Matrix is close to singular or badly scaled
11 views (last 30 days)
Show older comments
Dear Fellows,
I am trying to find out the answers for
if true
% inv([1.4326e+21,8.5958e+21;0.2001,2.2005])
end
and I got the error message and answer is
endWarning: Matrix is close to singular or badly scaled. Results may be inaccurate. RCOND = 1.661695e-23.
ans =
0.0000 -6.0009
-0.0000 1.0001
I know my matrix is badly scaled, but what I should do about this if I have to find the accurate inverse of this kind of matrix?
Cheers
0 Comments
Accepted Answer
Kye Taylor
on 31 May 2013
Edited: Kye Taylor
on 31 May 2013
So your inverse is
A = [1.4326e+21,8.5958e+21;0.2001,2.2005];
Ai = 1/det(A)*[2.2005,-8.5958e+21;-0.2001,1.4326e+21];
0 Comments
More Answers (2)
Matt J
on 31 May 2013
Edited: Matt J
on 31 May 2013
I know my matrix is badly scaled, but what I should do about this if I have to find the accurate inverse of this kind of matrix
What is the right hand side of the equation you're solving? If you're solving equations that look like this,
1.4326e+21 * x + 8.5958e+21 * y = 2e21
0.2001 * x + 2.2005 * y = 1
then the problem is simply that the coefficients in the first equation are measured in the wrong units. Rescale the system to this equivalent one
1.4326*x + 8.5958*y = 2
0.2001*x + 2.2005*y = 1
and now your system is much better conditioned.
1 Comment
Matt J
on 31 May 2013
As a footnote, you should not be using INV. You should be using MLDIVIDE or backslash.
Azzi Abdelmalek
on 31 May 2013
In your case the result is correct, you can check:
a*inv(a) % is an identity matrix
0 Comments
See Also
Categories
Find more on Matrix Indexing in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!