域上的矩陣(矩陣元素均爲0,1,2)滿足對角線元素 的可逆(模3意義下)矩陣有多少個,分別列舉出來
用Mathematica可以這麼處理:
Clear[boolmat]
boolmat[n_Integer /; n > 0] := Tuples[{0, 1, 2}, {n, n}]
s1 = Select[
boolmat[3], #[[1, 1]] + #[[2, 2]]+#[[3,3]] == 0 && Mod[Det[#], 3] != 0 &];
TraditionalForm /@ s1
輸出截圖(局部)
可以查看一共多少個:
一共336個,這爲後續分析打下基礎。
這裏面涉及3*3階行列式,我們可以自己編寫C代碼求行列式。C代碼和Mathematica可以相互結合。這個計劃以後琢磨。
#include<stdio.h>
int main(){
int a[3][3], i, j;
long determinant;
printf("Enter the 9 elements of matrix: ");
for(i = 0 ;i < 3;i++)
for(j = 0;j < 3;j++)
scanf("%d", &a[i][j]);
printf("\nThe matrix is\n");
for(i = 0;i < 3; i++){
printf("\n");
for(j = 0;j < 3; j++)
printf("%d\t", a[i][j]);
}
determinant = a[0][0] * ((a[1][1]*a[2][2]) - (a[2][1]*a[1][2])) -a[0][1] * (a[1][0]
* a[2][2] - a[2][0] * a[1][2]) + a[0][2] * (a[1][0] * a[2][1] - a[2][0] * a[1][1]);
printf("\nDeterminant of 3X3 matrix: %ld", determinant);
return 0;
}
注意到矩陣行列式自己定義的。
當然對於2階矩陣
#include <stdio.h>
#define SIZE 2 // Matrix size
int main()
{
int A[SIZE][SIZE];
int row, col;
long det;
/* Input elements in matrix A from user */
printf("Enter elements in matrix of size 2x2: \n");
for(row=0; row<SIZE; row++)
{
for(col=0; col<SIZE; col++)
{
scanf("%d", &A[row][col]);
}
}
/*
* det(A) = ad - bc
* a = A[0][0], b = A[0][1], c = A[1][0], d = A[1][1]
*/
det = (A[0][0] * A[1][1]) - (A[0][1] * A[1][0]);
printf("Determinant of matrix A = %ld", det);
return 0;
}
針對一般的 n*n 階(求解時候n 是具體·正整數)矩陣能否有效編寫呢?)
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
#define N 4
int determinantOfMatrix(int mat[N][N], int n);
void display(int mat[N][N], int row, int col);
int main(void)
{
int matrix[N][N] =
{
{6,1,1,1},
{4,-2,2,5},
{2,8,6,7},
{2,4,5,6},
};
display(matrix, N, N);
int determinant = determinantOfMatrix(matrix, N);
printf("Determinant of the matrix is: %d\n", determinant);
return 0;
}
int determinantOfMatrix(int mat[N][N], int n)
{
int det=0, p, h, k, i, j, temp[N][N];
if (n == 1)
{
return mat[0][0];
}
else if (n == 2)
{
det = (mat[0][0] * mat[1][1] - mat[0][1] * mat[1][0]);
return det;
}
else {
for (p = 0; p < n; p++) {
h = 0;
k = 0;
for (i = 1; i < n; i++) {
for (j = 0; j < n; j++) {
if (j==p) {
continue;
}
temp[h][k] = mat[i][j];
k++;
if (k == n-1) {
h++;
k = 0;
}
}
}
det = det + mat[0][p] * pow(-1, p) * determinantOfMatrix(temp, n-1);
}
return det;
}
}
void display(int mat[N][N], int row, int col)
{
int i;
int j;
for (i = 0; i < row; i++)
{
for (j = 0; j < col; j++)
{
printf("%d\t", mat[i][j]);
}
printf("\n");
}
}