Solve system of equations with some knowns and unknowns in the same matrix

Question

Yusuf 2022-10-25

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评论： David Goodmanson 2022-10-26

I am trying to solve for x in the equation Ax=b in matlab, but not all elements of x and b are unknown. I know the value of a few elements in x, and the rest are unknown, and I know the value of the elements in b for the x values that are unknown (ex. if I know x1, I dont know b1. If I dont know x2, I know b2, etc.). How can I automatically and efficiently solve for all of the unknowns in both vectors without manually writing the problem out as a system of equations each time?

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Answer 1

Bruno Luong 2022-10-25

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编辑：Bruno Luong 2022-10-25

在 MATLAB Online 中打开

% Random example
A = rand(5,5);
x = rand(5,1),
x = 5×1
    0.7978
    0.8825
    0.0743
    0.2101
    0.6719
b = A*x,
b = 5×1
    1.4535
    1.7942
    1.2279
    1.4185
    1.3270
[m,n] = size(A);
% put NaN at the position where x is unknown;
xunknown = randperm(n,3);
x(xunknown) = NaN;
% same for b
bunknown = randperm(m,2);
b(bunknown) = NaN;
% Reconstruct unknown x and then b
xunknown = isnan(x);
bunknown = isnan(b);
x(xunknown)=A(~bunknown,xunknown)\(b(~bunknown)-A(~bunknown,~xunknown)*x(~xunknown))
x = 5×1
    0.7978
    0.8825
    0.0743
    0.2101
    0.6719
b(bunknown)=A(bunknown,:)*x
b = 5×1
    1.4535
    1.7942
    1.2279
    1.4185
    1.3270

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Yusuf 2022-10-25

Thanks for the help. Going off my original question, if only x had knowns and unknowns (b is fully known), how would the solution change?

Bruno Luong 2022-10-25

The solution I give is applicable for all cases.

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Answer 2

David Goodmanson 2022-10-26

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编辑：David Goodmanson 2022-10-26

在 MATLAB Online 中打开

Hi Yusuf,

I'm later on an answer (and losing the race more often, to the extent that there is a race, which to be honest there sometimes is), but here is a slightly different way.

% data
n = 7;
m = 3;
A = rand(n,n);
i = sort(randperm(n,m))';   % index of unknown x
j = setdiff(1:n,i)';        % index of unknown b, the complementary index to i
x = zeros(n,1);
x(j) = rand(size(j))        % known x, zeros elsewhere
b = zeros(n,1);
b(i) = rand(size(i))        % known b, zeros elsewhere
%solve
G = zeros(n,n);
G(:,i) = A(:,i);
G(j,j) = -eye(length(j));
g = b-A(:,j)*x(j);
z = G\g
% z is a vector of all the unknowns,
% x(i) for the i indices
% b(j) for the j indices 
% check
x(i) = z(i);                % insert unknowns into orginal x vector                
b(j) = z(j);                % insert unknowns into orginal b vector
A*x-b                       % should be small   

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Bruno Luong 2022-10-26

Hi David;

My comments on yor method:

probably specific sparse construction of G when A is sparse
The eye and b on j components can be multiplied by a constants in both side, and might be some care needed to adjust the consant so as the condition of G is not bad (relative to that of A) when MATLAB internal factorization method such as QR during the "\'.

David Goodmanson 2022-10-26

Hi Bruno, that's a reasonable point about scaling part of the matrix I called G. I like the method because it gives all the unknowns in one go with backslash, different from the standard (and probably more foolproof) two-step process that you used.

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