I have a matrix A. Then I transform it by the matrix B. The result is Matrix
C. Given A and C, how do I find B ?

Instead of matrixes the input and output should be sets of 4 vectors each.

I imagine a macro like this:

FindMatrix (Ax,Ay,Az,Ap,Cx,Cy,Cz,Cp)

It would #declare the vectors Bx, By, Bz and Bp

Thanks in advance!

Rune
--
\ Include files, tutorials, 3D images, raytracing jokes,
/ The POV Desktop Theme, and The POV-Ray Logo Contest can
\ all be found at http://rsj.mobilixnet.dk (updated December 17)
/ Also visit http://www.povrayusers.org

Post a reply to this message

Rune wrote:
> 
> I have a matrix A. Then I transform it by the matrix B. The result is Matrix
> C. Given A and C, how do I find B ?

[A][B] = [C]

Premultiply both sides by the inverse of [A]

[A-1][A][B] = [A-1][C]

a matrix times its inverse is the indentity matrix, hence

[1][B] = [A-1][C]

any matrix times the identity matrix is the matrix itself, hence

[B] = [A-1][C]

Q.E.D.

However, this only works for square matrices (if my math isn't too
rusty).

> 
> Instead of matrixes the input and output should be sets of 4 vectors each.
> 
> I imagine a macro like this:
> 
> FindMatrix (Ax,Ay,Az,Ap,Cx,Cy,Cz,Cp)
> 
> It would #declare the vectors Bx, By, Bz and Bp

The macro coding is left to the student as an exercice. :)

-- 
Francois Labreque | Rimmer: "Let's go to red alert!"
    flabreque     | Kryten: "Are you sure, Sir?  You realize it
        @         |          actually means changing the bulb!"
   videotron.ca

Post a reply to this message

"Francois Labreque" wrote:
> [A][B] = [C]
>
> Premultiply both sides by the inverse of [A]

You've lost me. Too tricky for me.

> However, this only works for square matrices (if my math
> isn't too rusty).

AFAIK the matrixes in POV-Ray are square. the last column is always 0,0,0,1

> The macro coding is left to the student as an exercice. :)

But that's the tricky part! ...too!

I only know how to use matrixes in POV-Ray.
I don't have any general matrix knowledge at all. :(

Rune
--
\ Include files, tutorials, 3D images, raytracing jokes,
/ The POV Desktop Theme, and The POV-Ray Logo Contest can
\ all be found at http://rsj.mobilixnet.dk (updated December 17)
/ Also visit http://www.povrayusers.org

Post a reply to this message

Rune wrote in message <3a429879@news.povray.org>...
> I only know how to use matrixes in POV-Ray.
> I don't have any general matrix knowledge at all. :(


http://forum.swarthmore.edu/library/topics/matrices/

good luck :-)

ABX

Post a reply to this message

Rune <run### [at] inamecom> wrote:
:> Premultiply both sides by the inverse of [A]

: You've lost me. Too tricky for me.

  Just calculate the inverse matrix of A and then multiply A*C (matrix
multiplication, not item-by-item multiplication) and you'll get B.

  Calculating the inverse of a matrix is not a simple operation, though.

-- 
main(i,_){for(_?--i,main(i+2,"FhhQHFIJD|FQTITFN]zRFHhhTBFHhhTBFysdB"[i]
):_;i&&_>1;printf("%s",_-70?_&1?"[]":" ":(_=0,"\n")),_/=2);} /*- Warp -*/

Post a reply to this message

On 22 Dec 2000 04:50:45 -0500, Warp wrote:
>Rune <run### [at] inamecom> wrote:
>:> Premultiply both sides by the inverse of [A]
>
>: You've lost me. Too tricky for me.
>
>  Just calculate the inverse matrix of A and then multiply A*C (matrix
>multiplication, not item-by-item multiplication) and you'll get B.
>
>  Calculating the inverse of a matrix is not a simple operation, though.

I think there's a can't-miss formula for the inverse of a 3x3 matrix, but
I don't remember it off the top of my head.  And of course it doesn't work
for noninvertable matrices.

-- 
Ron Parker   http://www2.fwi.com/~parkerr/traces.html
My opinions.  Mine.  Not anyone else's.

Post a reply to this message

Ron Parker <ron### [at] povrayorg> wrote:
: I think there's a can't-miss formula for the inverse of a 3x3 matrix, but
: I don't remember it off the top of my head.

  But I think that povray matrices are 4x4 (and I think the original poster
was talking about povray matrices if I'm not mistaken).

-- 
main(i,_){for(_?--i,main(i+2,"FhhQHFIJD|FQTITFN]zRFHhhTBFHhhTBFysdB"[i]
):_;i&&_>1;printf("%s",_-70?_&1?"[]":" ":(_=0,"\n")),_/=2);} /*- Warp -*/

Post a reply to this message

On 22 Dec 2000 08:48:55 -0500, Warp wrote:
>Ron Parker <ron### [at] povrayorg> wrote:
>: I think there's a can't-miss formula for the inverse of a 3x3 matrix, but
>: I don't remember it off the top of my head.
>
>  But I think that povray matrices are 4x4 (and I think the original poster
>was talking about povray matrices if I'm not mistaken).

They're not really.  The fourth column is just a convenient way of specifying
the translate component.  You could easily take that out, invert, and put 
it back.

-- 
Ron Parker   http://www2.fwi.com/~parkerr/traces.html
My opinions.  Mine.  Not anyone else's.

Post a reply to this message

Rune wrote
I have a matrix A. Then I transform it by the matrix B. The result is Matrix

> C. Given A and C, how do I find B ?
>
> Instead of matrixes the input and output should be sets of 4 vectors each.
>
> I imagine a macro like this:
>
> FindMatrix (Ax,Ay,Az,Ap,Cx,Cy,Cz,Cp)
>
> It would #declare the vectors Bx, By, Bz and Bp

Tricky question.... well, not tricky, but complicated.
You can find B as the product of A^-1*C, so finding that inverse is tricky.

For a square matrix A, the inverse of A is 1/det(A)*adj(A), or in English, the
Adjoint of A divided by the determinant. (The determinant is a scalar value that
we can find from A).

So, A =
a b c 0
d e f  0
g h i 0
j k l 1.

(I know, a fixed width font would be good for that.)
To find the determinant of A, we can use a shortcut and use the fourth column,
so we only need to find the determinant of the 3 by 3 matrix a...i:
det(A) = a(ei - fh) - b(di - fg) + c(dh - eg)  and you better make sure that
this doesn't equal 0, if it does you can't find an inverse.

To find the adjoint we find the cofactors of A which is defined by
(-1)^(i+j)det(Mij)  Here i and j refer to the row and column of the matric
element (just to clarify a muddy situation)
Mij is the Minor of Aij, which means we remove row i and column j and we have a
smaller square matrix (3 by 3 in this scenario)
So to find the Minor of a in our original matrix, we are left with the 3 by 3
matrix
e f 0
h i 0
k l 1
and since we need the determinant of this three by three matrix, we take
advantage of the last column, and det(M11) = ei - fh. I hope you understand,
because I leave you to figure out the rest of the Minors, and remember that you
still have to mulitply the determinant of the Minor by (-1)^(i + j), basically,
this forms a checkerboard pattern of negatives, so the the determinant of the
Minor calculated for element a  is normal, the determinant for the Minor
calculated for values b,d,f,h,j, and l are multiplied by -1, and you will still
have minors down the fourth column.
Once you have 16 cofactors, divide each one by the determinate we calculated in
the beginning, and you should have your inverse matrix. If you want to check,
multiply A by its inverse and see if you get the identity matrix, which is a
matrix of 0's except the diagonal from top left to bottom right should be 1's.

This is probably more glib than it needs to be, so I won't be offended by
questions.

--
Josh English -- Lexiphanic Lethomaniac
eng### [at] spiritonecom
The POV-Ray Cyclopedia http://www.spiritone.com/~english/cyclopedia/

Post a reply to this message

Silly me, I already solved the problem of how to find the inverse of a matrix to
deal with a collision model for POV-Ray wth the help of a friend:

Here is the relevant code:
// The POV_Ray representation of the matrix that we need to find the inverse of.
#declare bX = <1,-0.05,0>;
#declare bY = <-0.5,0.5,0>;
#declare bZ = <0,0,1>;
#declare bC = <0,0,0>;

// THe 16 statements to fill out the 16 elements of the inverse
// // New system devised by David Jones
#declare dp = bX.x*bY.z - bX.z*bY.x;
#declare dq = bX.z*bY.y - bX.y*bY.z;
#declare dr = bX.x*bY.y - bX.y*bY.x;
#declare dt = bX.y*bZ.x - bZ.y*bX.x;

#declare dy = bZ.x*dq - bZ.y*dp + bZ.z*dr;
#declare dx = bC.x*dq - bC.y*dp + bC.z*dr;

#declare c11 = (bY.y * bZ.z - bY.z * bZ.y)/dy;
#declare c12 = (bX.z * bZ.y - bX.y * bZ.z)/dy;
#declare c13 = (bX.y * bY.z - bX.z * bY.y)/dy;
#declare c14 = 0;

#declare c21 = (bY.z * bZ.x - bY.x * bZ.z)/dy;
#declare c22 = (bX.x * bZ.z - bX.z * bZ.x)/dy;
#declare c23 = (bX.z * bY.x - bX.x * bY.z)/dy;
#declare c24 =0;

#declare c31 = (bY.x * bZ.y - bY.y * bZ.x)/dy;
#declare c32 = (bX.y * bZ.x - bZ.y * bX.x)/dy;
#declare c33 = (bX.x * bY.y - bX.y * bY.x)/dy;
#declare c34 = 0;

#declare c41 = -1*(bC.x * c11 + bC.y * c21 + bC.z * c31);
#declare c42 = -1*(bC.x * c12 + bC.y * c22 + bC.z * c23);
#declare c43 = -1*(bC.x * c13 + bC.y * c23 + bC.z * c33);
#declare c44 = 1;

This seemed to work after all of my tests, and it was a lot faster

Josh

Post a reply to this message