Pivot matrix function help

hi i wanna make a code that does the Pivot matrix thing from the Gauss method please

i need help with a pivot matrix gauss method code

function [A,b]=pivotea(A,b)

n = size(A, 1);

r = zeros(n, 1);

for i = 1 : 1 : n

r(i) = i;

end

x = zeros(n, 1);

for k = 1 : 1 : n

max = abs(A(r(k), r(k)));

max_pos = k;

for l = k : 1 : n

if abs(A(r(l), r(k))) > max

max = abs(A(r(l), r(k)));

max_pos = l;

end

end

temp_r = r;

r(k) = temp_r(max_pos);

r(max_pos) = temp_r(k);

for i = 1 : 1 : n

if i ~= k

zeta = A(r(i), k) / A(r(k), k);

for j = k : 1 : n

A(r(i), j) = A(r(i), j) - A(r(k), j) * zeta;

end

b(r(i)) = b(r(i)) - b(r(k)) * zeta;

end

end

end

for i = 1 : 1 : n

x(i) = b(r(i)) / A(r(i), i);

end

end

look i have this now problem is whenever i try pivot a matrix some spaces are left on 0 like here

Anueva =

0 -0.3636 0 0

-11.0000 0 0 -0.0000

0 0 0 -3.0486

0 0 61.7500 0

when it should be like

[Matt J]

It does not appear that you have begun anything. That makes it difficult for us to help you finish.

[Luis Gonzalez]

i need help with a pivot matrix gauss method code

function [A,b]=pivotea(A,b)

n = size(A, 1);

r = zeros(n, 1);

for i = 1 : 1 : n

r(i) = i;

end

x = zeros(n, 1);

for k = 1 : 1 : n

max = abs(A(r(k), r(k)));

max_pos = k;

for l = k : 1 : n

if abs(A(r(l), r(k))) > max

max = abs(A(r(l), r(k)));

max_pos = l;

end

end

temp_r = r;

r(k) = temp_r(max_pos);

r(max_pos) = temp_r(k);

for i = 1 : 1 : n

if i ~= k

zeta = A(r(i), k) / A(r(k), k);

for j = k : 1 : n

A(r(i), j) = A(r(i), j) - A(r(k), j) * zeta;

end

b(r(i)) = b(r(i)) - b(r(k)) * zeta;

end

end

end

for i = 1 : 1 : n

x(i) = b(r(i)) / A(r(i), i);

end

end

look i have this now problem is whenever i try pivot a matrix some spaces are left on 0 like here

Anueva =

0 -0.3636 0 0

-11.0000 0 0 -0.0000

0 0 0 -3.0486

0 0 61.7500 0

when it should be like

Answer 1

[Siraj]

Hi!

It seems that you are attempting to solve a system of linear equations using Gaussian elimination with partial pivoting. However, there are a couple of issues in the code. Firstly, all row swapping and elimination operations necessary to achieve the echelon form should be applied to the augmented matrix instead of the matrix "A". Secondly, to calculate the value of "x", the process should begin with the last row, a method known as back propagation or back substitution.

Please refer to the modified code below which uses an Augmented matrix "Aug" which serves as the main structure for pivoting operations (Row swapping and Gaussian Elimination).

A = [0.02 0.01 0 0; 1 2 1 0; 0 1 2 1; 0 0 100 200];
b = [0.02;1;4;800];
[reducedA, reduced_b] = pivot(A,b)

A = 4×4

0.0200 0.0100 0 0 1.0000 2.0000 1.0000 0 0 1.0000 2.0000 1.0000 0 0 100.0000 200.0000

b = 4×1

0.0200 1.0000 4.0000 800.0000

Aug = 4×5

0.0200 0.0100 0 0 0.0200 1.0000 2.0000 1.0000 0 1.0000 0 1.0000 2.0000 1.0000 4.0000 0 0 100.0000 200.0000 800.0000

x = 4×1

1 0 0 4

reducedA = 4×4

1.0000 2.0000 1.0000 0 0 1.0000 2.0000 1.0000 0 0 100.0000 200.0000 0 0 0 -0.0500

reduced_b = 4×1

1.0000 4.0000 800.0000 -0.2000

function [A,b]=pivot(A,b)
% Get the length of b
n = length(b);
% Initialize x with zeros
x = zeros(n,1);
% Create the augmented matrix
Aug = [A,b];
%Display
A
b
Aug

% Loop for partial pivoting and echelon form
for j = 1:n-1
    % partial pivoting
    % Find the maximum value and its position for partial pivoting
    [maxi, max_pos] = max(abs(Aug(j:n,j)));
    % Swap the rows for partial pivoting
    C = Aug(j,:);
    Aug(j,:) = Aug(max_pos+j-1,:);
    Aug(max_pos+j-1,:) = C;
    
    % echelon form
    for i = j+1:n
        % Calculate zeta for echelon form
        zeta = Aug(i,j)/Aug(j,j);
        % Perform row operations for echelon form
        Aug(i,:) = Aug(i,:) - zeta*Aug(j,:);
    end
end

% Back substitution
% Calculate x(n) using back substitution
x(n) = Aug(n,n+1)/Aug(n,n);
% Loop for back substitution
for k = n-1:-1:1
    % Calculate x(k) using back substitution
    x(k) = (Aug(k,n+1)-Aug(k,k+1:n)*x(k+1:n))/Aug(k,k);
end

% Display the result
x

% Separate A and b from the augmented matrix
A = Aug(:,1:end-1 );
b = Aug(:,end);
end

You can see the algorithm in action by visiting the following link:

https://web.mit.edu/10.001/Web/Course_Notes/GaussElimPivoting.html

You can explore various methods for solving a system of linear equations in MATLAB by referring to the following link:

https://www.mathworks.com/help/symbolic/solve-a-system-of-linear-equations.html#budihxp-2

To obtain the reduced row echelon form of a matrix in MATLAB, you can utilize the "rref" function. For more information about this function, please visit the following link:

https://www.mathworks.com/help/matlab/ref/rref.html

Hope this helps