给定一个矩阵 mat[row][column]，我们的任务是通过函数检查给定矩阵是否奇异并显示结果。
奇异矩阵是其行列式的矩阵为零，如果行列式不为零，则该矩阵是非奇异的。
因此，要确定该矩阵是奇异的还是非奇异的，我们需要首先计算行列式。矩阵的行列式可以计算为 -
$$M1[3][3]\:=\:\begin{bmatrix}a & b & c \d & e & f \g & h & i \end{bmatrix}$$
|m1| = a(e*i - f*h) - b(d*i - f*g) + c(d*h - e*g)
示例
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算法
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示例
实时演示
#include#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 "); else printf("Matrix is non-Singular
"); return 0; }
输出
如果运行上面的代码，它将生成以下输出 -
Matrix is non-Singular
