How to solve matrices with 6 unknowns?

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Mr. NailGuy 2017-11-1

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https://ww2.mathworks.cn/matlabcentral/answers/364384-how-to-solve-matrices-with-6-unknowns

评论： Mr. NailGuy 2017-11-1

I have two matrices as shown below. I need to equate the individual elements (e.g. Ad(1,1)=Af(1,1),Ad(1,2)=Af(1,2),etc.) which forms different set of equations (9 equations to be exact), then afterwards I need to solve these system of equations simultaneously to get the values of K11,K12,K13,K21,K22 and K23. Could you suggest any method to do this in Matlab?

Ad =

    -1     0     0
     0     0     1
     0    -6    -5

Af = [(-K11-1) (-K12+7) -K13;0 0 1;-K21 (-K22-2) (3-K23)]

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Answer 1

KSSV 2017-11-1

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https://ww2.mathworks.cn/matlabcentral/answers/364384-how-to-solve-matrices-with-6-unknowns#answer_288774

在 MATLAB Online 中打开

Ad = [ -1     0     0
     0     0     1
     0    -6    -5] ;
   syms K11 K12 K13 K21 K22 K23
   Af = [(-K11-1) (-K12+7) -K13;0 0 1;-K21 (-K22-2) (3-K23)] ;
   eqn = Af==Ad
   sol = solve(eqn)

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Mr. NailGuy 2017-11-1

在 MATLAB Online 中打开

Thanks for the response. Yeah I did define it, however, the final form of "eqn" will be as show below. As you can see, each cell of the matrix will now be coupled with different K11 and K21 which is not the case as what I posted previously :) sol.K11, etc don't give answers now when I executed it. Will appreciate your feedback on this.

 [               (1427*K11)/5 + (832*K21)/5 == 0,           (1427*K12)/5 + (832*K22)/5 + 1 == 1,                  (1427*K13)/5 + (832*K23)/5 == 0,             (1427*K14)/5 + (832*K24)/5 == 0]
[ 4933/10 - (15653*K21)/5 - (60167*K11)/10 == 0, - (60167*K12)/10 - (15653*K22)/5 - 164/5 == 0, - (60167*K13)/10 - (15653*K23)/5 - 27489/10 == 1, 706/5 - (15653*K24)/5 - (60167*K14)/10 == 0]
[                (208*K11)/5 + (129*K21)/5 == 0,                (208*K12)/5 + (129*K22)/5 == 0,                   (208*K13)/5 + (129*K23)/5 == 0,          (208*K14)/5 + (129*K24)/5 + 1 == 1]
[ 1159/10 - (2814*K21)/5 - 1073*K11 == -252/125,     - 1073*K12 - (2814*K22)/5 - 14/5 == -7/10,   - 1073*K13 - (2814*K23)/5 - 1287/2 == -441/100,    81/10 - (2814*K24)/5 - 1073*K14 == -18/5]

KSSV 2017-11-1

在 MATLAB Online 中打开

Abar = (1.0e+03)*[0.0000 0.0010 0 -0.0000; 0.4933 -0.0328 -2.7489 0.1412; -0.0000 -0.0000 0.0000 0.0010; 0.1159 -0.0028 -0.6435 0.0081] ;
Bbar = (1.0e+03)*[-0.2854 -0.1664; 6.0167 3.1306; -0.0416 -0.0258; 1.0730 0.5628] ;
Ad =[0 1 0 0; 0 0 1 0; 0 0 0 1; -2.0160 -0.7 -4.41 -3.6];
Kbar = Bbar\(Abar-Ad) ;