I have a certain matrix .... How do I know the transformation matrix multiplied by ... if the output is also given as a matrix

[James Tursa]

How many pairs of known (x,y,z) & (x',y',z') values do you have?

[ameer raheem]

3 pairs ....

[ameer raheem]

I want program to enter values(x,y,z) & (x',y',z')by function and find the value of transformation matrix that Multiplied by

[Roger Stafford]

You need four, not three, pairs of known (x,y,z) & (x',y',z') values. Otherwise, the transformation matrix is not uniquely determined. Given four pairs, you can use matlab's matrix right division (slash) to find the matrix.

Answer 1

[Matt J]

Let v1,v2,v3 be the 3 vectors [x';y';z'] and w1,w2,w3 be the 3 vectors [x;y;z]

Also define

 A=[v1-v2, v1-v3, v2-v3 ]/[w1-w2, w1-w3, w2-w3 ];
 b=(v1-A*w1);

Then the solution is

H = [A,b;[0 0 0 1]]

Answer 2

[Roger Stafford]

"Then the solution is" should read: "then one of infinitely many solutions is." What you have here are twelve equations and sixteen unknowns, so the problem is underdetermined with only three pairs. You could also say a solution is:

 [x1',x2',x3';y1',y2',y3';z1',z2',z3';1,1,1] / ... 
 [x1 ,x2 ,x3 ;y1 ,y2 ,y3 ;z1 ,z2 , z3;1,1,1]

for which I believe matlab gives a least squares solution chosen out of, as I say, infinitely many possible solutions.

[Matt J]

Was this meant to be a comment to my Answer?

I'm reading between the lines a bit here, but I don't think there are truly meant to be infinite solutions. Rather, I htink that the question was under-specified. The fact that the vectors multiplied here are of the form [x;y;z;1], plus the fact that only 3 input pairs are available for the problem, suggests that the OP is really trying to represent a 3D affine transformation

    [x';y';z'] = A*[x;y;z]+b

in 4D homogeneous coordinates. This means the final matrix has to be of the form

    H = [A,b;[0 0 0 1]]

and the idea is to solve for the 3x3 matrix A and 3x1 translation b.

[Roger Stafford]

Yes, Matt, you are probably right. It isn't the question Ameer asked, but it may be what was meant.