Given a two matrix we have to calculate the subtract of two matrix and stored the result into a resultant matrix.

Subtraction of two matrix is same like we subtract two integer.C=A-B.

**Example:**

```
input:
10 9
8 7
3 4
5 6
output:
7 5
3 1
```

**Logic **

`add the value of first matrix of same row column to second matrix of same row column.`

#include<bits/stdc++.h> using namespace std; int main() { int matrix1[10][10],matrix2[10][10] ,subtract[10][10]; int row, col, i, j; cout<<"Enter rows and columns: "<<endl; cin>>row>>col; cout<<" Enter matrix1 elements:"<<endl; for (i = 0; i < row; ++i) { for (j = 0; j < col; ++j) { cin>>matrix1[i][j]; } } cout<<" Enter matrix2 elements:"<<endl; for (i = 0; i < row; ++i) { for (j = 0; j < col; ++j) { cin>>matrix2[i][j]; } } for (i = 0; i < row; ++i) { for (j = 0; j < col; ++j) { subtract[i][j] = matrix1[i][j]+matrix2[i][j]; } } cout<<"subtraction of the matrix:"<<endl; for (i = 0; i < col; ++i) { for (j = 0; j < row; ++j) { cout<<subtract[i][j]<<" "; if (j == row-1) { cout<<endl; } } } }

**Output:**

Enter rows and columns: 2 2 Enter matrix1 elements: 10 9 8 9 Enter matrix2 elements:5 6 7 8 subtraction of the matrix: 5 3 1 1

**Time Complexity**=O(N*N)

**Space Complexity**=O(N*N);