Question

This is Using MATLAB:

I am trying to store the solution of this matrix because I want to do something with the result like find the norm of that answer however I am stuck and cannot seem to be able to. Help would be appreciated!

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MATLAB CODE:

close all

clear

clc

A = [1 -1 2 -1;

2 -2 2 -3;

1 1 1 0;

1 -1 4 5];

b = [-8 -20 -2 4]';

x = gauss_elim(A,b)

function x = gauss_elim(A, b)

[nrow, ~] = size(A);

nb = length(b);

x = zeros(1,nrow);

  

% Gaussian elimination

for i = 1:nrow-1

if A(i,i) == 0

t = min(find(A(i+1:nrow,i) ~= 0) + i);

if isempty(t)

disp ('Error: A matrix is singular');

return

end

temp = A(i,:); tb = b(i);

A(i,:) = A(t,:); b(i) = b(t);

A(t,:) = temp; b(t) = tb;

end

for j = i+1:nrow

m = -A(j,i) / A(i,i);

A(j,i) = 0;

A(j,i+1:nrow) = A(j,i+1:nrow) + m*A(i,i+1:nrow);

b(j) = b(j) + m*b(i);

end

end

  

% Back substitution

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

fprintf('\n\nHas exact solution:\n')

for i = nrow-1:-1:1

x(i) = (b(i) - sum(x(i+1:nrow) .* A(i,i+1:nrow))) / A(i,i);

end

end

Answer 1

I think you want to say that for different system you will get different solutions but you don't want to lose the previous solution you want to store it.
you can simply write a=x at the end

your x vector will be saved as a

if you want to save series of vectors in a matrix for different n number of systems.

I am modifying your code it will ask to enter matrix A and vector b, n number of times and then it will save your n solutions column wise in a matrix.

clc;

clear all;
n=input('enter number of times you want to enter A and b ');

for k=1:n
  
A=input('enter your matrix ');
b=input('enter b vector ');

x = gauss_elim(A,b);
mat(:,k)=x
end

function x = gauss_elim(A, b)

[nrow, ~] = size(A);

nb = length(b);

x = zeros(1,nrow);

  

% Gaussian elimination

for i = 1:nrow-1

if A(i,i) == 0

t = min(find(A(i+1:nrow,i) ~= 0) + i);

if isempty(t)

disp ('Error: A matrix is singular');

return

end

temp = A(i,:); tb = b(i);

A(i,:) = A(t,:); b(i) = b(t);

A(t,:) = temp; b(t) = tb;

end

for j = i+1:nrow

m = -A(j,i) / A(i,i);

A(j,i) = 0;

A(j,i+1:nrow) = A(j,i+1:nrow) + m*A(i,i+1:nrow);

b(j) = b(j) + m*b(i);

end

end

  

% Back substitution

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

fprintf('\n\nHas exact solution:\n')

for i = nrow-1:-1:1

x(i) = (b(i) - sum(x(i+1:nrow) .* A(i,i+1:nrow))) / A(i,i);

end

end