# how to solve the equation: Exdot=AX+Bu; with E is a singular matrix ? thank you

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khawla mrad il 2 Feb 2023
Modificato: Sam Chak il 2 Feb 2023
Exdot=AX+Bu; with E is a singular matrix ?
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Walter Roberson il 2 Feb 2023
see mass matrix options in odeset()

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### Risposte (1)

Sam Chak il 2 Feb 2023
Modificato: Sam Chak il 2 Feb 2023
If your system is linear, then try using dss() command. See example below:
If the system are nonlinear, then you need to specify 'MassSingular' and 'maybe', or 'yes' in the odeset().
A = [0 1; -1 -2]
A = 2×2
0 1 -1 -2
B = [0; 1]
B = 2×1
0 1
C = [1 0]
C = 1×2
1 0
D = 0;
E = [2 4; 4 8]
E = 2×2
2 4 4 8
rank(E) % Rank of matrix E is less than 2
ans = 1
inv(E)
Warning: Matrix is singular to working precision.
ans = 2×2
Inf Inf Inf Inf
% Descriptor state-space system
sys = dss(A, B, C, D, E)
sys = A = x1 x2 x1 0 1 x2 -1 -2 B = u1 x1 0 x2 1 C = x1 x2 y1 1 0 D = u1 y1 0 E = x1 x2 x1 2 4 x2 4 8 Continuous-time state-space model.
% Step Response
step(sys)
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John D'Errico il 2 Feb 2023
Modificato: John D'Errico il 2 Feb 2023
A comment: Do not test for singularity using det. This is simply a bad idea, and one that should NEVER be recommended to others, as that only propagates a terrible idea. Yes, teachers teach it. But they had to learn from someone else. And every time someone propagates a bad idea, it gets further entrenched in the minds of those who will read what you write.
If you want to test for a matrix being numerically singular, then use rank, or cond. If the condition number is sufficiently large (the limiting value depends on whether the matrix is single or double), then it is singular. If rank(A) is smaller than min(size(A)), then the array A is numerically singular.
If you don't believe me, then a simple example may suffice. Is the matrix A singular?
A = eye(1200);
Is the matrix B singular?
B = A/2;
Finally, is the matrix C singular?
C = A*2;
Surey you would agree that all are perfectly well behaved matrices, and as far from being singular as you could imagine. Test each of them using det.
det(A)
ans = 1
det(B)
ans = 0
det(C)
ans = Inf
Note that det(B) is returned as an EXACT zero. Not even just a small number.
det(B) == 0
ans = logical
1
And these are not the only reasons why det is a bad method to employ.
Remember that those who you will teach today will be teaching others tomorrow.
Sam Chak il 2 Feb 2023

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