Check If a Matrix is Invertible in C++

Here we will see how to check whether a matrix is invertible or not. If one matrix is M, then the inverted matrix M-1 will be −

$$M^-1=\frac{adj(M)}{|M\lvert}$$

So if the determinant of M is non-zero, then only we can get the inverse, otherwise, we will not get the inverse of it. So here we have to check the determinant is non-zero or not. Finding a determinant is one recursive process. We have to find submatrix, then find the determinant of it, then use that result for the final calculation. Let us see the code to get a better idea.

Example

 Live Demo

#include <iostream>
#define N 4
using namespace std;
void findCoFactor(int mat[N][N], int mat2[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) {
            mat2[i][j++] = mat[row][col];
            if (j == n - 1) {
               j = 0;
               i++;
            }
         }
      }
   }
}
int getDeterminant(int mat[N][N], int n) {
   int determinant = 0;
   if (n == 1)
      return mat[0][0];
   int temp[N][N];
   int sign = 1;
   for (int f = 0; f < n; f++) {
      findCoFactor(mat, temp, 0, f, n);
      determinant += sign * mat[0][f] * getDeterminant(temp, n - 1);
      sign = -sign;
   }
   return determinant;
}
bool isMatrixInvertible(int mat[N][N], int n) {
   if (getDeterminant(mat, N) != 0)
      return true;
   else
      return false;
}
int main() {
   int matrix[N][N] = {
      { 1, 0, 2, -1 },
      { 3, 0, 0, 5 },
      { 2, 1, 4, -3 },
      { 1, 0, 5, 0 }
   };
   if (isMatrixInvertible(matrix, N))
      cout << "The matrix is invetible";
   else
      cout << "The matrix is not invetible";
}

Output

The matrix is invetible