Doolittle

This method belongs to the LU factoring system. From matrix A, there are two matrices L and U, in this method the matrix L is triangular inferior and in the diagonal it has 1 (unitary), while U is a superior triangular matrix. Being the triangular matrices, we can calculate each element of the matrix A by means of...

Pseudocode.

Input A, b

n= # de filas de A

L=eye(n);

U=zeros(n);

for k=1:n

for j=k:n

suma=0;

for i=1:k-1

suma=suma+L(k,i)*U(i,j);

end

U(k,j)=A(k,j)-suma;

end

for i=k:n

suma=0;

for j=1:k-1

suma=suma+L(i,j)*U(j,k);

end

L(i,k)=(A(i,k)-suma)/U(k,k);

end

end

for i=1:n

suma=0;

for p=1:i-1

suma=suma+L(i,p)*d(p,1);

end

d(i,1)=(b(i,1)-suma)/L(i,i);

end

for i=n:-1:1

suma=0;

for p=i+1:n

suma=suma+U(i,p)*X(p,1);

end

X(i,1)=(d(i,1)-suma)/U(i,i);

end

Code.

Clc

A=input('ingrese la matriz: ');

b=input('ingrese la matriz de términos independientes: ');

n=size(A,1);

L=eye(n);

U=zeros(n);

for k=1:n

for j=k:n

suma=0;

for i=1:k-1

suma=suma+L(k,i)*U(i,j);

end

U(k,j)=A(k,j)-suma;

end

for i=k:n

suma=0;

for j=1:k-1

suma=suma+L(i,j)*U(j,k);

end

L(i,k)=(A(i,k)-suma)/U(k,k);

end

end

disp('U=')

disp(U)

disp('L=')

disp(L)

disp('L*U=')

disp(L*U)

for i=1:n

suma=0;

for p=1:i-1

suma=suma+L(i,p)*d(p,1);

end

d(i,1)=(b(i,1)-suma)/L(i,i);

end

disp('d=');

disp(d);

for i=n:-1:1

suma=0;

for p=i+1:n

suma=suma+U(i,p)*X(p,1);

end

X(i,1)=(d(i,1)-suma)/U(i,i);

end

disp(‘La solucion es::')

disp(X)

B=A*X