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Numerical & Statistical methods (2140706)   Darshan Ins Numerical & Statistical methods (2140706)   Darshan Ins

Numerical & Statistical methods (2140706) Darshan Ins - PowerPoint Presentation

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Numerical & Statistical methods (2140706) Darshan Ins - PPT Presentation

Unit3 Linear Algebric Equation 2140706 Numerical amp Statistical Methods Matrix Equation The matrix notation for following linear system of equation is as follow The above linear system is expressed in the matrix form ID: 574966

methods amp numerical row amp methods row numerical statistical 2140706 darshan institute engineering technology method matrix gauss equations solution

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Slide1

Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology

Unit-3

Linear

Algebric

Equation

2140706 – Numerical & Statistical MethodsSlide2

Matrix Equation

The matrix notation for following linear system of equation is as follow:

The above linear system is expressed in the matrix form as

 

Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology

2Slide3

Elementary Transformation or Operation on a Matrix

Operation

Meaning

or Interchange of and rowsMultiplication of all the elements of row by non zero scalar k. or Multiplication of all the elements of row by nonzero scalar k and added into row.OperationMeaningNumerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology3Slide4

Row Echelon Form of Matrix

To convert the matrix into

row echelon

form follow the following steps: Every zero row of the matrix occurs below the non zero rows.Arrange all the rows in strictly decreasing order.Make all the entries zero below the leading (first non zero entry of the row) element of 1st row.Repeat step-3 for each row.Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology4Slide5

Reduced Row Echelon Form of Matrix

To convert the matrix into

reduced row echelon

form follow the following steps: Convert given matrix into row echelon form.Make all leading elements 1(one).Make all the entries zero above the leading element 1(one) of each row.Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology5Slide6

Numerical Methods For Solution Of A Linear

Equation

Direct Methods

Iterative MethodsDirect MethodsThis method produce the exact solution after a finite number of steps but are subject to errors due to round-off and other factors.We will discuss two direct methods :Gauss Elimination methodGauss-Jordan methodNumerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology6Slide7

Indirect Method(Iterative Method)

In this method, an approximation to the true solution is assumed initially to start method. By applying the method

repeatedly,

better and better approximations are obtained. For large systems, iterative methods are faster than direct methods and round-off error are also smaller. Gauss seidel methodGauss jacobi methodNumerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology7Slide8

Gauss Elimination Method

To solve the given linear system using Gauss elimination method, follow the following steps:

Start with augmented matrix

Convert matrix into row echelon form.(we take leading element of each row is one(1).)Apply back substitution for getting equations.Solve the equations and find the unknown variables (i.e. solution). Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology8Slide9

Solve by Gauss Elimination method:

 

Solution:-

By Augmented matrix,  Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology9We have to make zero to these three elements. ( means multiply row 1 by and add it into row 2) ( means multiply row 1 by and add it into row 3)      Slide10

Now, solving equations by back-substitution

 

Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology

10     Now, these three elements are zerosSlide11

Solve by Gauss Elimination method:

 

Solution:-

By Augmented matrix,  Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology11We have to make zero to these three elements. ( means multiply row 1 by and add it into row 2) ( means multiply row 1 by and add it into row 3)      Slide12

Now, solving equations by back-substitution

 

Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology

12     Now, these three elements are zerosSlide13

Gauss Elimination Method With Partial Pivoting

To solve the given linear system using Gauss elimination method with partial pivoting, follow the following steps:

Find largest

absolute value(pivot element) in first column.Make the pivot element row to first row.Eliminate x1 below the pivot element.Again find pivot element in 2nd and 3rd row.Make the pivot element row to second row.Eliminate x2 below the pivot element.Apply back substitution for getting equations.Solve the equations and find the unknown variables (i.e. solution).Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology13Slide14

Solve by Gauss Elimination method with partial pivoting

 

Solution:-

By Augmented matrix,  Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology14Find Largest absolute value in first column and make that row into first row    Make these two elements to zeroSlide15

 

Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology

15Find Largest absolute value in second column in 2nd and 3rd row and interchange accordingly.Slide16

By back substitution,

 Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology16Slide17

Gauss-Jordan Method

This method is modification of the gauss elimination method. This method solves a given system of equation by transforming

the coefficient matrix into unit matrix

. Steps to solve Gauss-Jordan method:Write the matrix form of the system of equations.Write the augmented matrix.Reduce the coefficient matrix to unit matrix by applying elementary row transformations to the augmented matrix.Write the corresponding linear system of equations to obtain the solution.Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology17Slide18

Solve by Gauss Jordan method

 

Solution:-

By Augmented matrix,  Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology18Slide19

 Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology19Slide20

By back substitution,

 

Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology

20Slide21

Gauss Jacobi Method

This method is applicable to the system of equations in which leading diagonal elements of the coefficient matrix are dominant (large in magnitude) in their respective rows.

Consider the system of equations.

 Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology21Slide22

Where co-efficient matrix

must be diagonally dominant,

And the inequality is strictly greater than for at least one row.Solving the system for respectively , we obtain Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology22Slide23

We start with

in

equ

. Again substituting these value x1, y1, z1 in Eq. (2), the next approximation is obtained.This process is continued till the values of are obtained to desired degree of accuracy. Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology23Slide24

Solve by Gauss Jacobi method up to three iteration.

 

Solution:-

So, It is not diagonally dominant.We need to rearrange the equations. So, All Equations are diagonally dominant. Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology24Slide25

(Make subject

from diagonally dominant equations

.)

Here, Let the initial values are iteration, Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology25Slide26

Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology

26

Iteration

Iteration1230.851.021.001.00-0.97-1.001.251.031.00Ans:    Slide27

Gauss Seidel Method

This is a modification of Gauss-Jacobi method. In this method we

replace

the approximation by the corresponding new ones as soon as they are calculated.Consider the system of equations: Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology27Slide28

Where co-efficient matrix

must be diagonally dominant

, And the inequality is strictly greater than for at least one row.Solving the system for respectively, we obtain, Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology28Slide29

We start with

in

equ

. Now substituting in the second equ. Of Now substituting in the second equ. Of This process is continued till the values of are obtained to desired degree of accuracy. Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology29Slide30

Solve by gauss-Seidel method correct up two decimal places.

 

Solution:-

So, All Equations are diagonally dominant.By Gauss Seidel method, Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology30Slide31

Let the initial values are

In

iteration all values are almost same.

So, Numerical & Statistical methods (2140706) Darshan Institute Of Engineering & Technology31IterationIteration1231.51.912.891.162.122.981.142.132.9941.142.132.99