Matrix is close to singular or badly scaled

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

Answer 1

[Kye Taylor]

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];

Answer 2

[Matt J]

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.

[Matt J]

As a footnote, you should not be using INV. You should be using MLDIVIDE or backslash.

Answer 3

[Azzi Abdelmalek]

In your case the result is correct, you can check:

a*inv(a)  % is an identity matrix