Gauss column principal element elimination function

% Column principal element elimination solution Equation Group ax=b, realize pa=lu

function [X,deta] =gauss (A, B)

N=length (b); [P,q]=size (A);

If p~=q| | P~=n fprintf (' the dimension of the square is different, please lose again! '); % Error detection

End

% to increase the operating speed, give the initial value to L,U,X,C,D1

L=zeros (N,n);

U=zeros (N,n);

X=zeros (n,1);

C=zeros (1,n);

d1=0;

% Select the main element by column, and line Exchange, record line information

For i=1:n-1

Max=abs (A (i,i));

M=i;

For J=i+1:n

If Max<abs (A (j,i))

Max=abs (A (j,i));

M=j;

End

End

if (m~=i)

For K=i:n

C (k) =a (i,k);

A (i,k) =a (m,k);

A (M,k) =c (k);

End

D1=b (i);

B (i) =b (m);

B (m) =d1;

End

% to calculate the elimination of the element

For k=i+1:m

For J=i+1:n

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

End

B (k) =b (k)-B (i) *a (k,i)/A (i,i);

A (K,i) = 0;

End

% back-generation solution

X (n) =b (n)/A (n,n);

For I=n-1:-1:1

sum=0.0;

For J=i+1:n

Sum=sum+a (I,j) *x (j);

End

X (i) = (b (i)-sum)/A (i,i);

End

% calculates the value of the determinant

Deta=1;

For K=1:n

Deta=deta*a (K,K);

End

% output pa=lu information in the l,u

For I=1:n

For J=1:n

If I<j

U (i,j) =a (I,J);

ElseIf i==j

L (I,J) = 1;

U (i,j) =a (K,J);

Else

L (i,j) =a (I,J);

End

End

End

End

Solving the problem Equation Group (script file):

A=[-0.002 2 2

1 0.78125 0

3.996 5.5625 4];

B=[0.4;1.3816;7.4178];

%[l,u,x,deta]=mylu (A, B)

[X,deta]=gauss (A, B)

DISP ( the solution obtained by the Gaussian elimination method of the column principal element is:')

Six Operation Result:

x =

1.8167

0.0527

-0.0337

Deta =

798.0666

Gauss column principal element elimination function