Different Operations on Matrices (original) (raw)

Last Updated : 11 Jul, 2025

For an introduction to matrices, you can refer to the following article: Matrix Introduction
In this article, we will discuss the following operations on matrices and their properties:

Matrices Addition
Matrices Subtraction
Matrices Multiplication

Matrices Addition:

The addition of two matrices A m*n and Bm*n gives a matrix Cm*n. The elements of C are the sum of corresponding elements in A and B which can be shown as:

Key points:

The addition of matrices is commutative, which means A+B = B+A
The addition of matrices is associative, which means A+(B+C) = (A+B)+C
The order of matrices A, B, and A+B is always the same
If the order of A and B are different, A+B can’t be computed
The complexity of the addition operation is O(M*N) where M*N is the order of matrices

Implementation of the above approach:

C++ `

// C++ Program for matrix addition

#include using namespace std;

int main() {

int n = 2, m = 2;
int a[n][m] = { { 2, 5 }, { 1, 7 } };
int b[n][m] = { { 3, 7 }, { 2, 9 } };

int c[n][m];
for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
        c[i][j] = a[i][j] + b[i][j];
    }

for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++)
        cout << c[i][j] << " ";
    cout << endl;
}

}

Java

// Java program for addition // of two matrices import java.util.*; class GFG {

// Driver code public static void main(String[] args) { int n = 2, m = 2; int a[][] = { { 2, 5 }, { 1, 7 } };

int b[][] = { { 3, 7 }, { 2, 9 } };

// To store result
int c[][] = new int[n][m];

for (int i = 0; i < n; i++) {
  for (int j = 0; j < m; j++)
    c[i][j] = a[i][j] + b[i][j];
}

for (int i = 0; i < n; i++) {
  for (int j = 0; j < m; j++)
    System.out.print(c[i][j] + " ");
  System.out.print("\n");
}

} }

// This code is contributed by Aarti_Rathi

Python3

Python3 program for addition

of two matrices

N = 4

This function adds A[][]

and B[][], and stores

the result in C[][]

driver code

a = [ [2, 5], [1, 7]]

b= [ [3, 7], [2, 9]]

N = 2

c=a[:][:] # To store result

for i in range(N): for j in range(N): c[i][j] = a[i][j] + b[i][j]

for i in range(N): for j in range(N): print(c[i][j], " ", end='') print()

This code is contributed by Aarti_Rathi

C#

// C# program to rotate a // matrix by 90 degrees using System; class GFG {

// Driver Code static public void Main() { int N = 2; int M = 2;

// Test Case 1
int[,] a = { { 2, 5 }, { 1, 7 } };
int[,] b = { { 3, 7 }, { 2, 9 } };

int [,] c =new int[N,M];
for (int i = 0; i < N; i++)
  for (int j = 0; j < M; j++) {
    c[i,j] = a[i,j] + b[i,j];
  }

for (int i = 0; i < N; i++) {
  for (int j = 0; j < M; j++)
    Console.Write(c[i,j] + " ");
  Console.WriteLine();
}

} }

// This code is contributed by Aarti_Rathi

JavaScript

Complexity analysis:

Time Complexity: O(N*M)
Auxiliary Space: O(N*M)

Matrices Subtraction:

The subtraction of two matrices Am*n and Bm*n give a matrix Cm*n. The elements of C are difference of corresponding elements in A and B which can be represented as:

Key points:

Subtraction of matrices is non-commutative which means A-B ≠ B-A
Subtraction of matrices is non-associative which means A-(B-C) ≠ (A-B)-C
The order of matrices A, B, and A - B is always the same
If the order of A and B are different, A - B can’t be computed
The complexity of subtraction operation is O(M*N) where M*N is the order of matrices

Implementation of the above approach:

C++ `

// C++ Program for matrix subtraction

#include using namespace std;

int main() {

int n = 2, m = 2;
int a[n][m] = { { 2, 5 }, { 1, 7 } };
int b[n][m] = { { 3, 7 }, { 2, 9 } };

int c[n][m];
for (int i = 0; i < n; i++)
    for (int j = 0; j < n; j++) {
        c[i][j] = a[i][j] - b[i][j];
    }

for (int i = 0; i < n; i++) {
    for (int j = 0; j < n; j++)
        cout << c[i][j] << " ";
    cout << endl;
}

}

Java

public class GFG { // Java Program for matrix subtraction

public static void main(String[] args)
{

    int n = 2;
    int m = 2;
    int[][] a = { { 2, 5 }, { 1, 7 } };
    int[][] b = { { 3, 7 }, { 2, 9 } };

    int[][] c = new int[n][m];
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            c[i][j] = a[i][j] - b[i][j];
        }
    }

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            System.out.print(c[i][j]);
            System.out.print(" ");
        }
        System.out.print("\n");
    }
}

}

// This code is contributed by Aarti_Rathi

Python3

Python3 program for addition

of two matrices

N = 4

This function adds A[][]

and B[][], and stores

the result in C[][]

driver code

a = [[2, 5], [1, 7]]

b = [[3, 7], [2, 9]]

N = 2

c = a[:][:] # To store result

for i in range(N): for j in range(N): c[i][j] = a[i][j] - b[i][j]

for i in range(N): for j in range(N): print(c[i][j], " ", end='') print()

This code is contributed by Aarti_Rathi

C#

// C# program to rotate a // matrix by 90 degrees using System;

class GFG {

 // Driver Code
static public void Main()
{
    int N = 2;
    int M = 2;
   
    // Test Case 1
    int[,] a = { { 2, 5 }, { 1, 7 } };
    int[,] b = { { 3, 7 }, { 2, 9 } };

    int [,] c =new int[N,M];
    for (int i = 0; i < N; i++)
        for (int j = 0; j < M; j++) {
            c[i,j] = a[i,j] - b[i,j];
        }
 
    for (int i = 0; i < N; i++) {
        for (int j = 0; j < M; j++)
            Console.Write(c[i,j] + " ");
        Console.WriteLine();
    }
}

}

// This code is contributed by Aarti_Rathi

JavaScript

Complexity Analysis:

Time Complexity: O(N*M)
Auxiliary Space: O(N*M)

Matrices Multiplication:

The multiplication of two matrices Am*n and Bn*p give a matrix Cm*p. It means a number of columns in A must be equal to the number of rows in B to calculate C=A*B. To calculate element c11, multiply elements of 1st row of A with 1st column of B and add them (5*1+6*4) which can be shown as:

Key points:

Multiplication of matrices is non-commutative which means A*B ≠ B*A
Multiplication of matrices is associative which means A*(B*C) = (A*B)*C
For computing A*B, the number of columns in A must be equal to the number of rows in B
The existence of A*B does not imply the existence of B*A
The complexity of multiplication operation (A*B) is O(m*n*p) where m*n and n*p are the order of A and B respectively
The order of matrix C computed as A*B is m*p where m*n and n*p are the order of A and B respectively

Implementation of the above approach:

C++ `

// C++ Program for matrix Multiplication

#include using namespace std;

int main() {

int n = 2, m = 2;
int a[n][m] = { { 2, 5 }, { 1, 7 } };
int b[n][m] = { { 3, 7 }, { 2, 9 } };

int c[n][m];
int i, j, k;
for (i = 0; i < n; i++)
{
    for (j = 0; j < n; j++)
    {
        c[i][j] = 0;
        for (k = 0; k < n; k++)
            c[i][j] += a[i][k] * b[k][j];
    }
}

for (int i = 0; i < n; i++)
{
    for (int j = 0; j < n; j++)
        cout << c[i][j] << " ";
    cout << endl;
}

}

Java

// Java program for matrix multiplication import java.io.*;

class GFG { public static void main(String[] args) { int n = 2, m = 2;

int a[][] = { { 2, 5 }, { 1, 7 } };
int b[][] = { { 3, 7 }, { 2, 9 } };

int c[][] = new int[n][m];
int i, j, k;
for (i = 0; i < n; i++) {
  for (j = 0; j < n; j++) {
    c[i][j] = 0;
    for (k = 0; k < n; k++) {
      c[i][j] += a[i][k] * b[k][j];
    }
  }
}
for (i = 0; i < n; i++) {
  for (j = 0; j < n; j++) {
    System.out.print(c[i][j] + " ");
  }
  System.out.println();
}

} }

// This code is contributed by ishankhandelwals.

Python

Python code for matrix multiplication

a = [[2, 5], [1, 7]] b = [[3, 7], [2, 9]] c = [[0, 0], [0, 0]] for i in range(0, 2): for j in range(0, 2): c[i][j] = 0 for k in range(0, 2): c[i][j] = c[i][j]+(a[i][k]*b[k][j]) for i in range(0, 2): for j in range(0, 2): print(c[i][j])

    # This code is contributed by ishankhandelwals.

C#

// C# program to rotate a // matrix by 90 degrees using System;

class GFG {

// Driver Code static public void Main() { int N = 2; int M = 2;

// Test Case 1
int[,] a = { { 2, 5 }, { 1, 7 } };
int[,] b = { { 3, 7 }, { 2, 9 } };

int [,] c =new int[N,M];

for (int i = 0; i < N; i++)
{
  for (int j = 0; j < M; j++) {
    c[i,j]=0;
    for(int k=0;k<N;k++)
      c[i,j] = c[i,j]+(a[i,k] * b[k,j]);
  }
}

for (int i = 0; i < N; i++) {
  for (int j = 0; j < M; j++)
    Console.Write(c[i,j] + " ");
  Console.WriteLine();
}

} }

// This code is contributed by Aarti_Rathi

JavaScript

Complexity Analysis:

Time Complexity: O((N2)*M)
Auxiliary Space: O(N*M)