C Program to check if matrix is singular or not

CServer Side ProgrammingProgramming

Given a matrix as mat[row][column], our task is to check whether the given matrix is singular or not through a function and display the result.

Singular matrix is a matrix whose determinant is zero and if the determinant is not zero then the matrix is non-singular.

So to find whether the matrix is singular or non-singular we need to calculate determinant first. Determinant of a matrix can be calculated as −

$$M1[3][3]\:=\:\begin{bmatrix}a & b & c \\d & e & f \\g & h & i \end{bmatrix}$$

|m1| = a(e*i - f*h) - b(d*i - f*g) + c(d*h - e*g)

Example

Input-: mat[3][3]= { 4, 10, 1 },
   { 0, 2, 3 },
   { 1, 4, -3 }
Output-: matrix is non-singular
Input-: mat[3][3]= { 0, 0, 0 },
   { 10, 20, 30 },
   { 1, 4, -3 }
Output-: matrix is singular
Since the entire first row is 0 the determinant will be zero only

Algorithm

Start
In function cofactor(int matrix[N][N], int matrix2[N][N], int p, int q, int n)
{
   Step 1-> Declare and initialize i = 0, j = 0, row, col
   Step 2-> Loop For row = 0 and row < n and row++
   Loop For col = 0 and col < n and col++
      If row != p && col != q then,
      Set  matrix2[i][j++] as matrix[row][col]
         If j == n – 1 then,
            Set j = 0
            Increment i by 1
         End for
      End for
In function int check_singular(int matrix[N][N], int n)
   Step 1-> Declare and initialize int D = 0;
   Step 2-> If n == 1 then,
      Return matrix[0][0]
   Step 3-> Declare matrix2[N][N], sign = 1
   Step 4-> Loop For f = 0 and f < n and f++
      Call function cofactor(matrix, matrix2, 0, f, n)
         Set D += sign * matrix[0][f] * check_singular(matrix2, n - 1)
         Set sign = -sign
      End loop
   Step 5-> Return D
In main()
   Step 1-> Declare and initialize a matrix[N][N]
   Step 2-> If call check_singular(matrix, N) returns non 0 value then,
      Print "Matrix is Singular "
   Step 3-> Else
      Print "Matrix is non-Singular "
Stop

Example

 Live Demo

#include <stdio.h>
#define N 4
//to find the cofactors
int cofactor(int matrix[N][N], int matrix2[N][N], int p, int q, int n) {
   int i = 0, j = 0;
   int row, col;
   // Looping for each element of the matrix
   for (row = 0; row < n; row++) {
      for (col = 0; col < n; col++) {
         // Copying into temporary matrix only
         // those element which are not in given
         // row and column
         if (row != p && col != q) {
            matrix2[i][j++] = matrix[row][col];
            // Row is filled, so increase row
            // index and reset col index
            if (j == n - 1) {
               j = 0;
               i++;
            }
         }
      }
   }
   return 0;
}
/* Recursive function to check if matrix[][] is singular or not. */
int check_singular(int matrix[N][N], int n) {
   int D = 0; // Initialize result
   // Base case : if matrix contains single element
   if (n == 1)
   return matrix[0][0];
   int matrix2[N][N]; // To store cofactors
   int sign = 1; // To store sign multiplier
   // Iterate for each element of first row
   for (int f = 0; f < n; f++) {
      // Getting Cofactor of matrix[0][f]
      cofactor(matrix, matrix2, 0, f, n);
      D += sign * matrix[0][f] * check_singular(matrix2, n - 1);
      // terms are to be added with alternate sign
      sign = -sign;
   }
   return D;
}
// Driver program to test above functions
int main() {
   int matrix[N][N] = { { 4, 10, 1 },
   { 0, 2, 3 },
   { 1, 4, -3 } };
   if (check_singular(matrix, N))
      printf("Matrix is Singular\n");
   else
      printf("Matrix is non-Singular\n");
   return 0;
}

Output

If run the above code it will generate the following output −

Matrix is non-Singular
raja
Published on 18-Oct-2019 13:14:34
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