Answer to Question #298391 in C++ for Shan

Question

Write a program that calculates the inverse of a

matrix and proves that. “AA -1 = A -1 A = I”

Answer 1

#include<bits/stdc++.h>
using namespace std;
#define N 4
void getCofactor(int A[N][N], int temp[N][N], int p, int q, int n)
{
	int i = 0, j = 0;

	for (int row = 0; row < n; row++)
	{
		for (int col = 0; col < n; col++)
		{
			if (row != p && col != q)
			{
				temp[i][j++] = A[row][col];
				if (j == n - 1)
				{
					j = 0;
					i++;
				}
			}
		}
	}
}
int determinant(int A[N][N], int n)
{
	if (n == 1)
		return A[0][0];
	int temp[N][N]; // To store cofactors


	int sign = 1; // To store sign multiplier
	for (int f = 0; f < n; f++)
	{
		getCofactor(A, temp, 0, f, n);
		D += sign * A[0][f] * determinant(temp, n - 1);
		sign = -sign;
	}
	return D;
}
void adjoint(int A[N][N],int adj[N][N])
{
	if (N == 1)
	{
		adj[0][0] = 1;
		return;
	}
	int sign = 1, temp[N][N];


	for (int i=0; i<N; i++)
	{
		for (int j=0; j<N; j++)
		{
			getCofactor(A, temp, i, j, N);

			sign = ((i+j)%2==0)? 1: -1;
			adj[j][i] = (sign)*(determinant(temp, N-1));
		}
	}
}
bool inverse(int A[N][N], float inverse[N][N])
{
	// Find determinant of A[][]
	int det = determinant(A, N);
	if (det == 0)
	{
		cout << "Singular matrix, can't find its inverse";
		return false;
	}
	int adj[N][N];
	adjoint(A, adj);
	for (int i=0; i<N; i++)
		for (int j=0; j<N; j++)
			inverse[i][j] = adj[i][j]/float(det);


	return true;
}
template<class T>
void display(T A[N][N])
{
	for (int i=0; i<N; i++)
	{
		for (int j=0; j<N; j++)
			cout << A[i][j] << " ";
		cout << endl;
	}
}

int main()
{
	int A[N][N] = { {5, -2, 2, 7},
					{1, 0, 0, 3},
					{-3, 1, 5, 0},
					{3, -1, -9, 4}};
	int adj[N][N]; // To store adjoint of A[][]
	float inv[N][N]; // To store inverse of A[][]
	cout << "Input matrix is :\n";
	display(A);
	cout << "\nThe Adjoint is :\n";
	adjoint(A, adj);
	display(adj);
	cout << "\nThe Inverse is :\n";
	if (inverse(A, inv))
		display(inv);
	return 0;
}