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elica85

matrix multiplication

  • one year ago
  • one year ago

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  1. elica85
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    comp being slow, pls wait..

    • one year ago
  2. mathsmind
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    ok what is the matrix problem

    • one year ago
  3. elica85
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    let x be an arbitrary angle and C represents cosine, S represent Sine

    • one year ago
  4. mathsmind
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    and what are the dimensions of those matrices

    • one year ago
  5. elica85
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    matrix A= CX -SX 0 aCX SX CX 0 aSX 0 0 1 0 0 0 0 1 matrix B= CY -SY 0 bCY SY CY 0 bSY 0 0 1 0 0 0 0 1

    • one year ago
  6. elica85
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    and Y another arbitrary angle..

    • one year ago
  7. elica85
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    4x4 matrix

    • one year ago
  8. elica85
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    i think i get how to do the multiplication but since there are variables, i'm coming up with crazy figures for each new element in the new matrix. and this new matrix has to be multiplied by another 4x4

    • one year ago
  9. mathsmind
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    ok

    • one year ago
  10. elica85
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    i guess my problem is more towards trig...is there a simple expressions for each element?

    • one year ago
  11. mathsmind
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    i will solve for u as soon as i am done with the other student

    • one year ago
  12. elica85
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    thx, or you can just tell me if there's no simple expression cuz i can do all the work, it will just look messy

    • one year ago
  13. mathsmind
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    4x4 is a messy matrix, so should expect such a mess

    • one year ago
  14. elica85
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    ok. or are you familiar with matlab? from what i remember, you have to have values for matlab...

    • one year ago
  15. mathsmind
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    do u want to do this with matlab or using normal mathematical methods

    • one year ago
  16. elica85
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    it's just part of the actual problem so i don't think working it out by hand is important so if i can do it on matlab, it would be great

    • one year ago
  17. elica85
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    or calculator

    • one year ago
  18. mathsmind
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    ok i will do it for u with c++ or c sharp

    • one year ago
  19. mathsmind
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    so u just want to multiply those two matrices

    • one year ago
  20. elica85
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    i'm not familiar with c++ but there is another matrix, i will type it up..

    • one year ago
  21. elica85
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    Matrix C= CZ -SZ 0 cCZ SZ CZ 0 cSZ 0 0 1 0 0 0 0 1

    • one year ago
  22. elica85
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    and a,b,c are arbitrary constants. so in c++, you can keep the expression with variables?

    • one year ago
  23. mathsmind
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    do u want AxBxC

    • one year ago
  24. elica85
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    if that means the same thing as AxB=AB ABxC=result 4x4 matrix

    • one year ago
  25. mathsmind
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    ok

    • one year ago
  26. mathsmind
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    no it does not mean the same thing but i get what u want now

    • one year ago
  27. elica85
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    ok thx

    • one year ago
  28. mathsmind
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    am sorry of the delay just someone before u sent me a bunch of questions that i need to type the answers

    • one year ago
  29. mathsmind
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    so u can do others stuff and i will finish the work for u as soon as possible

    • one year ago
  30. elica85
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    that's fine, i will check later

    • one year ago
  31. mathsmind
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    sorry for keeping u waiting but i need to know do u want element by element multiplication or linear algebraic product

    • one year ago
  32. mathsmind
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    you have to be precise on that so i can give u the right answer

    • one year ago
  33. elica85
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    good question, i was actually trying to figure out on google. all the instructor said was to multiply them. what i found on google was this http://easycalculation.com/matrix/learn-matrix-multiplication.php which is what's going to make the result matrix look real messy. the problem is to find the transformation matrix of a mechanical grabber. so it would have multiple joints, point O as fixed, and point P as the position of the grabber. to find the position, find transformation matrices (which i've done) and the post multipy the matrices to get the result matrix. the matrix C was from point O to joint 1, matrix B is from joint 1 to joint 2, and matrix A is from joint 2 to point P

    • one year ago
  34. mathsmind
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    HI THERE I KNOW IT TOOK ME A LONG TIME BUT I PROMISED SOMEONE TO FINISH THERE 50 QUESTIONS

    • one year ago
  35. mathsmind
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    syms cosx sinx asinx acosx cosy siny bcosy bsiny cosz sinz csinz ccosz syms cosx sinx asinx acosx cosy siny bcosy bsiny cosz sinz csinz ccosz A=[cosx, -sinx,0, acosx;sinx,cosx,0,asinx;0,0,1,0;0,0,0,1]; B=[cosy, -siny,0, bcosy;siny,cosy,0,bsiny;0,0,1,0;0,0,0,1]; %C=[cosz, -sinz,0, ccosz;sinz,cosz,0,csinz;0,0,1,0;0,0,0,1]; %Linear algebraic product of the matrices... M1= A * B * C; M2= A*B; M3= B*C; M4= A*C; %Array arithmetic operations are carried out element by element M1= A * B * C; M2= A*B; M3= B*C; M4= A*C;

    • one year ago
  36. mathsmind
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    the result for linear algebraic multiplication for example A x B

    • one year ago
  37. mathsmind
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    A = [ cosx, -sinx, 0, acosx] [ sinx, cosx, 0, asinx] [ 0, 0, 1, 0] [ 0, 0, 0, 1]

    • one year ago
  38. mathsmind
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    B = [ cosy, -siny, 0, bcosy] [ siny, cosy, 0, bsiny] [ 0, 0, 1, 0] [ 0, 0, 0, 1]

    • one year ago
  39. mathsmind
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    A*B ans = [ cosx*cosy - sinx*siny, - cosx*siny - cosy*sinx, 0, acosx + bcosy*cosx - bsiny*sinx] [ cosx*siny + cosy*sinx, cosx*cosy - sinx*siny, 0, asinx + bsiny*cosx + bcosy*sinx] [ 0, 0, 1, 0] [ 0, 0, 0, 1]

    • one year ago
  40. mathsmind
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    so this is what u called messy result which seems to be exclusive of your intuitive expectation, however the element by element matrix multiplication would give a result hopefully to your expectation

    • one year ago
  41. mathsmind
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    A.*B ans = [ cosx*cosy, sinx*siny, 0, acosx*bcosy] [ sinx*siny, cosx*cosy, 0, asinx*bsiny] [ 0, 0, 1, 0] [ 0, 0, 0, 1]

    • one year ago
  42. mathsmind
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    Array arithmetic operations are carried out element by element, and can be used with multidimensional arrays. The period character (.) distinguishes the array operations from the matrix operations.

    • one year ago
  43. mathsmind
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    Sorry i was busy but i have to keep all my promises not matter what, so all the best and good luck.

    • one year ago
  44. mathsmind
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    Now with this basic method you can do whatever multiplication you want, and in any associative, distributive ... way you want.

    • one year ago
  45. elica85
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    thank you, i think i will have to ask which multiplication is needed since i don't know. i would hope it's the easier one though!

    • one year ago
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