矩阵的转置是通过交换行和列而获得的新矩阵。
在此程序中,要求用户输入行r和列c的数量 。在此程序中,它们的值应小于10。
然后,要求用户输入矩阵的元素( r*c
的顺序)。
然后,下面的程序计算矩阵的转置并将其打印在屏幕上。
查找矩阵转置的程序
#include
int main() {
int a[10][10], transpose[10][10], r, c, i, j;
printf("Enter rows and columns: ");
scanf("%d %d", &r, &c);
// Assigning elements to the matrix
printf("\nEnter matrix elements:\n");
for (i = 0; i < r; ++i)
for (j = 0; j < c; ++j) {
printf("Enter element a%d%d: ", i + 1, j + 1);
scanf("%d", &a[i][j]);
}
// Displaying the matrix a[][]
printf("\nEntered matrix: \n");
for (i = 0; i < r; ++i)
for (j = 0; j < c; ++j) {
printf("%d ", a[i][j]);
if (j == c - 1)
printf("\n");
}
// Finding the transpose of matrix a
for (i = 0; i < r; ++i)
for (j = 0; j < c; ++j) {
transpose[j][i] = a[i][j];
}
// Displaying the transpose of matrix a
printf("\nTranspose of the matrix:\n");
for (i = 0; i < c; ++i)
for (j = 0; j < r; ++j) {
printf("%d ", transpose[i][j]);
if (j == r - 1)
printf("\n");
}
return 0;
}
输出
Enter rows and columns: 2
3
Enter matrix elements:
Enter element a11: 1
Enter element a12: 4
Enter element a13: 0
Enter element a21: -5
Enter element a22: 2
Enter element a23: 7
Entered matrix:
1 4 0
-5 2 7
Transpose of the matrix:
1 -5
4 2
0 7