Retreiving equations from a matrix

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

end

But this doesnt work

I would like my answer to be

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

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

[Snicklefrits]

or equation22=B(2,2)

[Matt J]

What should eq(i,1) and eq(1,j) be? Why not just set eq=B if eq and B are pretty much the same.

[Snicklefrits]

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]

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

Answer 1

[Matt J]

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 = 

[Snicklefrits]

When i run this I get

eq =

[0, 0, 0]

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

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

[Snicklefrits]

Ill post my whole code

[Matt J]

Well, I already showed you what I get.

[Snicklefrits]

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.

Answer 2

[Sulaymon Eshkabilov]

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)

[Snicklefrits]

This worked as well thank you.