Inverse homogeneous transformation

HaniAlmousli

Hello, I have a question about inverse homogeneous transformation. while i am reading a book "Introduction to Robotics mechanics and control" there is an example which says that: We have two frames A and B, Frame A is deduced from A by rotating A 30 degree and translating it 4 by x and 3 by y so the matrix of the transformation of B relative to A is:
0.866 -0.5 0 4
0.5 0.866 0 3
0 0 1 0
0 0 0 1
The question is to find A relative to B so i said that the rotation will be -30 and the x=-4 and the y=-3 so it will be:
0.866 0.5 0 -4
-0.5 0.866 0 -3
0 0 1 0
0 0 0 1
but actually it is not the result they had a rule that changes the translation part(last column) by multiplying the matrix rotation of A relative to B by the translation vector and the result is inserted in the last column.
Please can any one help?
Hani Almousli......

dario_ramos

It depends on the order of the transformations.

Generally speaking, if you rotate first and translate afterwards, to reverse that, you need to revert the traslation and then revert the rotation.

It's easier if you think in terms of matrices. Say X is the original point, Y is the rotated and translated point (rotated first). Then, if T is the matrix which represents the translation and R represents the rotation:

Y = T R X

Premultiply both sides by T's inverse (remember, matrix product is not commutative):

(T^-1) Y = (T^-1) T R X

By definition, (T^-1)T = I, the identity

(T^-1) Y = R X

Now, premultiply by (R^-1) and you get:

(R^-1) (T^-1) Y = X

Therefore, (R^-1)(T^-1) is the matrix you're looking for