C程序检查矩阵是否为奇数
给定矩阵为mat [row] [column],我们的任务是通过函数检查给定矩阵是否为奇数并显示结果。
奇异矩阵是行列式为零的矩阵,如果行列式不为零,则矩阵是非奇异的。
因此,要确定矩阵是奇异的还是非奇异的,我们需要先计算行列式。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
StartIn 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
#include <stdio.h>#define N 4
//找到辅因子
int cofactor(int matrix[N][N], int matrix2[N][N], int p, int q, int n) {
int i = 0, j = 0;
int row, col;
//为矩阵的每个元素循环
for (row = 0; row < n; row++) {
for (col = 0; col < n; col++) {
//仅复制到临时矩阵
//那些没有给出的元素
//行和列
if (row != p && col != q) {
matrix2[i][j++] = matrix[row][col];
//行已填充,因此增加行
//索引并重置列索引
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
//基本情况:如果矩阵包含单个元素
if (n == 1)
return matrix[0][0];
int matrix2[N][N]; // To store cofactors
int sign = 1; // To store sign multiplier
//对第一行的每个元素进行迭代
for (int f = 0; f < n; f++) {
//获取矩阵的辅因子[0] [f]
cofactor(matrix, matrix2, 0, f, n);
D += sign * matrix[0][f] * check_singular(matrix2, n - 1);
//术语应添加备用符号
sign = -sign;
}
return D;
}
//驱动程序测试以上功能
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;
}
输出结果
如果运行上面的代码,它将生成以下输出-
Matrix is non-Singular
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