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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 a different dimension.

Implementation:

## C++

 // C++ program for subtraction of matrices#include using namespace std;#define N 4   // This function subtracts B[][] from A[][], and stores // the result in C[][] void subtract(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 codeint 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;     subtract(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 #define N 4  // This function subtracts B[][] from A[][], and stores// the result in C[][]void subtract(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;    subtract(A, B, C);      printf("Result matrix is \n");    for (i = 0; i < N; i++)    {        for (j = 0; j < N; j++)           printf("%d ", C[i][j]);        printf("\n");    }      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 subtract(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;        subtract(A, B, C);              System.out.print("Result matrix is \n");        for (i = 0; i < N; i++)        {            for (j = 0; j < N; j++)            System.out.print(C[i][j] + " ");            System.out.print("\n");        }    }}  // 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 0def subtract(A, B, C):          for i in range(N):        for j in range(N):            C[i][j] = A[i][j] - B[i][j]   # Driver CodeA = [ [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      subtract(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 subtract(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;    subtract(A, B, C);      Console.Write("Result matrix is \n");    for (i = 0; i < N; i++)    {        for (j = 0; j < N; j++)        {            Console.Write(C[i, j] + " ");        }        Console.Write("\n");    }}}  // This code is contributed by Shrikant13



## Javascript



Output

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

Note – The number at 0th row and 0th column of first matrix gets subtracted with number at 0th row and 0th column of second matrix. And its subtraction result gets initialized as the value of 0th row and 0th column of resultant matrix. Same subtraction process applied for all the elements

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?

Time complexity: O(n2).
Auxiliary space:O(n2). since n2 extra space has been taken.

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