import java.util.Scanner;

 public class Solution
 {
     public static void main(String[] args)
     {
         Scanner input = new Scanner(System.in);
         int[][] matrix1 = new int[3][3];
         int[][] matrix2 = new int[3][3];
         int[][] matrixSum = new int[3][3];
         System.out.print("Enter matrix1: ");
         for(int i = 0; i < 3; i++)
             for(int j = 0; j < 3; j++)
                 matrix1[i][j] = input.nextInt();
         System.out.print("Enter matrix2: ");
         for(int i = 0; i < 3; i++)
             for(int j = 0; j < 3; j++)
                 matrix2[i][j] = input.nextInt();
         input.close();

         System.out.println("The matrices are muliplied as follows");
         int[][] matrixProduct = muliplyMatrix(matrix1, matrix2);

         System.out.println(matrix1[0][0] + " " + matrix1[0][1] + " " + matrix1[0][2] +
                 "     " + matrix2[0][0] + " " + matrix2[0][1] + " " + matrix2[0][2] +
                 "     " + matrixSum[0][0] + " " + matrixSum[0][1] + " " + matrixSum[0][2]);
         System.out.println(matrix1[1][0] + " " + matrix1[1][1] + " " + matrix1[1][2] +
                 "  *  " + matrix2[1][0] + " " + matrix2[1][1] + " " + matrix2[1][2] +
                 "  =  " + matrixSum[1][0] + " " + matrixSum[1][1] + " " + matrixSum[1][2]);
         System.out.println(matrix1[2][0] + " " + matrix1[2][1] + " " + matrix1[2][2] +
                 "     " + matrix2[2][0] + " " + matrix2[2][1] + " " + matrix2[2][2] +
                 "     " + matrixSum[2][0] + " " + matrixSum[2][1] + " " + matrixSum[2][2]);
     }

     public static int[][] muliplyMatrix(int[][] a, int[][] b)
     {
         int[][] outcomeMatrix = new int[3][3];
         for(int i = 0; i < 3; i++)
         {
             for(int j = 0; j < 3; j++)
             {
                 outcomeMatrix[i][j] += a[i][j] * b[j][i];
             }
         }
         return outcomeMatrix;
     }
 }

HW7.6的更多相关文章

  1. HW7.18

    public class Solution { public static void main(String[] args) { int[][] m = {{1, 2}, {3, 4}, {5, 6} ...

  2. HW7.17

    import java.util.Scanner; public class Solution { public static void main(String[] args) { Scanner i ...

  3. HW7.16

    import java.util.Arrays; public class Solution { public static void main(String[] args) { int row = ...

  4. HW7.15

    public class Solution { public static void main(String[] args) { double[][] set1 = {{1, 1}, {2, 2}, ...

  5. HW7.14

    import java.util.Scanner; public class Solution { public static void main(String[] args) { Scanner i ...

  6. HW7.13

    import java.util.Scanner; public class Solution { public static void main(String[] args) { Scanner i ...

  7. HW7.12

    import java.util.Scanner; public class Solution { public static void main(String[] args) { double[] ...

  8. HW7.11

    import java.util.Scanner; public class Solution { public static void main(String[] args) { Scanner i ...

  9. HW7.10

    public class Solution { public static void main(String[] args) { int[][] array = new int[3][3]; for( ...

  10. HW7.9

    import java.util.Scanner; public class Solution { public static void main(String[] args) { Scanner i ...

随机推荐

  1. uva 10051

    将每一个分解为六个两面的 简单地dp 回溯输出路径..... #include <cstdio> #include <cstring> #include <cmath&g ...

  2. html5移动web开发实战必读书记

    原文  http://itindex.net/detail/50689-html5-移动-web 主题 HTML5 一.配置移动开发环境 1.各种仿真器.模拟器的下载安装 http://www.mob ...

  3. console中应用MFC类的方法

    1.添加#include <afx.h>或者<afxwin.h> 这时会报错1>c:\program files\microsoft visual studio 8\vc ...

  4. HDU4647+贪心

    /* 贪心. 题意:给定一些点 一些边 点和边都有价值.现在A B 选点.求A-B的maxVal 思路:分割边.边的1/2分给两个端点. 如果这两个点被同一个人取,则ok:否则 做减法也行,对题意无影 ...

  5. photoshop:模仿-广告放射背景

    模仿对象:图片大小960*400 过程: 1.新建文档,大小为:960*800 2.选择渐变工具,黑白从上往下渐变 3.滤镜->扭曲->波浪,参考设置 4.滤镜->扭曲->极坐 ...

  6. ScrollView can host only one direct child 解决

    主要是ScrollView内部只能有一个子元素,即不能并列两个子元素,所以需要把所有的子元素放到一个LinearLayout内部或RelativeLayout等其他布局方式让后再在这个layout外部 ...

  7. Spring个人总结

    编写Spring第一个程序 Spring是一种开源框架,通过使用它可以大大降低企业应用程序的复杂性.Spring是一种非常完善的框架,几乎涉及WEB开发中的每一层,但是在开发中通常使用Spring开发 ...

  8. [POJ 2420] A Star not a Tree?

    A Star not a Tree? Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 4058   Accepted: 200 ...

  9. [NOI2005] 维修数列

    1500: [NOI2005]维修数列 Time Limit: 10 Sec  Memory Limit: 64 MBSubmit: 8397  Solved: 2530 Description In ...

  10. Java [leetcode 24]Swap Nodes in Pairs

    题目描述: Given a linked list, swap every two adjacent nodes and return its head. For example, Given 1-& ...