Trying to write a code for solving a system of linear equations using Jacobi method, getting this error "In an assignment A(:) = B, the number of elements in A and B must be the same."

I think it must be something with my preallocation or perhaps the way I have my index set up? In order to comply with homework requirements I can only use one for loop.

clc; close all; clear;
load('set1_HW3');
load('set2_HW3');
A=diag(1./diag(A1_HW3));
A1=diag(diag(A1_HW3));
tol=10^-4;
x=zeros(size(A1_HW3,1),1);
for k=1:100
    x(k+1)=A*b1_HW3-A*(A1_HW3-A1)*x(k);
    if abs((x(k+1)-x(k))/x(k))<tol
        break
end

[Stephen23]

What sizes do A, b1_HW3 and A1_HW3 have?

[Abigail Grein]

A is 100x100, b1_HW3 is 100x1 and A1_HW3 is 100x100

Answer 1

[Walter Roberson]

We can see from your line

x=zeros(size(A1_HW3,1),1);

that size(A1_HW3,1) is not expected to be 1 -- that A1_HW3 is expected to have multiple rows.

You have A=diag(1./diag(A1_HW3)) . If A1_HW3 were a scalar at that point then A would end up being a scalar, but we have reason to believe that A1_HW3 is an array at that point, in which case A is going to end up being a square array.

You have

x(k+1) = A*b1_HW3-A*(A1_HW3-A1)*x(k)

we do not know how big b1_HW3 is at that point, but we can see that you are using the * algebraic matrix multiplication operator between A and b1_HW3 . If b1_HW3 is a scalar then the result of the * would be a matrix with the same size as A -- and since we know that A has multiple rows, we know that result cannot fit within the single location x(k+1) . If b1_HW3 is an array then with the * operator the size of the result will be size(A,1) by size(b1_HW3,2) -- and we know that size(A,1) is more than one so no matter what size(B1_HW3,2) is, we can see that the result of the * is going to have multiple rows and so is not going to fit within the single location x(k+1)

[Abigail Grein]

I see what you mean, I made a mistake allocating x with only one column. I want my output to be an array showing each iteration of solutions as columns, so it will be multiple columns. I guess I'm not sure how to allocate the correct number of columns because I don't know when the values will converge.

[Walter Roberson]

Your loop k goes to at most 100, so if there were never any "break" then the output could end up 100 columns. Use that for pre-allocation. And then when you get an early break, remove the columns that were not used.

[Abigail Grein]

I made x 100x100 and got the same error code. It must be something with my index then? Something to do with the program not knowing that x(1)=zeroes?

[Walter Roberson]

maxk = 100;
x = zeros(size(A1_HW3,1),maxk);
for k = 1:maxk
    x(:,k+1)=A*b1_HW3-A*(A1_HW3-A1)*x(:,k);
    if all( abs((x(:,k+1)-x(:,k))./x(:,k)) < tol )
        break
    end
end
x(:,k+1:end) = [];   %remove unusued