How can I make a system identify if a solution has no solutions?

Hello,

I was wondering how I can make this backsubstitution program identify whether a solution is infinite or has no solutions?

Thank you

function x = backsub(U,b)
if det(A)<=0.000001 %For infinite solution
    n = length(b);
    syms t
    x=sym(zeros(n,1))
    x(n)=sym('t')
    b=(sym(b))
    for i = n:-1:1
        x(i)=b(i);
        x(n)=sym('t')
        if i<n
            for j = n:-1:i+1            
                x(i)=x(i)-A(i,j)*x(j);            
            end
        end    
        x(i)=x(i)/A(i,i);
    end    
else %For unique solution
    n = length(b);
    x = zeros(size(b));
    for i = n:-1:1
        x(i)=b(i);
        if i<n
            for j = n:-1:i+1
                x(i)=x(i)-A(i,j)*x(j);
            end
        end
        x(i)=x(i)/A(i,i);
        x=double(x)
    end
end
end

Answer 1

[Keyur Mistry]

I understand you want to identify that the given system has ‘infinite solution’ or ‘no solution’. For the same you can consider using command ‘rank’ on the system matrix ‘U’.

I hope this is useful for you to find the solution.

[Jonas Freiheit]

Thank you, Another approach i've done earlier was

if abs(det(A))<=0.00005

then possibly inf or no sol then

if b(n)~=0

then no solutions

[Keyur Mistry]

For more clarifiacton rank(U) and rank([U b]) can be compared to check if 'b' is inside the image space of 'U' or not. This is to identify if there is ‘infinite solution’ or ‘no solution’.

[Jonas Freiheit]

Interesting.

Thanks