C++程式:檢查矩陣是否可逆


矩陣的行列式可以用來判斷它是否可逆。如果行列式不為零,則矩陣可逆。因此,如果行列式為零,則矩陣不可逆。例如:

The given matrix is:

4 2 1
2 1 1
9 3 2
The determinant of the above matrix is: 3
So the matrix is invertible.

下面是一個檢查矩陣是否可逆的程式。

示例

 線上演示

#include<iostream>
#include<math.h>
using namespace std;
int determinant( int matrix[10][10], int n) {
   int det = 0;
   int submatrix[10][10];
   if (n == 2)
   return ((matrix[0][0] * matrix[1][1]) - (matrix[1][0] * matrix[0][1]));
   else {
      for (int x = 0; x < n; x++) {
         int subi = 0;
         for (int i = 1; i < n; i++) {
            int subj = 0;
            for (int j = 0; j < n; j++) {
               if (j == x)
               continue;
               submatrix[subi][subj] = matrix[i][j];
               subj++;
            }
            subi++;
         }
         det = det + (pow(-1, x) * matrix[0][x] * determinant( submatrix, n - 1 ));
      }
   }
   return det;
}
int main() {
   int n, d, i, j;
   int matrix[10][10];
   cout << "Enter the size of the matrix:\n";
   cin >> n;
   cout << "Enter the elements of the matrix:\n";
   for (i = 0; i < n; i++)
   for (j = 0; j < n; j++)
   cin >> matrix[i][j];
   cout<<"The entered matrix is:"<<endl;
   for (i = 0; i < n; i++) {
      for (j = 0; j < n; j++)
      cout << matrix[i][j] <<" ";
      cout<<endl;
   }
   d = determinant(matrix, n);
   cout<<"Determinant of the matrix is "<< d <<endl;
   if( d == 0 )
   cout<<"This matrix is not invertible as the determinant is zero";
   else
   cout<<"This matrix is invertible as the determinant is not zero";
   return 0;
}

輸出

Enter the size of the matrix: 3
Enter the elements of the matrix:
1 2 3
2 1 2
1 1 4
The entered matrix is:
1 2 3
2 1 2
1 1 4
Determinant of the matrix is -7
This matrix is invertible as the determinant is not zero

在上面的程式中,矩陣的大小和元素在`main()`函式中給出。然後呼叫`determinant()`函式。它返回矩陣的行列式,並將其儲存在`d`中。如果行列式為0,則矩陣不可逆;如果行列式不為0,則矩陣可逆。以下程式碼片段演示了這一點。

cout << "Enter the size of the matrix:\n";
cin >> n;
cout << "Enter the elements of the matrix:\n";
for (i = 0; i < n; i++)
for (j = 0; j < n; j++)
cin >> matrix[i][j];
cout<<"The entered matrix is:"<<endl;
for (i = 0; i < n; i++) {
   for (j = 0; j < n; j++)
   cout << matrix[i][j] <<" ";
   cout<<endl;
}
d = determinant(matrix, n);
cout<<"Determinant of the matrix is "<< d <<endl;
if( d == 0 )
cout<<"This matrix is not invertible as the determinant is zero";
else
cout<<"This matrix is invertible as the determinant is not zero";

在`determinant()`函式中,如果矩陣的大小為2,則直接計算行列式並返回其值。如下所示:

if (n == 2)
return ((matrix[0][0] * matrix[1][1]) - (matrix[1][0] * matrix[0][1]));

如果矩陣的大小不是2,則遞迴計算行列式。使用迴圈變數`x`、`i`和`j`使用了3個巢狀迴圈。這些迴圈用於計算行列式,並遞迴呼叫`determinant()`函式來計算內部行列式,然後將其與外部值相乘。以下程式碼片段演示了這一點。

for (int x = 0; x < n; x++) {
   int subi = 0;
   for (int i = 1; i < n; i++) {
      int subj = 0;
      for (int j = 0; j < n; j++) {
         if (j == x)
         continue;
         submatrix[subi][subj] = matrix[i][j];
         subj++;
      }
      subi++;
   }
   det = det + (pow(-1, x) * matrix[0][x] * determinant( submatrix, n - 1 ))
}

更新於:2020年6月24日

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