Description

Lab Assignment 6
Student and ID : Pushkar Patel(201851094)

1 Solve the following linear system    
1 1 1 4
A =  2 1 3 b =  7 
   
3 4 −2 9
1.1 Using Gauss elimination
1.2 Using LU decomposition
1.3 Using Gauss elimination+partial pivoting
2 Solve the following linear system
 1 2 2 1   1 
 2 2 4 2   0 
A =  b =  
 1 3 2 5   2 

2 6 5 8 4
2.1 Using Jacobi method of iteration 2.2 Using Gauss-Siedel method
1

Matlab script to calculate solution of Linear Equations
Table of Contents
………………………………………………………………………………………………………………………. 1
Using Gauss Elemination ……………………………………………………………………………………….. 1 Using LU Decomposition ……………………………………………………………………………………….. 1 Using Gauss elimination + partial pivoting …………………………………………………………………… 1
clc; clear; close all;
A = [1 1 1; 2 1 3; 3 4 -2]; b =[4;7;9];
Using Gauss Elemination
x = solutionofLinearEquations(A,b); disp(“Solution using Gauss Elemination:”); disp(x);
Solution using Gauss Elemination:
1
2
1
Using LU Decomposition
x = solutionofLinearEquations(A,b,2); disp(“Solution using LU Decomposition:”); disp(x);
Solution using LU Decomposition:
1
2
1
Using Gauss elimination + partial pivoting
x = solutionofLinearEquations(A,b,3);
disp(“Solution using Gauss elimination + partial pivoting:”); disp(x);
1
Matlab script to calculate solution of Linear Equations

Solution using Gauss elimination + partial pivoting:
1.0000
2.0000
1.0000
Published with MATLAB® R2019b
2

Solution of linear equations
Table of Contents
Function define ……………………………………………………………………………………………………. 1 Gauss Elimination ………………………………………………………………………………………………… 1 LU Decomposition ……………………………………………………………………………………………….. 2
Gauss elimination + partial pivoting …………………………………………………………………………… 2
Function define
function fval = solutionofLinearEquations(a,b,choice) if ~exist(‘choice’,’var’) % third parameter does not exist, so default it to something choice = 1; % By Default method is Gauss Elemination end switch choice case 1 fval = GaussElemination(a, b); case 2 fval = LUdecomposition(a,b); case 3 fval = partialpivoting(a, b); end end Not enough input arguments.
Error in solutionofLinearEquations (line 11) fval = GaussElemination(a, b);
Gauss Elimination
function fval = GaussElemination(A,b)
%get augumented matrix
Ab =[A,b];
%Row Operation % Rj =Rj-k(i,j)*Ri where k(i,j) = A(j,i)/A(i,i) n = length(A); %A(1,1) as pivot element for i =2:n
k = Ab(i,1)/Ab(1,1); Ab(i,:)= Ab(i,:)-k*Ab(1,:); end %A(2,2) as pivot element i =n; k = Ab(i,2)/Ab(2,2);
Solution of linear equations

Ab(i,:)= Ab(i,:)-k*Ab(2,:);
%A(3,3 ) as pivot element
%Back-Subsituation fval = zeros(n,1); %x(3)=Ab(3,4)/Ab(3,3); for i =n :-1:1 %x(2)= (Ab(2,4)-Ab(2,3)*x(3))/Ab(2,2); fval(i)= (Ab(i,end)-Ab(i,i+1:n)*fval(i+1:n))/Ab(i,i); %x(1) = (Ab(1,4)-(Ab(1,3)*x(3)+Ab(1,2)*x(2)))/Ab(1,1); %x(1) = (Ab(1,4)-(Ab(1,1+1:n)*x(1+1:n))/Ab(1,1); end end
LU Decomposition
function fval = LUdecomposition(A,b)
%get augumented matrix Ab =[A,b]; n = length(A); L =eye(n);
%Row Operation
% Rj =Rj-k(i,j)*Ri where k(i,j) = A(j,i)/A(i,i)
%A(1,1) as pivot element for i =2:n k = Ab(i,1)/Ab(1,1); L(i, 1)=k; Ab(i,:)= Ab(i,:)-k*Ab(1,:);
end
%A(2,2) as pivot element i =n; k = Ab(i,2)/Ab(2,2); L(i,2)=k;
Ab(i,:)= Ab(i,:)-k*Ab(2,:);
%A(3,3 ) as pivot element
U = Ab(1:n,1:n); y = inv(L)*b; fval = inv(U)*y ; end
Gauss elimination + partial pivoting
function fval = partialpivoting(A,b)
%get augumented matrix
Solution of linear equations

Ab =[A,b]; n = length(A);
%A(1,1) as pivot element % Ensure A(1,1) is largest element in column-1 col1 = Ab(:,1); [dummy,idx]= max(col1); dummy =Ab(1,:); Ab(1,:)=Ab(idx,:); Ab(idx,:)= dummy;
for i =2:n k = Ab(i,1)/Ab(1,1); Ab(i,:)= Ab(i,:)-k*Ab(1,:);
end
%A(2,2) as pivot element % Ensure A(2,2) is largest element in column-2 col2 = Ab(2:end,2); [dummy,idx]= max(col2); dummy =Ab(2,:); Ab(2,:)=Ab(idx,:); Ab(idx,:)= dummy;
i =3; k = Ab(i,2)/Ab(2,2); Ab(i,:)= Ab(i,:)-k*Ab(2,:);
%A(3,3 ) as pivot element % Back-Subsituation
fval = zeros(n,1);
%x(3)=Ab(3,4)/Ab(3,3); for i =n :-1:1 fval(i)= (Ab(i,end)-Ab(i,i+1:n)*fval(i+1:n))/Ab(i,i); end
end
Published with MATLAB® R2019b

Matlab script to calculate solution of Linear Equations
Table of Contents
………………………………………………………………………………………………………………………. 1 Using Gauss Seidel Method …………………………………………………………………………………….. 1 Using Jacobi Method …………………………………………………………………………………………….. 1
clc; clear; close all;
A = [1 2 2 1; 2 2 4 2;1 3 2 5;2 6 5 8]; b =[1;0;2;4];
Using Gauss Seidel Method
x = IterativemethodofLS(A,b,1,1e-3,100); disp(“Solution using Gauss Seidel Method :”); disp(x);
Solution using Gauss Seidel Method :
1.0e+30 *
-5.0703
-1.2677
7.6056
-2.5352
Using Jacobi Method
x = IterativemethodofLS(A,b,2,1e-3,100); disp(“Solution using Jacobi Method :”); disp(x);
Solution using Jacobi Method :
1.0e+54 *
-1.1841
-0.9701
-0.8970
-0.4485
1
Matlab script to calculate solution of Linear Equations

Published with MATLAB® R2019b
2

Solution of linear equations using Iterative Method
Table of Contents
Function define ……………………………………………………………………………………………………. 1 Gauss Seidel Method …………………………………………………………………………………………….. 1
Jacobi Method …………………………………………………………………………………………………….. 2
Function define
function fval= IterativemethodofLS(a,b,choice,tol,maxItr) switch choice case 1 fval = gaussSeidel(a, b,tol,maxItr); case 2 fval = Jacobi(a,b,tol,maxItr); end end Not enough input arguments.
Error in IterativemethodofLS (line 4) switch choice
Gauss Seidel Method
function sol= gaussSeidel(A,b,tol,maxitr)
% Here co-efficient matrix A must be ‘strictly diagonally dominant matrix’ % tol is maximum bearable tolerance in answer % maxitr is limit of iterations
n=length(A);
Xnext=zeros(n,1); % assuming initial approximation as zero vector
for loop=1:maxitr Xcurr=Xnext;
for i=1:n temp=0; Solution of linear equations using Iterative Method

for j=1:n
if(i~=j)
temp=temp+(A(i,j)*Xnext(j)); end
end
Xnext(i)=(b(i)-temp)/A(i,i); end
error=Xnext – Xcurr;
err=norm(error);
if err<=tol sol=Xnext; break; end
end sol=Xnext; end
Jacobi Method
function fval= Jacobi(A,b,tol,maxitr)
% Here co-efficient matrix A must be ‘strictly diagonally dominant matrix’ %tol is maximum bearable tolerance in answer % maxitr is limit of iterations
n=length(A);
Xcurr =zeros(n,1); % assuming initial approximation as zero vector Xnext=zeros(n,1);
for loop=1:maxitr
for i=1:n temp=0;
for j=1:n % % if(i~=j) % temp=temp+(A(i,j)*Xcurr(j)); % This loop calculates #k#j a(k,j)*x(j)
end % % Solution of linear equations using Iterative Method

end %
Xnext(i)=(b(i)-temp)/A(i,i); end
error=Xnext-Xcurr; err=norm(error);
if err<=tol fval=Xnext; break; end Xcurr=Xnext;
end fval=Xnext; end
Published with MATLAB® R2019b