Solution of system of linear equation by LU factorisation.
#include<stdio.h>
#include<malloc.h>
void LUfactorisation(double *A[10],double *L[10],double *U[10], int ROW);
void Solution1(double *A[10],double *L[10],double *U[10],double *b,int ROW);
int main()
{
double *A[10],*L[10],*U[10],*b;
int ROW,i,j;
printf("\n Enter the order of the matrix:");
scanf("%d",&ROW);
for(i=0;i<=ROW;i++)
{
A[i]=(double *)malloc((ROW+1)*sizeof(double));
L[i]=(double *)malloc((ROW+1)*sizeof(double));
U[i]=(double *)malloc((ROW+1)*sizeof(double));
}
b=(double *)malloc((ROW+1)*sizeof(double *));
printf("\n Enter the elements of the matrix A:");
for(i=1;i<=ROW;i++)
for(j=1;j<=ROW;j++)
{
printf("\n Enter the element at position A[%d][%d]:",i,j);
scanf("%lf",A[i]+j);
}
printf("\n Enter the elements of the coloumn matrix b:");
for(i=1;i<=ROW;i++)
{
printf("\n Enter the element at position b[%d]:",i);
scanf("%lf",(b+i));
}
printf("\n The matrix A is:\n");
for(i=1;i<=ROW;i++)
{
for(j=1;j<=ROW;j++)
printf("%lf ",*(A[i]+j));
printf("\n");
}
for(i=1;i<=ROW;i++)
for(j=1;j<=ROW;j++)
{
if(i>j)
*(U[i]+j)=0.0;
else
*(L[i]+j)=0.0;
}
for(i=1;i<=ROW;i++)
*(L[i]+i)=1.0;
LUfactorisation(A,L,U,ROW);
printf("\n The matrix L is:\n");
for(i=1;i<=ROW;i++)
{
for(j=1;j<=ROW;j++)
printf("%lf ",*(L[i]+j));
printf("\n");
}
printf("\n The matrix U is:\n");
for(i=1;i<=ROW;i++)
{
for(j=1;j<=ROW;j++)
printf("%lf ",*(U[i]+j));
printf("\n");
}
Solution1(A,L,U,b,ROW);
return 0;
}
void Solution1(double *A[10],double *L[10],double *U[10],double *b,int ROW)
{
double *Y,*X,temp=0.0;
int i,j,k,l;
Y=(double *)malloc((ROW+1)*sizeof(double));
X=(double *)malloc((ROW+1)*sizeof(double));
for(i=1;i<=ROW;i++)
{
if(i==1)
Y[i]=b[i]/L[i][i];
else
{
temp = 0.0;
k=i-1;
l=1;
while(l<=k)
{
temp = temp +(L[i][l]*Y[l]);#include<stdio.h>
#include<malloc.h>
void LUfactorisation(double *A[10],double *L[10],double *U[10], int ROW);
void Solution1(double *A[10],double *L[10],double *U[10],double *b,int ROW);
int main()
{
double *A[10],*L[10],*U[10],*b;
int ROW,i,j;
printf("\n Enter the order of the matrix:");
scanf("%d",&ROW);
for(i=0;i<=ROW;i++)
{
A[i]=(double *)malloc((ROW+1)*sizeof(double));
L[i]=(double *)malloc((ROW+1)*sizeof(double));
U[i]=(double *)malloc((ROW+1)*sizeof(double));
}
b=(double *)malloc((ROW+1)*sizeof(double *));
printf("\n Enter the elements of the matrix A:");
for(i=1;i<=ROW;i++)
for(j=1;j<=ROW;j++)
{
printf("\n Enter the element at position A[%d][%d]:",i,j);
scanf("%lf",A[i]+j);
}
printf("\n Enter the elements of the coloumn matrix b:");
for(i=1;i<=ROW;i++)
{
printf("\n Enter the element at position b[%d]:",i);
scanf("%lf",(b+i));
}
printf("\n The matrix A is:\n");
for(i=1;i<=ROW;i++)
{
for(j=1;j<=ROW;j++)
printf("%lf ",*(A[i]+j));
printf("\n");
}
for(i=1;i<=ROW;i++)
for(j=1;j<=ROW;j++)
{
if(i>j)
*(U[i]+j)=0.0;
else
*(L[i]+j)=0.0;
}
for(i=1;i<=ROW;i++)
*(L[i]+i)=1.0;
LUfactorisation(A,L,U,ROW);
printf("\n The matrix L is:\n");
for(i=1;i<=ROW;i++)
{
for(j=1;j<=ROW;j++)
printf("%lf ",*(L[i]+j));
printf("\n");
}
printf("\n The matrix U is:\n");
for(i=1;i<=ROW;i++)
{
for(j=1;j<=ROW;j++)
printf("%lf ",*(U[i]+j));
printf("\n");
}
Solution1(A,L,U,b,ROW);
return 0;
}
void Solution1(double *A[10],double *L[10],double *U[10],double *b,int ROW)
{
double *Y,*X,temp=0.0;
int i,j,k,l;
Y=(double *)malloc((ROW+1)*sizeof(double));
X=(double *)malloc((ROW+1)*sizeof(double));
for(i=1;i<=ROW;i++)
{
if(i==1)
Y[i]=b[i]/L[i][i];
else
{
temp = 0.0;
k=i-1;
l=1;
while(l<=k)
{
temp = temp +(L[i][l]*Y[l]);
l++;
}
Y[i]=(b[i]-temp)/L[i][i];
} // end of else
} // end of for
for(i=ROW;i>=1;i--)
{
if(i==ROW)
X[ROW]=Y[ROW]/U[ROW][ROW];
else
{
temp = 0.0;
k=i+1;
while(k<=ROW)
{
temp = temp +(U[i][k]*X[k]);
k++;
}
X[i]=(Y[i]-temp)/U[i][i];
} // end of else
} // end of outer for
printf("\n Solution of the given linear equation using LU factorisation:\n");
for(i=1;i<=ROW;i++)
printf(" x[%d] = %lf\n",i,X[i]);
return;
}
void LUfactorisation(double *A[10],double *L[10],double *U[10], int ROW)
{
int i,j,k,l;
double temp=0.0;
for(j=1;j<=ROW;j++)
{
for(i=1;i<=ROW;i++)
{
if(i<=j)
{
temp=0.0;
k=i-1;
l=1;
while(l <=k && i!=1)
{
temp = temp + (L[i][l]*U[l][j]);
l++;
}
U[i][j]=A[i][j]-temp;
} // end of if
else
{
temp=0.0;
k=j-1;
l=1;
while(l<=k)
{
temp = temp + L[i][l]*U[l][j];
l++;
}
L[i][j]=(A[i][j]-temp)/U[j][j];
} // end of else
} // end of inner for loop
} // end of outer for loop
return;
}
l++;
}
Y[i]=(b[i]-temp)/L[i][i];
} // end of else
} // end of for
for(i=ROW;i>=1;i--)
{
if(i==ROW)
X[ROW]=Y[ROW]/U[ROW][ROW];
else
{
temp = 0.0;
k=i+1;
while(k<=ROW)
{
temp = temp +(U[i][k]*X[k]);
k++;
}
X[i]=(Y[i]-temp)/U[i][i];
} // end of else
} // end of outer for
printf("\n Solution of the given linear equation using LU factorisation:\n");
for(i=1;i<=ROW;i++)
printf(" x[%d] = %lf\n",i,X[i]);
return;
}
void LUfactorisation(double *A[10],double *L[10],double *U[10], int ROW)
{
int i,j,k,l;
double temp=0.0;
for(j=1;j<=ROW;j++)
{
for(i=1;i<=ROW;i++)
{
if(i<=j)
{
temp=0.0;
k=i-1;
l=1;
while(l <=k && i!=1)
{
temp = temp + (L[i][l]*U[l][j]);
l++;
}
U[i][j]=A[i][j]-temp;
} // end of if
else
{
temp=0.0;
k=j-1;
l=1;
while(l<=k)
{
temp = temp + L[i][l]*U[l][j];
l++;
}
L[i][j]=(A[i][j]-temp)/U[j][j];
} // end of else
} // end of inner for loop
} // end of outer for loop
return;
}
