Iteration of a matrix equality

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fert 2017-5-14

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https://ww2.mathworks.cn/matlabcentral/answers/340230-iteration-of-a-matrix-equality

编辑： fert 2017-5-15

采纳的回答： John D'Errico

I have been trying to solve an equation of the matrices with the iteration.

A*PNode=del_node ; [Eq. 1]
del_node=(L(i)/D(i)) ; [Eq. 2]

The sizes of the matrices are respectively; A [5x6], PNode [6x1], del_node [5x1], L [1X6] and D [1x6].

The first and the last elements of the PNode should be 2 and 6 respectively. How can I find the possible converged values for the D and PNode matrices assuming that A and L are given?

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Answer 1

John D'Errico 2017-5-14

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You cannot do so. Too many unknowns.

PNode has 6 unknowns, but two are given, so only 4 true unknowns.

But you cannot solve for anything without the 6 unknown values of D. So up to 10 unknowns.

Oh, and del_node is completely unknown too. 5 more of them, so 15 unknowns in total.

The second set of equations, relating del_node to D, is of little help, since we can use that to essentially eliminate del_node.

One other point, the second set of equations is meaningless, since you tell us there are 5 elements of del_node. But the elements of del_node are simple ratios or L(i) and D(i). And since there are 6 values in L and D, that means you are trying to stuff 6 points of "stuff" into a 5 pound bag.

Regardless, you cannot solve the problem. No mathematics will give you an answer.

You need to spend some time re-thinking this problem.

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John D'Errico 2017-5-14

编辑：John D'Errico 2017-5-14

So IF you provide D, AND L, AND we assume that D and L are compatible in size, so they MUST be of length 5, NOT 6.

Then you have an over determined problem. Thus 4 unknowns, and 5 equations.

I don't have A so I can't generate the actual solution. Nor do you need to use symbolic variables at all. People seem to get trapped in the idea that if you don't know something, then it MUST be symbolic. That is not true.

So in your example, elements 1 and 5 are known. I'll pick some random matrix A, just to get some solution out.

A = rand(5,6);
rhs = [10;20;45;58;87];
PNode = NaN(6,1);
PNode([1 5]) = [2;6];
PNode([2 3 4 6]) = A(:,[2 3 4 6])\(rhs - A(:,[1 5])*[2;6])
PNode =
            2
       27.605
        102.9
       -16.17
            6
      -46.669

The result is of course random crap, since A was random crap. Again, since the problem is now over determined, there will in general be no analytical solution.

[A*PNode,rhs]
ans =
494           10
364           20
455           45
972           58
778           87

See that I never needed to use syms anywhere in the solution. And of course, you don't need to hard code the indices of those variables. Could you have done it using syms? Of course. But why bother?

Alternatively, you could have used the lsqlin function to solve this problem. Or, you could download my lse tool, which could have solved the problem too. But backslash, as I used it, is entirely adequate.

John D'Errico 2017-5-15

Notice the 1e+05* on top. Use the proper display format to be able to see all of the numbers.

format short g

is a good start. You really did get the same result as I did.

fert 2017-5-15

编辑：fert 2017-5-15

Now the hard part, I am basically trying to get the values of the u(and hence D) and PNodes with the While Loop. I am using a contemporary matrix kkk which must be equal to the del_pipe with the residual of the diff. Though I am getting the error of dimensions disagreement.Can you also give a hand on this matter? Also assuming you are an Italian, grazie millie Dottore:)

A = transpose([1 0 0 0 0;0 1 0 0 0; 0 -1 1 0 0; -1 0 0 1 0; 0 0 -1 -1 -1; 0 0 0 0 1]);
f=0.002;g=9.81; rho=1000;del_t=9;cp=4.18;
L =[12 11 8 6 4];Q_node=[210 157 281 150 50];u=ones(1,5);D=ones(1,5);
PNode=zeros(6,1);PNode([1 5])=[2;6];
del_pipe=ones(5,1);kkk=ones(5,1);
diff=del_pipe-kkk;
jj=0;
%%converging the del_pipe and the kkk matrices in order to find the optimum D and PNode
while diff<1e-5
    jj=jj+1;
    for i=1:1:5
        for j=1:1:6
            D(i)=sqrt((Q_node(i)*4)/(3*u(i)*cp*rho*9));
            del_node(i)=-1*(rho*f*L(i)/D(i))*u(i)^2/2;
            del_pipe=transpose(del_node);
            kkk(i)=(A(i,j)*PNode(j))+kkk(i);
            diff=abs(del_pipe-kkk);
        end
    end
end