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

13 views (last 30 days)
Jonas Freiheit on 19 Sep 2020
Commented: Jonas Freiheit on 23 Sep 2020
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

Keyur Mistry on 22 Sep 2020
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 on 23 Sep 2020
Interesting.
Thanks

R2020a

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