NUmeric : Naive Gauss code

Sample equation for Naïve Gauss

3x₁-.01x₂-0.2x₃=7.85

0.1x₁+7x₂-0.3x₃=-19.3

0.3x₁-0.2x₂+10x₃=71.43

 

 

//perfect for all gauss method

#include<stdio.h>

int main()

{

   // freopen("input.txt","r",stdin);

    float a[10][10],x3,x2,x1;

    float p;

    int c,r,i,j;

    printf("Enter row=");

    scanf("%d",&r);

    printf("Enter collom=");

    scanf("%d",&c);

//.............................................

    for(i=0;i<r;i++)

    {

        for(j=0;j<c;j++)

        {

            printf("a[%d][%d]=",i,j);

            scanf("%f",&a[i][j]);

        }

    }

    //...........................................

    printf("\nHere is the matrix =\n");

for (i=0;i<r;i++)

{

    for(j=0;j<c;j++)

    {

        printf("%.2f  ",a[i][j]);

    }

    printf("\n");

}

//...............................................

//2nd row

printf(" \n\n\n");

p=a[1][0];

for(i=1;i<2;i++)

{

    for(j=0;j<4;j++)

    {

        a[i][j]=(a[i][j]-a[0][j]*(p/a[0][0]));

        //printf("%.2f  ",a[i][j]);

    }

}

//...............................................

printf("\nAFTER r2=r2-r1*(10/00) =\n");

for (i=0;i<r;i++)

{

    for(j=0;j<c;j++)

    {

        printf("%.2f  ",a[i][j]);

    }

    printf("\n");

}

//...............................................

printf("\n");

//3rd row

float q=a[2][0];

for(i=2;i<3;i++)

{

    for(j=0;j<4;j++)

    {

        a[i][j]=(a[i][j]-a[0][j]*(q/a[0][0]));

       // printf("%.2f  ",a[i][j]);

    }

}

//...............................................

printf("\nAFTER r3=r3-r1*(20/00) =\n");

for (i=0;i<r;i++)

{

    for(j=0;j<c;j++)

    {

        printf("%.2f  ",a[i][j]);

    }

    printf("\n");

}

//...............................................

printf("\n");

float z=a[2][1];

for(i=2;i<3;i++)

{

    for(j=0;j<4;j++)

    {

        a[i][j]=(a[i][j]-a[1][j]*(z/a[1][1]));

       // printf("%.2f  ",a[i][j]);

    }

}

//...............................................

printf("\nAFTER r3=r3-r2*(21/11) =\n");

for (i=0;i<r;i++)

{

    for(j=0;j<c;j++)

    {

        printf("%.2f  ",a[i][j]);

    }

    printf("\n");

}

//...............................................

x3=a[2][3]/a[2][2];

printf("\n here x3 is = %f ",x3);

x2=((a[1][3]-a[1][2]*x3)/a[1][1]);

printf("\n here x2 is = %f ",x2);

x1=(a[0][3]-(a[0][1]*x2)-(a[0][2]*x3))/a[0][0];

printf("\n here x1 is = %f ",x1);

return 0;

}