Error "Matrix dimention must agree"

Here is the function:

function [L,U] = eluinv(A)

[~,n]=size(A);

[L,U] = lu(A);

format compact

if closetozeroroundoff(A,7) == closetozeroroundoff(L*U,7)

disp('Yes, I have got LU factorization')

end

if closetozeroroundoff(rref(U),7) == closetozeroroundoff(rref(A),7)

disp('U is an echelon form of A')

else

disp('Something is wrong')

end

if rank(A) ~= max(size(A))

sprintf('A is not invertible')

invA=[];

return

else

leye = [L eye(size(L,1))];

ueye = [U eye(size(U,1))];

invLech = rref(leye);

invUech = rref(ueye);

invL = invLech(:,(n+1):size(invLech,2));

invU = invUech(:,(n+1):size(invUech,2));

invA = invU*invL;

if isequal(closetozeroroundoff(invA-inv(A)),zeros(size(A)))

disp('Yes, LU factorization works for calculating the inverses')

else

disp('LU factorization does not work for me?!')

end

end

Answer 1

[Cris LaPierre]

The error message is telling you what the issue is. You are trying to compare two matrices that do not have the same number of elements.

The left side of your comparison is 3x3 while the right side is 4x3. See this documentation page for more.

This is because the result of lu is two different sized matrices.

A = [1 1 4; 0 -4 0; -5 -1 -8; 2 3 -1];
[L,U] = lu(A)

L = 4×3

-0.2000 -0.2000 -0.5714 0 1.0000 0 1.0000 0 0 -0.4000 -0.6500 1.0000

U = 3×3

-5.0000 -1.0000 -8.0000 0 -4.0000 0 0 0 -4.2000

Answer 2

[Tianlan Yang]

Thank you. Could you please look another question I post?