Gauss Elimination Method Using Partial Pivoting and Dynamic Memory Allocation Using c/c++

We can find the solution of linear equation of any order using Gauss Elimination Method.Partial Pivoting is Method apply to eliminate 0 at the diagonal of matrix . So we can find the solution of any order linear equation EFFICIENTLY.This program may helpful on various way ie finding the inverse of matrix or finding the determinant of matrix of any order (n*n) or on solving linear equation of n unknown values.
  Here memory of  two dimensional matrix is allocated dynamically using malloc / calloc which is the best method of allocating memory in C programming language.
Source Code Of Gauss Elimination Method Using Partial Pivoting and Dynamic Memory Allocation Using c/c++ 

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IDE  CODE :: BLOCK 10.05
partial_pivoting.c
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#include<stdlib.h>
#include<stdio.h>
#include<math.h>
int main()
{
        float **a,*temp,app,sum,mult;
        int i,j,k,n,p;
//take no of terms
        printf("Enter n : ");scanf("%d",&n);
//memory allocation
        a = (float**)malloc(n*sizeof(float*));
        for(i = 0; i < n; i++)
            a[i] = (float*)malloc(n*sizeof(float));
        temp = (float*)malloc(n*sizeof(float));
//take input from user
        printf("Enter the elements of augmended matrix rowwise\n");
        for(i=0;i<n;i++)
            for(j=0;j<=n;j++)
                scanf("%f",&a[i][j]);
//generation of scalar matrix
       for(i=0;i<(n);i++){
            app = a[i][i];
//initialization of p
            p = i;
//find largest no of the columns and row no. of largest no.
            for(k = i+1; k < n; k++)
            if(fabs(app) < fabs(a[k][i])){
               app = a[k][i] ;
               p = k;
            }
//swaping the elements of diagonal row and row containing largest no
            for(j = 0; j <= n; j++)
            {
                temp[j] = a[p][j];
                a[p][j] = a[i][j];
                a[i][j] = temp[j];
            }
//calculating triangular matrix
            for(j=i+1;j<n;j++){
                mult = a[j][i]/a[i][i];
                for(k=0;k<=n;k++)
                    a[j][k] -= mult*a[i][k];
            }
       }
//for calculating value of z,y,x via backward substitution method
        for(i=n-1;i>=0;i--)
        {
            sum = 0;
            for(j=i+1;j<n;j++)
                sum += a[i][j]*temp[j];
            temp[i] = (a[i][n]-sum)/a[i][i];
        }
        printf("****The matrix is : ***\n");
        for(i=0;i<n;i++){
            for(j=0;j<=n;j++)
                printf("%.2f\t",a[i][j]);
            printf("\n");
        }
//display solution
        printf("-------------The solution is ----------\n");
        for(i=0;i<n;i++)
        printf("X[%d] = %.2f\n",i+1,temp[i]);
//free allocated memory
        for(i = 0; i < n; i++)
            free(a[i]);
        free(a);
        free(temp);
    return 0;
}
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OUTPUT
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Enter n : 3
Enter the elements of augmende
0       4       5       8
7       8       9       2
2       3       5       1
****The matrix is : ***
7.00    8.00    9.00    8.00
0.00    4.00    5.00    2.00
-0.00   0.00    1.54    -1.64
-------------The solution is----------
X[1] = -2.09
X[2] = 2.81
X[3] = -.65