Why there is an error in my code after I type "[L,U] = eluinv(A);"? ( the error is that too many output arguments ) Thank you!
2 views (last 30 days)
Show older comments
Tianlan Yang
on 18 Mar 2021
Commented: Tianlan Yang
on 18 Mar 2021
Here is the function:
function [] = eluinv(A)
[~,n]=size(A);
[L,U] = lu(A);
format compact
%have to round to supress extra zeroes
if (closetozeroroundoff(A) == closetozeroroundoff(round(L * U)))
disp('Yes, I have got LU factorization')
end
%comparing the reduced row echelon forms of each
if closetozeroroundoff(rref(U)) == closetozeroroundoff(rref(A))
disp('U is an echelon form of A')
else
disp('Something is wrong')
end
%if is invertible, code will continue and inversion will occur
if rank(A) ~= min(size(A))
sprintf('A is not invertible')
invA=[];
return
else
eyeForL = [L eye(n)];
eyeForU = [U eye(n)];
aux = rref(eyeForL);
aux2 = rref(eyeForU);
invL = aux(:,(n+1):size(aux,2));
invU = aux2(:,(n+1):size(aux2,2));
invA = invU*invL;
P = inv(A);
if (closetozeroroundoff(invA - P) == zeros(size(A)))
disp('Yes, LU factorization works for calculating the inverses')
else
disp('LU factorization does not work for me?')
end
end
0 Comments
Accepted Answer
Walter Roberson
on 18 Mar 2021
function [] = eluinv(A)
You declared that eluinv() does not return any values.
[L,U] = eluinv(A)
When you call a function and assign the output to variable names, MATLAB never looks inside the function to see if it can find variables with the same name to return the values of. Output in MATLAB is strictly positional. The first output location will be assigned the value of the first variable mentioned in the [] on the left side of the = in the function statement.
More Answers (0)
See Also
Categories
Find more on Linear Algebra in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!