How to find which equation that have "Warning: Solution does not exist because the system is inconsistent"
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I have big matrix and want to solve it with "\".
Ax=B so I use
[A,B]=equationsToMatrix(k,q)
x=A\B
but the matrix cannot be solved, the solution of x = Inf
How to know which one equation that makes x dont have solutions?
Answers (1)
If you get inf on output, there are few possibilities:
- that there are infinite entries in the matrix (but these would typically lead to NaN)
- that the matrix has a quite high condition number, leading to overflows in the calculation (NaN might well show up as well). This can include the case where the matrix is singular algebraically but the determinant is not exact zero due to floating point round-off
Example:
format long g
M = imag(hilbert(magic(5))) / 1e5
rank(M)
cond(M)
M\(1:5).'
If you do not have actual infinite entries then there isn't "one equation" that makes it "not have a solution". Consider for example
[1 2 3
2 4 6
3 5 9]
Which "one equation" makes this not have an answer? You might at first say that it is the [2 4 6] because that is a constant multiple times an earlier row, but you could just as well say that the problem is the [1 2 3] for being a constant multiple of [2 4 6] .
When you use EquationsToMatrix() then you are working with systems of symbolic equations that are probably not inherently ordered. In such cases it could not be considered to be the "fault" of the second matrix that the system is not consistent; if you had happened to enter the equations in a different order then it would have been a different row "at fault".
It is fair to ask which rows are linearly dependent on earlier rows, understanding that does not mean that such rows are "wrong" (could be the other rows fault). I am not sure what a robust solution for that is, but it looks to me as if you can proceed to
[P,r] = rref(M.');
dependent_rows = setdiff(1:size(M,1), r)
This particular example happens to indicate that the last row is the first dependent row, but if you do something like
N = 20;
A = rand(N);
RRR = randperm(N,3)
A(RRR(1),:) = A(RRR(2),:) * 5 - A(RRR(3),:) * 19;
[P,r] = rref(A.');
dependent_rows = setdiff(1:N, r)
dependent_rows == max(RRR)
We were successful in identifying a dependent row. We constructed row 6 as dependent on row 3 and 18, but mathematically that is equivalent to saying that row 18 is dependent on rows 3 and 6 -- which further emphasizes my point earlier that you have to be careful about which equation you blame.
7 Comments
This is the code for the matrix.
format rational
A_0=[-4 0 1;1 -8 2;0 4 -11]
B=[2 1 4;3 1 0;4 0 2]
A=[-15 2 3;2 -11 0;1 0 -14]
C=[1 2 0;0 1 4;2 3 2] %check if sum of every row must equal to 0
O=zeros(3);
Q=[A_0 C O;
B A C;
O B A]
sum(Q,2)
R=zeros(3);
X=0;
Z=0;
Err=0;
for i=1:18
fprintf('i=%g',i);
R=((-C)-((R)^2)*B)*A^-1
%num2str(R,'%7.3f'); %I'm not using it because it may affect the result (?)
Err=R-X
if isequal(R,X)
Y=R
break
end
if Err<=10^(-5)
K=R
break
end
X=R
%num2str(X,'%7.3f');
end
After getting R, i can get q1,q2,q3. But in this example, always pop up warning that the system is inconsistent. I try to search if the matrix has dependent row but the command cannot be done, another warning.
q=sym('q',[1 3])
I=eye(3);
e=ones(3,1);
w=q*(A_0+R*B)
%w=vpa(w,5)
a=q*inv(I-R)*e
%a=vpa(a,5)
aa=a-1
%aa=vpa(aa,5)
j=solve(aa)
%j=vpa(j,5)
k=subs(w,sym('q1'),j)
%k=vpa(k,5)
size(k)
[C,S]=equationsToMatrix(k,q)
%C=vpa(C,5)
%S=vpa(S,5)
rank(C)
rank(S)
cond(C)
rref(C)
ji=C\S
sum=0;
for i=1:3
sum = sum +ji(i);
end
sum
size(sum);
q1=1-sum
Torsten
on 14 Dec 2022
The first column of the matrix C is the 3x1 null vector. Thus your system C*q = S has no solution.
I have example that all element on the first column is zero
ll=sym('l',[1 3])
dd=[0 -2 2;0 -1 1;0 3/2 -3/2]
ak=ll*[1;1;1];
akk=ak-1
ja=solve(akk)
ee=ll*dd
ek=subs(ee,sym('l1'),ja)
[DD,SS]=equationsToMatrix(ek,ll)
size(DD)
size(SS)
rank(DD)
rank(SS)
jak=DD\SS
summe=0
for i=1:3
summe = summe +jak(i);
end
summe
size(summe)
l1=1-summe
For this example, i still get l1 to l3. I dont get the inconsistent warning.
Torsten
on 14 Dec 2022
Yes,
x + y = 2
2*x + 2*y = 4
has a solution, but
x + y = 2
2*x + 2*y = 5
has none.
DoinK
on 14 Dec 2022
Torsten
on 14 Dec 2022
if rank(C) = rank([C,S]), you can solve the system.
if rank(C) = rank([C,S]) -1, the system is inconsistent.
DoinK
on 27 Dec 2022
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on 27 Dec 2022
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