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
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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