Solve the set of linear equations
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The document contains code to solve a system of linear equations using Gauss elimination and Gauss-Jordan elimination methods. It defines a matrix A and vector b representing the linear system. The Gauss elimination code performs row operations to reduce the matrix to row echelon form, then back substitution to solve for the unknowns. It outputs the reduced matrix and solution vector. The Gauss-Jordan code combines elimination and back substitution into a single process to directly compute the inverse matrix and solution. It verifies the solutions match between the two methods.
![www.engineeringwithsandeep.com| Student Assignment
Gauss Elimination
% solve Gauss elimination method consider
% 2a+2b-c+d=4
% 4a+3b-c+2d=6
% 8a+5b-3c+4d=12
%3a+3b-2c+2d=6
% checking A^-1*b
A=[2 2 -1 1;4 3 -1 2;8 5 -3 4;3 3 -2 2];
b=[4 6 12 6]';
[n, n] = size(A);
[n, k] = size(b);
x = zeros(n,k); %%initial value
for i = 1:n-1
m = -A(i+1:n,i)/A(i,i);
A(i+1:n,:) = A(i+1:n,:) + m*A(i,:);
b(i+1:n,:) = b(i+1:n,:) + m*b(i,:);
end;
disp("Matrix before back substitution ")
A
% Use back substitution to find unknowns
x(n,:) = b(n,:)/A(n,n);
for i = n-1:-1:1
x(i,:) = (b(i,:) -
A(i,i+1:n)*x(i+1:n,:))/A(i,i);
end
disp("Value of a ,b ,c & d ")
x
disp("Checking result")
disp("inverse of A")
val=inv(A)
disp("Value of a,b,c &d ")
val*b
Ouput
>> Gauss
Matrix before back substitution
A =
2.0000 2.0000 -1.0000 1.0000
0 -1.0000 1.0000 0
0 0 -2.0000 0
0 0 0 0.5000
Value of a ,b ,c & d
x =
1
1
-1
-1
Checking result
inverse of A
val =
0.5000 1.0000 0.2500 -1.0000
0 -1.0000 -0.5000 0
0 0 -0.5000 0
0 0 0 2.0000
Value of a,b,c &d
ans =
1
1
-1
-1](https://image.slidesharecdn.com/linearsystem-shivamchoubey-210526150250/85/Solve-the-set-of-linear-equations-2-320.jpg)
![www.engineeringwithsandeep.com| Student Assignment
Gauss Jordan Method
function [x,err]=gauss_jordan_elim(A,b)
A=[2 2 -1 1;4 3 -1 2;8 5 -3 4;3 3 -2 2];
b=[4 6 12 6]';
[n,m]=size(A);
Aa=[A,b];
for i=1:n
[Aa(i:n,i:n+1),err]=gauss_pivot(Aa(i:n,i:n+1));
if err == 0
Aa(1:n,i:n+1)=gauss_jordan_step(Aa(1:n,i:n+1),i);
end
end
x=Aa(:,n+1);
A=0;](https://image.slidesharecdn.com/linearsystem-shivamchoubey-210526150250/85/Solve-the-set-of-linear-equations-3-320.jpg)
![www.engineeringwithsandeep.com| Student Assignment
Continue code
function A1=gauss_jordan_step(A,i)
[n,m]=size(A);
A1=A;
s=A1(i,1);
A1(i,:) = A(i,:)/s;
k=[[1:i-1],[i+1:n]];
for j=k
s=A1(j,1);
A1(j,:)=A1(j,:)-A1(i,:)*s;
end % end of for loop
function [A1,err]=gauss_pivot(A)
[n,m]=size(A);
A1=A;
err = 0;
if A1(1,1) == 0
check = logical(1);
i = 1;
end
gauss_jordon
ans =
1
1
-1
-1
Checking
inv(A)
ans =
1.0000 -0.5000 0.5000 -1.0000
1.0000 0.5000 -0.5000 0
-1.0000 1.5000 -0.5000 0
-4.0000 1.5000 -0.5000 2.0000
>> inv(A)*b
ans =
1
1
-1
-1](https://image.slidesharecdn.com/linearsystem-shivamchoubey-210526150250/85/Solve-the-set-of-linear-equations-4-320.jpg)
Solve the set of linear equations
- 1. www.engineeringwithsandeep.com| Student Assignment Assignment No. 04 Solve the set of linear equations 04.04.2021 Shivam Choubey Student No: CFDB30321004
- 2. www.engineeringwithsandeep.com| Student Assignment Gauss Elimination % solve Gauss elimination method consider % 2a+2b-c+d=4 % 4a+3b-c+2d=6 % 8a+5b-3c+4d=12 %3a+3b-2c+2d=6 % checking A^-1*b A=[2 2 -1 1;4 3 -1 2;8 5 -3 4;3 3 -2 2]; b=[4 6 12 6]'; [n, n] = size(A); [n, k] = size(b); x = zeros(n,k); %%initial value for i = 1:n-1 m = -A(i+1:n,i)/A(i,i); A(i+1:n,:) = A(i+1:n,:) + m*A(i,:); b(i+1:n,:) = b(i+1:n,:) + m*b(i,:); end; disp("Matrix before back substitution ") A % Use back substitution to find unknowns x(n,:) = b(n,:)/A(n,n); for i = n-1:-1:1 x(i,:) = (b(i,:) - A(i,i+1:n)*x(i+1:n,:))/A(i,i); end disp("Value of a ,b ,c & d ") x disp("Checking result") disp("inverse of A") val=inv(A) disp("Value of a,b,c &d ") val*b Ouput >> Gauss Matrix before back substitution A = 2.0000 2.0000 -1.0000 1.0000 0 -1.0000 1.0000 0 0 0 -2.0000 0 0 0 0 0.5000 Value of a ,b ,c & d x = 1 1 -1 -1 Checking result inverse of A val = 0.5000 1.0000 0.2500 -1.0000 0 -1.0000 -0.5000 0 0 0 -0.5000 0 0 0 0 2.0000 Value of a,b,c &d ans = 1 1 -1 -1
- 3. www.engineeringwithsandeep.com| Student Assignment Gauss Jordan Method function [x,err]=gauss_jordan_elim(A,b) A=[2 2 -1 1;4 3 -1 2;8 5 -3 4;3 3 -2 2]; b=[4 6 12 6]'; [n,m]=size(A); Aa=[A,b]; for i=1:n [Aa(i:n,i:n+1),err]=gauss_pivot(Aa(i:n,i:n+1)); if err == 0 Aa(1:n,i:n+1)=gauss_jordan_step(Aa(1:n,i:n+1),i); end end x=Aa(:,n+1); A=0;
- 4. www.engineeringwithsandeep.com| Student Assignment Continue code function A1=gauss_jordan_step(A,i) [n,m]=size(A); A1=A; s=A1(i,1); A1(i,:) = A(i,:)/s; k=[[1:i-1],[i+1:n]]; for j=k s=A1(j,1); A1(j,:)=A1(j,:)-A1(i,:)*s; end % end of for loop function [A1,err]=gauss_pivot(A) [n,m]=size(A); A1=A; err = 0; if A1(1,1) == 0 check = logical(1); i = 1; end gauss_jordon ans = 1 1 -1 -1 Checking inv(A) ans = 1.0000 -0.5000 0.5000 -1.0000 1.0000 0.5000 -0.5000 0 -1.0000 1.5000 -0.5000 0 -4.0000 1.5000 -0.5000 2.0000 >> inv(A)*b ans = 1 1 -1 -1