Given two integers **N** and **K**, the task is to find an **N** x **N** square matrix such that sum of every row and column should be equal to **K**. **Note** that there can be multiple such matrices possible. Print any one of them.**Examples:**

Input:N = 3, K = 15Output:

2 7 6

9 5 1

4 3 8Input:N = 3, K = 7Output:

7 0 0

0 7 0

0 0 7

**Approach:** An **N x N** matrix such that each left diagonal element is equal to **K** and rest elements are **0** will satisfy the given condition. In this way, the sum of the elements of the each row and column will be equal to **K**.

Below is the implementation of the above approach:

## C++

`// C++ implementation of the approach` `#include <bits/stdc++.h>` `using` `namespace` `std;` `// Function to print the` `// required matrix` `void` `printMatrix(` `int` `n, ` `int` `k)` `{` ` ` `for` `(` `int` `i = 0; i < n; i++) {` ` ` `for` `(` `int` `j = 0; j < n; j++) {` ` ` `// Print k for the left` ` ` `// diagonal elements` ` ` `if` `(i == j)` ` ` `cout << k << ` `" "` `;` ` ` `// Print 0 for the rest` ` ` `else` ` ` `cout << ` `"0 "` `;` ` ` `}` ` ` `cout << ` `"\n"` `;` ` ` `}` `}` `// Driver code` `int` `main()` `{` ` ` `int` `n = 3, k = 7;` ` ` `printMatrix(n, k);` ` ` `return` `(0);` `}` |

## Java

`// Java implementation of the approach` `import` `java.util.*;` `class` `GFG` `{` `// Function to print the required matrix` `static` `void` `printMatrix(` `int` `n, ` `int` `k)` `{` ` ` `for` `(` `int` `i = ` `0` `; i < n; i++)` ` ` `{` ` ` `for` `(` `int` `j = ` `0` `; j < n; j++)` ` ` `{` ` ` `// Print k for the left` ` ` `// diagonal elements` ` ` `if` `(i == j)` ` ` `System.out.print(k + ` `" "` `);` ` ` `// Print 0 for the rest` ` ` `else` ` ` `System.out.print(` `"0 "` `);` ` ` `}` ` ` `System.out.print(` `"\n"` `);` ` ` `}` `}` `// Driver code` `public` `static` `void` `main(String[] args)` `{` ` ` `int` `n = ` `3` `, k = ` `7` `;` ` ` `printMatrix(n, k);` `}` `}` `// This code is contributed by Princi Singh` |

## Python3

`# Python3 implementation of the approach` `# Function to print the` `# required matrix` `def` `printMatrix(n, k) :` ` ` `for` `i ` `in` `range` `(n) :` ` ` `for` `j ` `in` `range` `(n) :` ` ` `# Print k for the left` ` ` `# diagonal elements` ` ` `if` `(i ` `=` `=` `j) :` ` ` `print` `(k, end ` `=` `" "` `);` ` ` `# Print 0 for the rest` ` ` `else` `:` ` ` `print` `(` `"0"` `, end ` `=` `" "` `);` ` ` ` ` `print` `();` `# Driver code` `if` `__name__ ` `=` `=` `"__main__"` `:` ` ` `n ` `=` `3` `; k ` `=` `7` `;` ` ` `printMatrix(n, k);` `# This code is contributed by AnkitRai01` |

## C#

`// C# implementation of the approach` `using` `System;` ` ` `class` `GFG` `{` `// Function to print the required matrix` `static` `void` `printMatrix(` `int` `n, ` `int` `k)` `{` ` ` `for` `(` `int` `i = 0; i < n; i++)` ` ` `{` ` ` `for` `(` `int` `j = 0; j < n; j++)` ` ` `{` ` ` `// Print k for the left` ` ` `// diagonal elements` ` ` `if` `(i == j)` ` ` `Console.Write(k + ` `" "` `);` ` ` `// Print 0 for the rest` ` ` `else` ` ` `Console.Write(` `"0 "` `);` ` ` `}` ` ` `Console.Write(` `"\n"` `);` ` ` `}` `}` `// Driver code` `public` `static` `void` `Main(String[] args)` `{` ` ` `int` `n = 3, k = 7;` ` ` `printMatrix(n, k);` `}` `}` `// This code is contributed by PrinciRaj1992` |

## Javascript

`<script>` `// javascript implementation of the approach` `// Function to print the required matrix` `function` `printMatrix(n , k)` `{` ` ` `for` `(i = 0; i < n; i++)` ` ` `{` ` ` `for` `(j = 0; j < n; j++)` ` ` `{` ` ` `// Print k for the left` ` ` `// diagonal elements` ` ` `if` `(i == j)` ` ` `document.write(k + ` `" "` `);` ` ` `// Prvar 0 for the rest` ` ` `else` ` ` `document.write(` `"0 "` `);` ` ` `}` ` ` `document.write(` `"</br>"` `);` ` ` `}` `}` `// Driver code` `var` `n = 3, k = 7;` `printMatrix(n, k);` `// This code is contributed by 29AjayKumar` `</script>` |

**Output:**

7 0 0 0 7 0 0 0 7

Attention reader! Don’t stop learning now. Get hold of all the important DSA concepts with the **DSA Self Paced Course** at a student-friendly price and become industry ready.