Matrix is close to singular or badly scaled

84 ビュー (過去 30 日間)
xueqi
xueqi 2013 年 5 月 31 日
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

採用された回答

Kye Taylor
Kye Taylor 2013 年 5 月 31 日
編集済み: Kye Taylor 2013 年 5 月 31 日
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];

その他の回答 (2 件)

Matt J
Matt J 2013 年 5 月 31 日
編集済み: Matt J 2013 年 5 月 31 日
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 件のコメント
Matt J
Matt J 2013 年 5 月 31 日
As a footnote, you should not be using INV. You should be using MLDIVIDE or backslash.

サインインしてコメントする。


Azzi Abdelmalek
Azzi Abdelmalek 2013 年 5 月 31 日
In your case the result is correct, you can check:
a*inv(a) % is an identity matrix

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by