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hi, i'm trying to solve for a system of linear equation in form of matrices and i have made a function to perform the backward subtitute after doing all the pivoting and solve for the case where it has infinite solution only. this is what i got so far:

function x = backsub_syms(U,b)

n = length(b);

syms x [1 n]

u = rank(U);

p = n - u;

if (p == 1)

syms t

x(1,n)= t;

else if (p>1)

????????????????

end

end

for i=n-1:-1:1

m =1/U(i,i).*(b(i)-sum(U(i,i+1:end).*x(1,i+1:end)));

x(1,i)= m;

end

x=x(1,:)';

end

it is working perfectly fine only when the 1 free variable (p = 1) so in case p is larger than 1, i dont know how to approach this since this can only use symbolic variable to calculate. can some one give me an idea or a hint to complete the '??????' part

Keyur Mistry
on 21 Sep 2020

I understand that you want to generalize your code for ‘p>1’. Find following hints to achieve it.

As there will be ‘p’ numbers of free variables, symbolic variable array can be defined ‘t1’,’t2’,…’tp’ as below

syms t [1 p];

for i=n:-1:n-p+1

x(1,i) = t(1,p-n+i); %if first non-zero entry in each row of U is diagonal element

end

It is necessary to find first non-zero entry for each row of ‘U’ to get correct solution. For the same find below code.

for i=n-p:-1:1

j=1;

d=U(i,j);

while d==0

j=j+1;

d=U(i,j);

end

m =1/d.*(b(i)-sum(U(i,j+1:end).*x(1,j+1:end)));

x(1,j) = m;

end

I hope these are useful hints to find your solution.

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