Matlab gives wrong determinant value of 2x2 Matrix without warning.

Nitin Chhabra (view profile)

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.

Derek O'Connor (view profile)

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

Andreas Goser (view profile)

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

Titus Edelhofer (view profile)

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

Titus Edelhofer (view profile)

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

Aldin (view profile)

on 19 Mar 2012
Yes, you have right!

Aldin (view profile)

on 19 Mar 2012

It depends of your MATLAB version. My version is MATLAB 7.9.0 (R2009b) - worked a charm. By me the result of matrix is 2.

Andreas Goser

Andreas Goser (view profile)

on 19 Mar 2012
Which may be more the effect of 32 bit vs. 64 bit MATLAB or a different processor or a different BLAS library...

Derek O'Connor (view profile)

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.