution set is

the line through the point (2,3,0) that is parallel to the vector (1, -2, 1).

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CONCLUSION

Gaussian elimination is a method of solving a linear system (consisting

of equations in unknowns) by bringing the augmented matrix. In the

Gaussian Elimination Method, Elementary Row Operations (E.R.O.'s) are

applied in a specific order to transform an augmented matrix into triangular

echelon form as efficiently as possible.

This is the essence of the method: Given a system of m equations in n variables

or unknowns, pick the first equation and subtract suitable multiples of it from

the remaining m-1 equations. In each case choose the multiple so that the

subtraction cancels or eliminates the same variable, say x1. The result is that the

remaining m-1 equations contain only n-1 unknowns (x1 no longer appears).

Now set aside the first equation and repeat the above process with the remaining

m-1 equations in n-1 unknowns.

Continue repeating the process. Each cycle reduces the number of variables and

the number of equations. The process stops when either:

There remains one equation in one variable. In that case, there is a uniquesolution and back-substitution is used to find the values of the other

variables.

There remain variables but no equations. In that case there is no uniquesolution.

There remain equations but no variables (i.e. the lowest row(s) of theaugmented matrix contain only zeros on the left side of the vertical line).

This indicates that either the system of equations is inconsistent or

redundant. In the case of inconsistency the information contained in the

equations is contradictory. In the case of redundancy, there may still be a

unique solution and back-substitution can be used to find the values of the

other variables.

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REFERENCES

1. HIGHER ENGINEERING MATHEMATICSBY B.V. RAMANA.

2. MODERN APPROACH TO MATHEMATICSBY N.K. NAG.

3. S. CHAND MATHEMATICS FOR CLASS XIIBY S. CHAND PUBLICATIONS.

4. www.google.com
http://www.google.com/http://www.google.com/

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