Asked by Nitin Chhabra
on 19 Mar 2012

Hi,

When I am taking the determinant of 2x2 reciprocal matrix with matlab ( which should be 1 ) , matlab is giving wrong results without any warning for overflow. Here is what I am doing :

m =

63245986 102334155

102334155 165580141

>> det(m)

ans =

1.5249

>> m(1)*m(4)-m(2)*m(3)

ans =

2

If I mannually do the calculation , I am getting the right result ie ans=1 . Please check and tell the reason for the same.

Answer by Derek O'Connor
on 20 Mar 2012

@Nitin,

The "issue" here is not round-off. It is the ill-condition of your matrix. Floating point arithmetic and its rounding merely certify that your matrix is ill-conditioned; it does not cause your matrix to be ill-conditioned. Even if you use infinite precision (exact) arithmetic, your matrix is still ill-conditioned. There is no floating point "work-around".

Your calculation does not "suffer from large roundoff error". The large (forward) error is caused by a small round-off error which is then magnified by the huge condition number: Ef = Cond(A)*Er.

I would follow Nick Trefethen's advice and ask a different question (use a different matrix):

If the answer is highly sensitive to perturbations, you have probably asked the wrong question.

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Answer by Andreas Goser
on 19 Mar 2012

I can explain, but can't tell what do do else. If you look at

m(1)*m(4)

ans =

1.047227927956403e+16

m(2)*m(3)

ans =

1.047227927956402e+16

Then you see those numbers are large and relatively close to each other. What you see is simply a numerical effect.

Titus Edelhofer
on 19 Mar 2012

To add one point which also helps the explanation: from a numerical point of view this matrix is "bad":

norm(m)

ans =

2.2883e+08

Any numerical calculation, not only det, will suffer from large roundoff error.

Titus

Titus Edelhofer
on 19 Mar 2012

Second comment: to get the exact result you will need to compute the determinant symbolically:

M = sym(m)

M =

[ 63245986, 102334155]

[ 102334155, 165580141]

det(M)

ans =

1

Aldin
on 19 Mar 2012

Yes, you have right!

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Answer by Aldin
on 19 Mar 2012

Andreas Goser
on 19 Mar 2012

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Answer by Derek O'Connor
on 19 Mar 2012

The products above are too big to fit in 32-bit integers. You can get the correct result by switching to 64-bit integers:

>> m64 = int64(A)

>> detm64 = m64(1)*m64(4)-m64(2)*m64(3)

detm64 = 1

Look at the L-U decomposition of A:

>> [L U P] =lu(A)

L =

1.000000000000000e+000 0

6.180339887498949e-001 1.000000000000000e+000

U =

1.023341550000000e+008 1.655801410000000e+008

0 -1.490116119384766e-008

P =

0 1

1 0

This shows that U(2,2) is numerically zero relative to U(1,1) and U(1,2). Thus U is numerically singular, and so is A.

Now look at the lower bound estimate of the 1-norm condition of A:

lbcond1A = condest(A) = 4.707074327130931e+016

This shows that A is very ill-conditioned. There is no hope of getting accurate results with double precision because log10(condest(A)) is about 17, the number of digits of precision lost when calculating the solution of Ax = b.

This example shows that there is no connection between det(A) and the condition of a matrix.

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Answer by Nitin Chhabra
on 20 Mar 2012

Hi Andreas/Titus/Derek,

You are correct that values are numerically zero relative to other values and I am getting roundoff errors. But I require some work-around for this round off errors ( numerical issues ) in case of complex matix(a+jb).

with regards,

Nitin

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