检查矩阵是否奇异的C程序

2023年 8月 27日 50.6k 0

检查矩阵是否奇异的C程序

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

登录后复制

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 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 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

登录后复制

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;
}

登录后复制

输出

如果运行上面的代码,它将生成以下输出 -

Matrix is non-Singular

登录后复制

以上就是检查矩阵是否奇异的C程序的详细内容,更多请关注每日运维网(www.mryunwei.com)其它相关文章!

相关文章

JavaScript2024新功能:Object.groupBy、正则表达式v标志
PHP trim 函数对多字节字符的使用和限制
新函数 json_validate() 、randomizer 类扩展…20 个PHP 8.3 新特性全面解析
使用HTMX为WordPress增效:如何在不使用复杂框架的情况下增强平台功能
为React 19做准备:WordPress 6.6用户指南
如何删除WordPress中的所有评论

发布评论