Retreiving equations from a matrix

Question

Snicklefrits 2021-11-2

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1577435-retreiving-equations-from-a-matrix

评论： Snicklefrits 2021-11-2

采纳的回答： Matt J

I have a symbolic matrix that has equation in it.

I want to extract the equations.

I have the code from the inside nodes.

B =

[50, 100, 100, 200]

[50, T23 - 4*T22 + T32 + 150, T22 - 4*T23 + T33 + 300, 200]

[50, T22 - 4*T32 + T33 + 350, T23 + T32 - 4*T33 + 500, 200]

[50, 300, 300, 200]

for i=2:n;

for j=2:m;

eq(i,j)=B(i,j)

end

But this doesnt work

I would like my answer to be

eq(2,2)=B(2,2)

eq(2,3)=B(2,3)

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Snicklefrits 2021-11-2

I want them to be individual equation to use the solve function. Is there any easier method to solve for the symbolic functions inside the matrix?

Snicklefrits 2021-11-2

I would want eq(2,1) to equal the equation of B(2,1) so I can use Solve(eq,T1 T2 T3 T4)

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Answer 1

Matt J 2021-11-2

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1577435-retreiving-equations-from-a-matrix#answer_822345

编辑：Matt J 2021-11-2

在 MATLAB Online 中打开

syms B eq T23 T22 T32 T33

B =[50, 100, 100, 200;

50, T23 - 4*T22 + T32 + 150, T22 - 4*T23 + T33 + 300, 200;

50, T22 - 4*T32 + T33 + 350, T23 + T32 - 4*T33 + 500, 200;

50, 300, 300, 200];

[n,m]=size(B);

eq(2:n,2:m)=B(2:end,2:end)

eq =

eq(2,2)

ans =

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Matt J 2021-11-2

Well, I already showed you what I get.

Snicklefrits 2021-11-2

What if i wanted to list all of the those values ie (B22,B23,B32,B33) like I just type with commas in between them So I can use

eqns = [B(2,2),B(2,3),B(3,2),B(3,3)

]

[A,b] = equationsToMatrix(eqns)

Y = inv(A)

Y*b

This code works. Thanks for you suggestion of just using the B matrix but I just need a way to call these values and seperate them by comas so I can use inverse matrix to solve it.

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Answer 2

Sulaymon Eshkabilov 2021-11-2

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1577435-retreiving-equations-from-a-matrix#answer_822350

在 MATLAB Online 中打开

Here is the corrected code:

clearvars
syms T22 T32 T33 T23
%% Note how your equations are set up
B =[50,                     100,                     100, 200;
50, T23 - 4*T22 + T32 + 150, T22 - 4*T23 + T33 + 300, 200;
50, T22 - 4*T32 + T33 + 350, T23 + T32 - 4*T33 + 500, 200; 
50,                     300,                     300, 200];
for ii = 1:4
    for jj=1:4
        EQN(ii,jj)=B(ii,jj)
    end
end
% Test
EQN(2,2)
EQN(3,2)