Affine Transformation matrix estimation gives inconsistent result with MATLAB

I wanted to compare the result from the tform obtained using "fitgeotrans" with the result obtained from least-squares parameters estimation of the affine transformation. I get different results in the translation but correct results in the rotation. Below is an example of the code, maybe I am missing something in my code or maybe I missing something on how Matlab does its computation albeit I know it uses a variant of ransac, but the kepoints are not unmatched in this simple example.

clear all;
I = checkerboard(40);
J = imrotate(I,30);
imshowpair(I,J,'montage')

fixedPoints = [41 41; 281 161; 281 281];
movingPoints = [56 175; 324 160;384 263];

tform1 = fitgeotrans(movingPoints,fixedPoints,'NonreflectiveSimilarity');
tform1.T   

ans = 3×3

0.8667 0.4988 0 -0.4988 0.8667 0 79.7088 -138.8091 1.0000

Jregistered1 = imwarp(J,tform1,'OutputView',imref2d(size(I)));

matchedOriginal = fixedPoints;
matchedDistorted =  movingPoints;

if size(matchedDistorted) == size(matchedOriginal)
    Amat = zeros(2*size(matchedOriginal,1),6);
    bvec= zeros(2*size(matchedDistorted,1),1); 
    j = 1;
    for i=1:size(matchedOriginal,1)        
        Amat(j,:) = [matchedOriginal(i,:) 1 0 0 0];        
        Amat(j+1,:) = [0 0 0 matchedOriginal(i,:) 1];
        bvec(j,1) = matchedDistorted(i,1);        
        bvec(j+1,1)= matchedDistorted(i,2); 
        j = j+2 ;
    end
    affineParam = Amat\bvec;
end

tform2 = affine2d([affineParam(1,1) ,affineParam(2,1) ,0 ; ...
    affineParam(4,1) , affineParam(5,1),0 ; ...
    affineParam(3,1) ,affineParam(6,1) , 1]);
tform2.T

ans = 3×3

0.8667 0.5000 0 -0.4917 0.8583 0 -0.0333 159.9667 1.0000

Jregistered2 = imwarp(J,tform2,'OutputView',imref2d(size(I)));

centroidOrig = [sum(fixedPoints(:,1))/size(fixedPoints,1), sum(fixedPoints(:,2))/size(fixedPoints,1)];
centroidDistorted = [sum(movingPoints(:,1))/size(movingPoints,1), sum(movingPoints(:,2))/size(movingPoints,1)];
% To check the translation
tvec = centroidDistorted' - tform2.T(1:2,1:2)*centroidOrig';

figure(1)
imshowpair(I,Jregistered1)

figure(2)
imshowpair(I,Jregistered2)

Answer 1

[Matt J]

These were mixed up:

matchedOriginal = movingPoints;
matchedDistorted =  fixedPoints;

Also, you loaded your affineParam(i) elements in the wrong order into the matrix

tform2 = affine2d([affineParam(1) ,affineParam(4) ,0 ; ...
    affineParam(2) , affineParam(5),0 ; ...
    affineParam(3) ,affineParam(6) , 1]);
tform2.T;

With those corrections, the images agree much better.

figure(1)
imshowpair(I,Jregistered1)

figure(2)
imshowpair(I,Jregistered2)

[Sam Callahan]

Thanks. I get that the affine transform is transposed in Matlab. But I don't get why are these mixed up. Is it because I am reversing the warp ?!

matchedOriginal = movingPoints;
matchedDistorted =  fixedPoints;

[Matt J]

The movingPoints are the ones to which the transform is applied. In your equations,

Amat*affineParams=bvec

the left hand side is where the parameters are applied, so Amat must be built from the movingPoint data, not the fixedPoint data.

Answer 2

[Matt J]

Your algebraic method estimates a general 6DOF affine transform, whereas your call to fitgeotrans requests a non-reflective similarity transform. Because similarity transforms are a restricted special case of affine transforms, you are doing a less constrained estimation than what fitgeotrans is doing.

[Sam Callahan]

Shouldn't theoretically speaking 3 pairs be sufficient to estimate the 6 parameters of affine. Even if the affine structure is general, shouldn't the least squares estimation result in matrix parameters that show accurately only the rotation and translation. I actually tried 6 pairs and got exactly the same translation value from the least-squares. Is there anything fitgeotrans is doing that I am not aware of.....maybe pixel manipulation to register the images, which is something I am not doing....am I erroneously misinterpreting the translation value from the least squares ?