Difference between revisions of "Scilab/C4/Linear-equations-Gaussian-Methods/English"

Welcome to the Spoken Tutorial on “Solving System of Linear Equations using Gauss Elimination and Gauss-Jordan Methods”

• Solve system of linear equations using Scilab
• Develop Scilab code to solve linear equations

• Ubuntu 12.04 as the operating system
• and Scilab 5.3.3 version

• should have basic knowledge of Scilab
• and should know how to solve Linear Equations.

To learn Scilab, please refer to the relevant tutorials available on the Spoken Tutorial website.

• Finite collection of
• linear equations
• of the same set of variables

Given a system of equations

• A x equal to b
• with m equations and
• n unknowns

• We write the coefficients of the variables a one to a n
• along with the constants b one to b n of the system of equations
• in one matrix called the augmented matrix

We do so by performing row wise manipulation of the matrix.

This defines how many digits should be displayed in the answer.

Point to “ “

The message should be placed within double quotes.

Highlight B

Highlight b

if n ~= n1

error('gaussianelimination - Matrix A must be square');

• if the matrices are compatible with each other and
• if A is a square matrix.

If n and n one are not equal, then we display a message that Matrix A must be square.

elseif n ~= m1

error('gaussianelimination - incompatible dimension of A & b');

C=[A b];

This matrix C is called augmented matrix.

n=size(A,1);

for k=1:n-1

for i=k+1:n

factor=A(i,k)/A(k,k);

for j=k+1:n

A(i,j)=A(i,j)-factor*A(k,j);

end

b(i)=b(i)-factor*b(k);

end

end

This code converts the augmented matrix to upper triangular matrix form.

x(n)=b(n)/A(n,n);

for i=n-1:-1:1

sum=0;

for j=i+1:n

sum=sum+A(i,j)*x(j);

end

x(i)=(b(i)-sum)/A(i,i);

end

• Finally we perform back substitution.
• Once the upper triangular matrix is obtained, we take the last row and find the value of the variable in that row.
• Then once one variable is solved, we take this variable to solve the other variables.
• Thus the system of linear equations is solved.

So we enter the values of matrix A.

[3.41 1.23 -1.09;2.71 2.14 1.29;1.89 -1.91 -1.89]

two point seven one space two point one four space one point two nine semi colon

one point eight nine space minus one point nine one space minus one point eight nine close square bracket.

Press Enter

[4.72;3.1;2.92]

So we type

open square bracket four point seven two semi colon three point one semi colon two point nine one close square bracket.

Press Enter

naivegaussianelimination(A,b)

naive gaussian elimination open paranthesis A comma b close paranthesis

Press Enter

• The first step is to form the augmented matrix.
• To do this place the coefficient matrix A and the right hand side matrix b together in one matrix.
• Then we perform row operations to convert matrix A to diagonal form.
• In diagonal form, only the elements a i i are non-zero. Rest of the elements are zero.

We do this to get diagonal element equal to one.

The resulting value of the elements of each row of the right hand side matrix gives the value of each variable.

format

'e'

A=input("Enter the coeffiecient matrix : ")

b=input("Enter the right-hand side matrix : ")

function [x] = GaussJordanElimination( A, b )

[m, n] = size( A )

[r, s] = size( b )

if ( m <> r ) then

error("Error: matrix A and vector b are incompatible sizes")

end

for k = 1 : 1 : m

indices = [ 1 : 1 : k-1, k+1 : 1 : m ]

// For all rows below and above the pivot, subtract a multiple

// of the pivoting row to get a zero

for i = indices

multiplier = C(i, k) / C(k,k)

for j = k+1 : n

C(i, j) = C(i, j) - multiplier * C(k, j)

end

end

end

Here pivot refers to the first non-zero element of a column.

x = zeros( m, s )

for i = 1 : 1 : m

for j = 1 : 1 : s

x(i, j) = C(i, m+j) / C(i, i)

end

end

• we divide the right hand side part of augmented matrix
• by the corresponding diagonal element
• to get the value of each variable.

x(i, j) = C(i, m+j) / C(i, i)

return x

endfunction

[0.7, 1725;0.4352,-5.433]

So we type

open square bracket zero point seven comma one seven two five semi colon

zero point four three five two comma minus five point four three three close square bracket.

Press Enter

[1739;3.271]

So we type

open squre bracket one seven three nine semi colon

three point two seven one close square bracket

Press Enter

GaussJordanElimination(A,b)

Gauss Jordan Elimination open paranthesis A comma b close paranthesis

Press Enter

In this tutorial, we have learnt to:

• Develop Scilab code for solving system of linear equations
• Find the value of the unknown variables of a system of linear equations

Title: About the Spoken Tutorial Project

• Watch the video available at http://spoken-tutorial.org/What_is_a_Spoken_Tutorial

• It summarises the Spoken Tutorial project

• If you do not have good bandwidth, you can download and watch it

• It summarises the Spoken Tutorial project
• If you do not have good bandwidth, you can download and watch it

Title: Spoken Tutorial Workshops

The Spoken Tutorial Project Team

• Conducts workshops using spoken tutorials

• Gives certificates for those who pass an online test

• For more details, please write to contact@spoken-tutorial.org

• Conducts workshops using spoken tutorials

• Gives certificates for those who pass an online test

• For more details, please write to contact at spoken hyphen tutorial dot org

Title: Acknowledgement

• Spoken Tutorial Project is a part of the Talk to a Teacher project

• It is supported by the National Mission on Education through ICT, MHRD, Government of India

• More information on this Mission is available at

• http://spoken-tutorial.org/NMEICT-Intro

• It is supported by the National Mission on Education through ICT, MHRD, Government of India
• More information on this Mission is available at
• spoken hyphen tutorial dot org slash NMEICT hyphen Intro