There is an error in my code
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Here is the function:
function [L,U] = eluinv(A)
[~,n]=size(A);
[L,U] = lu(A)
LU = roundingfixer(L*U);
if (A == LU)
disp('LU factorization is valid')
else
disp('A mistake was made somewhere.')
end
if (rank(U) == rank(A))
disp('U is an echelon form of A')
else
disp('A mistake was made somewhere.')
end
if (rank(A) == n)
% Use rref to row reduce the matrices
invL = [L eye(n)];
invU = [U eye(n)];
invL = rref(invL);
invU = rref(invU);
% Cut off the first half of the matrix
invL = invL(:,(n+1:n*2));
invU = invU(:,(n+1:n*2));
invA = roundingfixer(invU * invL)
P = roundingfixer(inv(A))
if (invA == P)
disp('LU works for calculating the inverse for A')
else
disp('Error in calculating the inverse for A')
end
else
sprintf('A is not invertible.')
invA = [];
end
end
roundingfixer.m:
function B = roundingfixer(A)
[m,n]=size(A);
A=closetozeroroundoff(A,7)
for i=1:m
for j=1:n
A(i,j) = round(A(i,j),8);
end
end
B=A;
end
답변 (1개)
Look at line 17: invL = [L eye(n)];
You are concatenating L and eye(n) horizontally. The error message is again telling you what the problem is. See this documentation page for more.
A = [1; 2]
B = [3 4 5]
C = [A B]
댓글 수: 8
Tianlan Yang
2021년 3월 18일
Cris LaPierre
2021년 3월 18일
Figure out why your sizes are not what you expected, and update your code accordingly.
Tianlan Yang
2021년 3월 18일
Cris LaPierre
2021년 3월 18일
Horizontal concatenation means the number of rows (first dimension) must be the same. This is why a 3x4 and a 3x3 can be horizontally concatenated (both have 3 rows), but a 4x3 and a 3x3 cannot (one has 4, the other has 3, so a dimension mismatch error occurs).
Tianlan Yang
2021년 3월 18일
Cris LaPierre
2021년 3월 18일
I'm afraid I don't undertand what it is you are trying to do, so am not in a position to recommend a solution.
Tianlan Yang
2021년 3월 18일
Cris LaPierre
2021년 3월 18일
I am not aware of a function that will do this for you. I believe you will need to write your code in a way to either ensure the two matrices have the same number of rows.
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