anonymous
  • anonymous
matrix multiplication
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
comp being slow, pls wait..
anonymous
  • anonymous
ok what is the matrix problem
anonymous
  • anonymous
let x be an arbitrary angle and C represents cosine, S represent Sine

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anonymous
  • anonymous
and what are the dimensions of those matrices
anonymous
  • anonymous
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
anonymous
  • anonymous
and Y another arbitrary angle..
anonymous
  • anonymous
4x4 matrix
anonymous
  • anonymous
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
anonymous
  • anonymous
ok
anonymous
  • anonymous
i guess my problem is more towards trig...is there a simple expressions for each element?
anonymous
  • anonymous
i will solve for u as soon as i am done with the other student
anonymous
  • anonymous
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
anonymous
  • anonymous
4x4 is a messy matrix, so should expect such a mess
anonymous
  • anonymous
ok. or are you familiar with matlab? from what i remember, you have to have values for matlab...
anonymous
  • anonymous
do u want to do this with matlab or using normal mathematical methods
anonymous
  • anonymous
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
anonymous
  • anonymous
or calculator
anonymous
  • anonymous
ok i will do it for u with c++ or c sharp
anonymous
  • anonymous
so u just want to multiply those two matrices
anonymous
  • anonymous
i'm not familiar with c++ but there is another matrix, i will type it up..
anonymous
  • anonymous
Matrix C= CZ -SZ 0 cCZ SZ CZ 0 cSZ 0 0 1 0 0 0 0 1
anonymous
  • anonymous
and a,b,c are arbitrary constants. so in c++, you can keep the expression with variables?
anonymous
  • anonymous
do u want AxBxC
anonymous
  • anonymous
if that means the same thing as AxB=AB ABxC=result 4x4 matrix
anonymous
  • anonymous
ok
anonymous
  • anonymous
no it does not mean the same thing but i get what u want now
anonymous
  • anonymous
ok thx
anonymous
  • anonymous
am sorry of the delay just someone before u sent me a bunch of questions that i need to type the answers
anonymous
  • anonymous
so u can do others stuff and i will finish the work for u as soon as possible
anonymous
  • anonymous
that's fine, i will check later
anonymous
  • anonymous
sorry for keeping u waiting but i need to know do u want element by element multiplication or linear algebraic product
anonymous
  • anonymous
you have to be precise on that so i can give u the right answer
anonymous
  • anonymous
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
anonymous
  • anonymous
HI THERE I KNOW IT TOOK ME A LONG TIME BUT I PROMISED SOMEONE TO FINISH THERE 50 QUESTIONS
anonymous
  • anonymous
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;
anonymous
  • anonymous
the result for linear algebraic multiplication for example A x B
anonymous
  • anonymous
A = [ cosx, -sinx, 0, acosx] [ sinx, cosx, 0, asinx] [ 0, 0, 1, 0] [ 0, 0, 0, 1]
anonymous
  • anonymous
B = [ cosy, -siny, 0, bcosy] [ siny, cosy, 0, bsiny] [ 0, 0, 1, 0] [ 0, 0, 0, 1]
anonymous
  • anonymous
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]
anonymous
  • anonymous
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
anonymous
  • anonymous
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]
anonymous
  • anonymous
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.
anonymous
  • anonymous
Sorry i was busy but i have to keep all my promises not matter what, so all the best and good luck.
anonymous
  • anonymous
Now with this basic method you can do whatever multiplication you want, and in any associative, distributive ... way you want.
anonymous
  • anonymous
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!

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