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Matrix is close to singular or badly scaled

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xueqi
xueqi on 31 May 2013
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

Accepted Answer

Kye Taylor
Kye Taylor on 31 May 2013
Edited: Kye Taylor on 31 May 2013
There is a closed form expression for the inverse of a 2-by-2 matrix here (see equation 4).
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];

More Answers (2)

Matt J
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
Matt J on 31 May 2013
As a footnote, you should not be using INV. You should be using MLDIVIDE or backslash.

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Azzi Abdelmalek
Azzi Abdelmalek on 31 May 2013
In your case the result is correct, you can check:
a*inv(a) % is an identity matrix

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