java输入两个矩阵AN×N与BN×N（2<N<10）进行运算

我可以回答这个问题。对于两个矩阵AN×N和BN×N，可以进行矩阵乘法运算，得到一个新的矩阵CN×N。具体的计算方法是，对于矩阵C中的每一个元素C(i,j)，都是矩阵A第i行和矩阵B第j列对应元素的乘积之和。矩阵乘法的运算规则可以参考相关的数学教材或者网上的资料。

相关问题

用java编写一个程序，输入两个矩阵AN×N与BN×N（2<N<10）进行如下运算： 1. 加 2. 减 3. 乘

好的，这是一个计算问题，我可以回答。以下是Java代码示例：

import java.util.Scanner;

public class MatrixOperation {
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
System.out.print("Enter the size of the matrix (N): ");
int n = input.nextInt();
int[][] a = new int[n][n];
int[][] b = new int[n][n];
System.out.println("Enter the elements of matrix A:");
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
a[i][j] = input.nextInt();
}
}
System.out.println("Enter the elements of matrix B:");
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
b[i][j] = input.nextInt();
}
}
System.out.println("Choose an operation:");
System.out.println("1. Addition");
System.out.println("2. Subtraction");
System.out.println("3. Multiplication");
int choice = input.nextInt();
int[][] result = new int[n][n];
switch (choice) {
case 1:
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
result[i][j] = a[i][j] + b[i][j];
}
}
break;
case 2:
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
result[i][j] = a[i][j] - b[i][j];
}
}
break;
case 3:
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
for (int k = 0; k < n; k++) {
result[i][j] += a[i][k] * b[k][j];
}
}
}
break;
default:
System.out.println("Invalid choice!");
return;
}
System.out.println("Result:");
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
System.out.print(result[i][j] + " ");
}
System.out.println();
}
}
}