I'm using an affine transformation matrix to transform 2D coordinates from screen (magnitude 10e3) to small parts of fractal sets (magnitude as little as 10e-15).

I also need to map the other way round, so I simply invert the matrix. It works when the magnitudes are not too far apart, but it fails when the determinant gets too close to zero to be represented with standard double numbers.

The matrices I use are of the form:

a c e
b d f
0 0 1

And the inversion algorithm I have is:

var dt = a * d - b * c;
return new Matrix(d/dt, -b/dt, -c/dt, a/dt, (c * f - d * e) / dt, -(a * f - b * e) / dt);

Is there an alternate way of inverting the matrix? I'm not very good at mathematics, so I would need a simple solution, and one I could find the code, or an algorithm, for.

An example of such a matrix:

 3.26091378894248e-9   -1.882689453850103e-9   -0.7172216437740687
-1.882689453814925e-9  -3.2609137888815494e-9  -0.23371832131832268
 0                      0                       1
  • $\begingroup$ I don't see a problem, doubles can represent numbers as small as 10^-308. $\endgroup$
    – user1703
    Commented Mar 4, 2016 at 9:00
  • 1
    $\begingroup$ Perhaps your problem might stem from the order of ops in your calculation. You might have something like (BigValue + SmallValue) - BigValue, and instead of getting "SmalValue" you get 0. $\endgroup$
    – Simon F
    Commented Mar 4, 2016 at 18:29
  • 1
    $\begingroup$ Just occurred to me that you seem to, effectively, have a rotation, R, followed a translation T. If you can keep these separate, to get Inverse(R*T) you can just do Inverse(T)*Inverse(R), both of which are trivial. Would that help? $\endgroup$
    – Simon F
    Commented Mar 4, 2016 at 18:46
  • $\begingroup$ Simon, you're right about the problem stemming from multiplying numbers of different magnitudes. The matrix is more complex than an R*T; it's also a scale, and can be a multiplication of any affine.. I managed to fix the problem, however, see below. Thanks for your help. $\endgroup$
    – solendil
    Commented Mar 4, 2016 at 21:10
  • $\begingroup$ You should ask on the math forum ! $\endgroup$ Commented Mar 4, 2016 at 21:16

1 Answer 1


I found a solution to my specific problem. Instead of computing the determinant and hitting the precision wall, I use the Gauss-Jordan method step by step.

In my specific case of affine transformation matrices and the range of values I use, I don't hit any precision problem this way.


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