matlab gives wrong answer to determinant of a 3x3 matrix

A=[1,4,7;11,5,-1;0,2,4]

det(A)

= -3.3509e-15

which is wrong and the result is actually equal to 0.

why does matlab calculate it wrong?

[Subhamoy Saha]

It is not wrong answer. Actually it is very small number which arises due to infinitesimal fraction remains. You can check the result round(det(A))=0

[Ozan Mirzanli]

thank you

[John D'Errico]

@Subhamoy Saha - Sadly, that is poor advice. USING ROUND IS A BAD IDEA HERE.

Ahat=[1,4,7;11,5,-1;0.01,2,4];
det(Ahat)
ans =
        -0.390000000000004

Changing one element of A by a small amount yields a non-zero determinant. The perturbed matrix does not in fact have a zero determinant. However, if you round the result, then you do get zero, a result that is completely incorrect.

round(det(Ahat))
ans =
     0

Had you instead suggested to use a tolerance, testing to see if the determinant is close to zero, then it would have been acceptable, if you also then explained about tolerances and how to do the test. But round is a TERRIBLE idea. Please do not suggest that as a solution.

Best of course is to be very careful with determinants always, since as you see, changing one element of A from 0 to 0.01 was in fact sufficient to change the determinant by a fairly significant amount.

In general, I strongly recommend that you never use a determinant to do any computation unless you know enoungh about the mathematics to know why a determinant is a usually bad thing to use. Sadly, many teachers seem not to appreciate that concept. Instead, they teach students about determinants, and then just go on to the next topic in the book.

[Ozan Mirzanli]

thank you very much John!

[John D'Errico]

As a followup, why are determinants bad things? Because they have nasty scaling probems. Consider the following determinants:

A = eye(1000);
B = 3*A;
C = A/3;

det(A)
ans =
     1
     
det(B)
ans =
   Inf
   
det(C)
ans =
     0

While this is a bit of an extreme example, note that all three matrices are diagonal matrices, and are about as non-singular as you can possibly imagine. However, simply by multiplying the matrix by a small constant, I can make the determinants as large or as small as I want to, even overflowing or underflowing the dynamic range of a double.

The point is, you should never test for a determinant being zero, and even testing for small determinants is a dangerous thing, because I can make them as arbitrarily small or large as I wish. And since that arbitrary scaling of the matrix has absolutely nothing to do with the singularity status of said matrix, using a determinant is a bad idea. Yet, we are taught to use determinants to do exactly that.

Instead, there are far safer tools to infer if a matrix is singular or nearly so, starting with svd, cond, rank, rref, qr, etc.

[Subhamoy Saha]

Dear @John, I completely agree with you. It's my fault. Actually I meant to check the tolerance and yes rounding is irrelevant here. However, in context of the question, I must say the refernces provided by @the cyclist and @Samuele are relevant because the question asked why matlab gives error and not why the use of determinant=0 should be avoided. But yes, whatever you suggested is valuable and I accept it was not known to me before. Thank you.

Answer 1

[the cyclist]

MATLAB gives the correct result to within the limits of double-precision floating-point arithmetic.

This is a very common question on the forum. You could start reading about it in this answer.

[Ozan Mirzanli]

thank you

Answer 2

[Samuele Sandrini]

To calculate the determinant Matlab doesn't use an analytical formula but to calculate it in a simpler way, first it passes from the LU factorization which is susceptible to floating-point round-off errors (Limitations and Algorithms of function det).

Note:

The LU factorization allows to write the matrix A as the product of two matrices (L, U) where L is lower triangular and U is upper triangular.

Therefore:

in this way it is easier calculate the determinant because L and U are triangular.

Alternative:

If you want to verify that the matrix A is singular you can use the condition number (cond) and if it assumes very high values it means that the matrix is "close" to the singularity.

Otherwise, if you need to invert the matrix A you can use the operator "\" which uses more efficient methods (you could reading about it in this answer) .