Ini Algoritmanya
1. untuk eliminasi maju
DO k = 1,n-1
DO i = k+1,n
factor = a[i,k] / a[k,k]
DO j = k+1 to n
a[i,j] =
a[i,j] - factor * a[k,j]
END DO
b[i] = b[i] - factor * b[k]
END DO
END DO
2. untuk eliminasi kebalikannya
x[n] = b[n] / a[n,n]
DO i = n-1,1,-1
sum = 0
DO j = 1+1,n
sum = sum + a[i,j] * x[j]
END DO
x[i] = (b[i] - sum)/a[i,i]
END DO
listing programnya kira-2 jiga kieu
dalam bahaasa CCCCCCC
#include<stdio.h>
#include<conio.h>
main()
{
int i,a[4][4],b[4],j,k;
float x[4],sum,factor;
clrscr();
printf("assigment nilai");
a[0][0]=3;a[0][1]=1; a[0][2]=2; a[0][3]=-1;b[0]=16;
a[1][0]=1;a[1][1]=-4;a[1][2]=-1;a[1][3]=0;b[0]=9;
a[2][0]=1;a[2][1]=-2;a[2][2]=1; a[2][3]=1;b[0]=15;
a[3][0]=0;a[3][1]=1; a[3][2]=-3;a[3][3]=5;b[0]=12;
for(i=0;i<4;i++)
for(j=0;j<4;j++)
{
printf("\na[%d,%d]
= %d ",i+1,j+1,a[i][j]);
}
//eliminasi maju
for(k=0;k<4-1;k++)
for(i=k+1;i<4;i++)
{
factor =
a[i][k] / a[k][k];
for(j=k+1;j<4;j++)
a[i][j]=a[i][j] - factor * a[k][j];
b[i] = b[i] -
factor * b[k];
}
//eleminasi mundur
x[3]=b[3]/a[3][3];
for(i=3;i>-1;i--)
{
sum =0;
for(j=i+1;j<4;j++)
sum = sum +
a[i][j] * x[j];
x[i] = (b[i]-sum)/a[i][i];
}
for(i=0;i<4;i++)
{
printf("\nx[%d] = %f ",i+1,x[i]);
}
getche();
return 0;
}