Solution Vector for Matrix from Upper Bandwith, transformed vector, and upper triangular result

back_subst_banded.m

Modify the backward substitution algorithm to take advantage of the upper bandwidth of an upper triangular system. No arithmetic should be performed on any element known to be zero.

Inputs:

U The upper triangular result

d The transformed right-hand-side vector

ub The upper bandwidth, i.e., the number of stripes above the matrix diagonal that have non-zero elements.

Outputs:

x The solution vector

. . . .. So far this is what I have:

Function x= back_subst_banded(U, d, ub)

% U= upper triangular result from gaussbandfunc, d= transformed

%right hand vector for gaussbandfunc, ub= upper bandwith

format short

 p = (1:ub)';          % initialize the pivoting vector

s = max(abs(A'));     % compute the scale of each row

for k = 1:(ub-1)

    r = abs(A(p(k),k)/s(p(k)));

    kp = k;

    for i = (k+1):bb

      t = abs(A(p(i),k)/s(p(i))); 

      if t > r,  r = t;  
kp = i; 

      end

    end

    l = p(kp); 
 p(kp) = p(k); 
 p(k) = l;   % interchange p(kp) and p(k) 

    for i = (k+1):ub

      A(p(i),k) = A(p(i),k)/A(p(k),k);

      for j = (k+1):ub

        A(p(i),j) = A(p(i),j)-A(p(i),k)*A(p(k),j);

      end

    end

end

d = zeros(bb,1);    

d(1) = b(p(1));

for i = 2:ub

    d(i) = b(p(i));

    for j = 1:(i-1)

      d(i) = d(i)-A(p(i),j)*d(j);

    end

end

d

. . . . .

. . Im not sure how to use the upper bandwith to get my solution vector here