FINDING DETERMINANT OF MATRIX AFTER TAKING COVARIANCE
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Z=cov(x);
disp(Z);
ans:cov
0.3333 0.4167 0.2500
0.4167 0.5833 0.3750
0.2500 0.3750 0.2500
G=det(Z);
disp(G);
detr
2.2407e-018 determinant value obtained manually and in calci is -1.5625*(10^-6). using det after covariance producing wrong ans. help me how to take a matrix (3*3), then take covariance and for that want to take determinant.... help me in this
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回答(3 个)
Oleg Komarov
2011-8-28
disp(Z)
try:
format long
disp(pi)
format short
disp(pi)
3 个评论
Oleg Komarov
2011-8-28
Your cov matrix is stored with precision up to the 16 decimal places, after that floating point approximation "kicks in" (not MATLAB dependendant).
Unless you specify more clearly what constitutes problem for you, I would say yes, no problem. Use cov and then det.
Honglei Chen
2011-8-29
As Oleg mentioned, I think this is a precision thing. I'm curious what is the original number you put in as x. Based on the number you are given, I tried to put in the fraction as the input to det and it works fine
>> det([1/3 5/12 1/4;5/12 7/12 3/8;1/4 3/8 1/4])
ans =
1.1565e-18
Mike Hosea
2011-9-7
I can give you a quick manual calculation of det(cov(x)). It's just 0. The problem is that determinant is not well-behaved in finite precision arithmetic. If you matrix is badly scaled and the exact answer is zero, your calculated determinant will be entirely composed of numerical noise. All of these answers are noise:
>> x = cov(rand(3))
x =
0.147692200988288 0.095764562838736 0.073054502962085
0.095764562838736 0.063238321621117 0.049927796895473
0.073054502962085 0.049927796895473 0.041859114554002
>> det(x)
ans =
5.736287575505627e-21
>> det(100*x)
ans =
5.454390014652211e-15
>> det(1e6*x)
ans =
0.008592762115391
>> det(1e10*x)
ans =
6.898406910429394e+09
I'm not aware of any good reasons to calculate the determinant of a matrix numerically. For example, use the RANK function to determine whether a matrix is singular. -- Mike
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