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

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

R2020a

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