Home > Backend Development > C++ > body text

C program to check if matrix is ​​singular

WBOY
Release: 2023-08-25 23:09:06
forward
1419 people have browsed it

C program to check if matrix is ​​singular

Given a matrix mat[row][column], our task is to check whether the given matrix is ​​singular through a function and display the result.

A singular matrix is ​​a matrix whose determinant is zero. If the determinant is not zero, the matrix is ​​non-singular.

So, to determine whether the matrix is ​​singular or non-singular, we need to first calculate the determinant. The determinant of the 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
Copy after login

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 &ndash; 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
Copy after login

Example

Real-time demonstration

#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</p><p>");
   else
      printf("Matrix is non-Singular</p><p>");
   return 0;
}
Copy after login

Output

If you run the above code, it The following output will be generated -

Matrix is non-Singular
Copy after login

The above is the detailed content of C program to check if matrix is ​​singular. For more information, please follow other related articles on the PHP Chinese website!

Related labels:
source:tutorialspoint.com
Statement of this Website
The content of this article is voluntarily contributed by netizens, and the copyright belongs to the original author. This site does not assume corresponding legal responsibility. If you find any content suspected of plagiarism or infringement, please contact admin@php.cn
Popular Tutorials
More>
Latest Downloads
More>
Web Effects
Website Source Code
Website Materials
Front End Template