Why do I keep getting this error message? Error using ./ Matrix dimensions must agree.
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hello, I keep getting an error message about the matrix dimension. I think if i get the write sign to use for the division, the error will be fixed. Plese help me out if you can. thanks.
firstmatrix=[-2 -4.1 2 3; -2 1.1 2 3.3; 4.1 2 5 8.1; 1 3 2.2 1.5];
secondmatrix=[1 3 9 2; 5 8 4 3; 10 16 8 6; 2 3 1 8];
y=[-1; 2; -2; 4];
prompt='choose firstmatrix or secondmatrix';
A=input(prompt);
d=det(A);
if d~=0
eigenvalue=eig(A);
[V,D]=eig(A);
x=y./A;
else
disp('d')
disp('Calculation Complete')
end
답변 (2개)
Asad (Mehrzad) Khoddam
2020년 11월 12일
if you want to solve the equations, use this:
x=A\y;
댓글 수: 9
Toritsereju Okotie
2020년 11월 12일
Asad (Mehrzad) Khoddam
2020년 11월 12일
What is the dimensions of A?
Asad (Mehrzad) Khoddam
2020년 11월 12일
And
Can you print A after prompt?
A=input(prompt);
disp(A)
Toritsereju Okotie
2020년 11월 12일
Asad (Mehrzad) Khoddam
2020년 11월 12일

Toritsereju Okotie
2020년 11월 12일
Asad (Mehrzad) Khoddam
2020년 11월 12일
What is your answer to the input(prompt);
Toritsereju Okotie
2020년 11월 12일
Asad (Mehrzad) Khoddam
2020년 11월 12일
use
disp(A)
to make sure that A is correct
Jon
2020년 11월 12일
0 개 추천
I can run your code without getting any error messages. Can you please be more specific about how to reproduce the error. Also what version of MATLAB are you running? Maybe an older one that does not handle implicit expansion?
Looking at your code though I wonder about what you are trying to do. In particular it seems like you never use the result of the eigenvalue calculation. You could compute x = y./A without using the results of the eigenvalue calculation. Also, that statement (x = y./A) seems strange, y is a length 4 vector and A is a 4 by 4 matrix. While MATLAB will handle this with explicit expansion, but I wonder if that is really what you want. In any case dimensioning errors would not be caused by the algebraic sign of one of the entries.
If you could try to further explain what your end objective is for this code people may be able to help you further.
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