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Program for subtraction of matrices

The below program subtracts of two square matrices of size 4*4, we can change N for different dimension.

C++

// C++ program for subtraction of matrices
#include <bits/stdc++.h>
using namespace std;
#define N 4 
  
// This function subtracts B[][] from A[][], and stores 
// the result in C[][] 
void multiply(int A[][N], int B[][N], int C[][N]) 
    int i, j; 
    for (i = 0; i < N; i++) 
        for (j = 0; j < N; j++) 
            C[i][j] = A[i][j] - B[i][j]; 
  
// Driver code
int main() 
    int A[N][N] = { {1, 1, 1, 1}, 
                    {2, 2, 2, 2}, 
                    {3, 3, 3, 3}, 
                    {4, 4, 4, 4}}; 
  
    int B[N][N] = { {1, 1, 1, 1}, 
                    {2, 2, 2, 2}, 
                    {3, 3, 3, 3}, 
                    {4, 4, 4, 4}}; 
  
    int C[N][N]; // To store result 
    int i, j; 
    multiply(A, B, C); 
  
    cout << "Result matrix is " << endl; 
    for (i = 0; i < N; i++) 
    
        for (j = 0; j < N; j++) 
        cout << C[i][j] << " "
        cout << endl; 
    
  
    return 0; 
  
// This code is contributed by rathbhupendra

C

#include <stdio.h>
#define N 4
  
// This function subtracts B[][] from A[][], and stores
// the result in C[][]
void multiply(int A[][N], int B[][N], int C[][N])
{
    int i, j;
    for (i = 0; i < N; i++)
        for (j = 0; j < N; j++)
            C[i][j] = A[i][j] - B[i][j];
}
  
int main()
{
    int A[N][N] = { {1, 1, 1, 1},
                    {2, 2, 2, 2},
                    {3, 3, 3, 3},
                    {4, 4, 4, 4}};
  
    int B[N][N] = { {1, 1, 1, 1},
                    {2, 2, 2, 2},
                    {3, 3, 3, 3},
                    {4, 4, 4, 4}};
  
    int C[N][N]; // To store result
    int i, j;
    multiply(A, B, C);
  
    printf("Result matrix is ");
    for (i = 0; i < N; i++)
    {
        for (j = 0; j < N; j++)
           printf("%d ", C[i][j]);
        printf(" ");
    }
  
    return 0;
}

Java

// Java program for subtraction of matrices
  
class GFG
{
     static final int N=4;
  
    // This function subtracts B[][] 
    // from A[][], and stores
    // the result in C[][]
    static void multiply(int A[][], int B[][], int C[][])
    {
        int i, j;
        for (i = 0; i < N; i++)
            for (j = 0; j < N; j++)
                C[i][j] = A[i][j] - B[i][j];
    }
      
    // Driver code
    public static void main (String[] args)
    {
        int A[][] = { {1, 1, 1, 1},
                        {2, 2, 2, 2},
                        {3, 3, 3, 3},
                        {4, 4, 4, 4}};
      
        int B[][] = { {1, 1, 1, 1},
                        {2, 2, 2, 2},
                        {3, 3, 3, 3},
                        {4, 4, 4, 4}};
                          
        // To store result
        int C[][]=new int[N][N]; 
  
        int i, j;
        multiply(A, B, C);
      
        System.out.print("Result matrix is ");
        for (i = 0; i < N; i++)
        {
            for (j = 0; j < N; j++)
            System.out.print(C[i][j] + " ");
            System.out.print(" ");
        }
    }
}
  
// This code is contributed by Anant Agarwal.

Python3

# Python 3 program for subtraction 
# of matrices
  
N = 4
  
# This function returns 1
# if A[][] and B[][] are identical
# otherwise returns 0
def multiply(A, B, C):
      
    for i in range(N):
        for j in range(N):
            C[i][j] = A[i][j] - B[i][j] 
  
# Driver Code
A = [ [1, 1, 1, 1],
      [2, 2, 2, 2],
      [3, 3, 3, 3],
      [4, 4, 4, 4]]
  
B = [ [1, 1, 1, 1],
      [2, 2, 2, 2],
      [3, 3, 3, 3],
      [4, 4, 4, 4]]
                      
C = A[:][:] # To store result
      
multiply(A, B, C)
  
print("Result matrix is")
for i in range(N):
    for j in range(N):
        print(C[i][j], " ", end = '')
    print()
      
# This code is contributed
# by Anant Agarwal.

C#

// C# program for subtraction of matrices 
using System;
  
class GFG
{
static int N = 4;
  
// This function subtracts B[][] 
// from A[][], and stores 
// the result in C[][] 
public static void multiply(int[][] A, 
                            int[][] B, 
                            int[, ] C)
{
    int i, j;
    for (i = 0; i < N; i++)
    {
        for (j = 0; j < N; j++)
        {
            C[i, j] = A[i][j] - B[i][j];
        }
    }
}
  
  
// Driver code 
public static void Main(string[] args)
{
    int[][] A = new int[][]
    {
        new int[] {1, 1, 1, 1},
        new int[] {2, 2, 2, 2},
        new int[] {3, 3, 3, 3},
        new int[] {4, 4, 4, 4}
    };
  
    int[][] B = new int[][]
    {
        new int[] {1, 1, 1, 1},
        new int[] {2, 2, 2, 2},
        new int[] {3, 3, 3, 3},
        new int[] {4, 4, 4, 4}
    };
  
    // To store result 
  
    int[, ] C = new int[N, N];
  
    int i, j;
    multiply(A, B, C);
  
    Console.Write("Result matrix is ");
    for (i = 0; i < N; i++)
    {
        for (j = 0; j < N; j++)
        {
            Console.Write(C[i, j] + " ");
        }
        Console.Write(" ");
    }
}
}
  
// This code is contributed by Shrikant13

/div>

PHP

<?php
// This function subtracts B[][] 
// from A[][], and stores the 
// result in C[][]
function multiply(&$A, &$B, &$C)
{
    $N = 4;
    for ($i = 0; $i < $N; $i++)
        for ($j = 0; $j < $N; $j++)
            $C[$i][$j] = $A[$i][$j] - 
                         $B[$i][$j];
}
  
// Driver code
$N = 4;
$A = array(array(1, 1, 1, 1),
           array(2, 2, 2, 2),
           array(3, 3, 3, 3),
           array(4, 4, 4, 4));
  
$B = array(array(1, 1, 1, 1),
           array(2, 2, 2, 2),
           array(3, 3, 3, 3),
           array(4, 4, 4, 4));
  
multiply($A, $B, $C);
  
echo "Result matrix is ";
for ($i = 0; $i < $N; $i++)
{
    for ($j = 0; $j < $N; $j++)
    {
        echo $C[$i][$j];
        echo " ";
    }
        echo " ";
}
  
// This code is contributed
// by Shivi_Aggarwal
?>

Output:

Result matrix is
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0

The program can be extended for rectangular matrices. The following post can be useful for extending this program.

How to pass a 2D array as a parameter in C?

The time complexity of the above program is O(n2).

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above



This article is attributed to GeeksforGeeks.org

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