Multiplying Difference Size Matrices

Question

I am attempting to use the inverse mapping approach to multiply an inv(3x3) matrix to an image(540x420) to rotate it! I cannot use any toolbox functions and am attempting to do a double FOR loop. I am confused about how this works if anyone could help me out it would be greatly appreciated!

NOTE: A is the image! I have a bilinear interpretation code after this that will fix the image accordingly. I am just a having a hard time wrapping my head around this (3x3) * (540x420) matrix.

[xold,yold,k] = size(A);
xnew = xold;
ynew = yold;
Rot = [cosd(angle) -sind(angle) 0 ; sin(angle) cos(angle) 0; 0 0 1];
InverseRot = double(inv(Rot));
E = zeros(xnew, ynew,k);
E = double(E);
A = double(A);
for i=1:ynew
    for j=1:xnew
       E(i,:) =  %This is where I'm confused
       
    end
end

Answer 1

You're not multiplying the inverse rotation matrix by the image, you're multiplying it by the homogeneous coordinates of a point in the new image to get the corresponding homogeneous coordinates in the original image.

newpt = [xnew ynew 1].';
origpt = InverseRot*newpt;
xold = origpt(1);
yold = origpt(2);

Notes:

• Your rotation matrix mixes the degree and radian versions of the trig functions. That will mess up your results.
• Rotations without translations are about the origin. Your image is not centered at the origin. You'll need to do something about that.
• The inverse of a rotation matrix is the same as its transpose. If you fix the note above, you'll end up with an affine transformation matrix rather than a straight rotation matrix. This also has an easier way to calculate the inverse. If the affine transformation matrix is [R | t; 0 | 1], the inverse is [RT | -RT*t; 0 | 1] where RT is the transpose of R.
• The coordinates in the original image that you get after transformation are not likely to be integers, so you won't be able to simply assign the value of a pixel in the old image to a pixel in the new image. As you said, you'll have to find some way to interpolate the value and assign that to the point in the new image.