Numerical Mathematics : Naive Gauss code (C Programming )

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;  
 }