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Linear Algebra Error: Matrix is close to singular or badly scaled.

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Shrishti Yadav
Shrishti Yadav 2023 年 9 月 2 日
回答済み: Star Strider 2023 年 9 月 2 日
Hi,
I have a simple localization code that calculates the x,y,z coordinates for a wave heard at 4 receivers. The equations are set up using matrices but it is giving me an error I don't understand. I checked if the matrix reaches singularity and it does not. So I am not sure what the error is pointing to. I am attaching the code here:
% Sensor/receiver locations/ vertices of a tetrahedron
% si(xi,yi,zi)
r1 = [0.125 0.125 0.125];
r2 = [0.25 0.125 0.125];
r3 = [0.1875 0.25 0.125];
r4 = [0.1875 0.1875 0.25];
% Calculate the Time of Arrival at the receivers
c = 343; % m/s
t = zeros(1, 4);
sound_loc = [10 20 10]; % Replace x, y, and z with your sound location coordinates
vertices = [r1; r2; r3; r4]; % Replace vertex1 to vertex4 with your 3D vertices
for j = 1:4
t(j) = pdist([sound_loc; vertices(j, :)], 'euclidean') / c;
end
% Setting up matrices to solve for location of source
A = [2*r1(1) 2*r1(2) 2*r1(3) 2*c*t(1) -1; 2*r2(1) 2*r2(2) 2*r2(3) 2*c*t(2) -1; 2*r3(1) 2*r3(2) 2*r3(3) 2*c*t(3) -1; 2*r4(1) 2*r4(2) 2*r4(3) 2*c*t(4) -1 ];
%u = [x y z -c*t r^2-c^2*t^2];
r1_sq = sqrt(r1(1)^2 + r1(2)^2 + r1(3)^2);
r2_sq = sqrt(r2(1)^2 + r2(2)^2 + r2(3)^2);
r3_sq = sqrt(r3(1)^2 + r3(2)^2 + r3(3)^2);
r4_sq = sqrt(r4(1)^2 + r4(2)^2 + r4(3)^2);
b = [r1_sq-c^2*t(1)^2; r2_sq-c^2*t(2)^2; r3_sq-c^2*t(3)^2; r4_sq-c^2*t(4)^2 ];
% Calculating u : u = [x y z -c*t r^2-c^2*t^2];
% u = (A^T*A)^-1*A^T*b
t = 0;
u1 = (transpose(A)*b);
u2 = (transpose(A)*A);
u = inv(u2) * u1;
  3 件のコメント
Bruno Luong
Bruno Luong 2023 年 9 月 2 日
編集済み: Bruno Luong 2023 年 9 月 2 日
@Star Strider "u2 = A\A"
where it comes from??? what you want to show here?
Shrishti Yadav
Shrishti Yadav 2023 年 9 月 2 日
編集済み: Shrishti Yadav 2023 年 9 月 2 日
The original equation was: Au = b with A as an mxn matrix and u is nx1 and b is nx1
so taking the transpose on both sides and then get only u on the left hand side, we get the following after multiplying both sides by the inverse of the transpose of A * A:
u = inv(A^TA)*A^T*b
So it should be ok in terms of singularity. The goal was to calculate u correctly. that was it. i didn't know if the way i am multiplying it out makes an error.

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採用された回答

Star Strider
Star Strider 2023 年 9 月 2 日
I am not certain what you want to do.
Consider using the pinv function.

その他の回答 (1 件)

Bruno Luong
Bruno Luong 2023 年 9 月 2 日
編集済み: Bruno Luong 2023 年 9 月 2 日
@Shrishti Yadav "I checked if the matrix reaches singularity and it does not. "
Your A matrix has size 4 x 5, the the rank is maximum 4.
The matrix u2 = A'*A is 5 x 5 with maximum rank <= 4 so it must be singular, despite what you claim.
In short you try to solve for 5 unknown with 4 equations. The system is then underdetermined and MATLAB warns you that.

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