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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.

Keyur Mistry
on 22 Sep 2020

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