Program for scalar multiplication of a matrix
Last Updated :
01 Mar, 2023
Given a matrix and a scalar element k, our task is to find out the scalar product of that matrix.
Examples:
Input : mat[][] = {{2, 3}
{5, 4}}
k = 5
Output : 10 15
25 20
We multiply 5 with every element.
Input : 1 2 3
4 5 6
7 8 9
k = 4
Output : 4 8 12
16 20 24
28 32 36
The scalar multiplication of a number k(scalar), multiply it on every entry in the matrix. and a matrix A is the matrix kA.
C
#include <stdio.h>
#define N 3
void scalarProductMat(int mat[][N], int k)
{
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
mat[i][j] = mat[i][j] * k;
}
int main()
{
int mat[N][N] = { { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
int k = 4;
scalarProductMat(mat, k);
printf("Scalar Product Matrix is : \n");
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++)
printf("%d ", mat[i][j]);
printf("\n");
}
return 0;
}
|
C++
#include <bits/stdc++.h>
using namespace std;
#define N 3
void scalarProductMat(int mat[][N], int k)
{
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
mat[i][j] = mat[i][j] * k;
}
int main()
{
int mat[N][N]
= { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };
int k = 4;
scalarProductMat(mat, k);
printf("Scalar Product Matrix is : \n");
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++)
cout<< mat[i][j]<<" ";
cout << endl;
}
return 0;
}
|
Java
import java.io.*;
class GFG {
static final int N = 3;
static void scalarProductMat(int mat[][],
int k)
{
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
mat[i][j] = mat[i][j] * k;
}
public static void main (String[] args)
{
int mat[][] = { { 1, 2, 3 },
{ 4, 5, 6 },
{ 7, 8, 9 } };
int k = 4;
scalarProductMat(mat, k);
System.out.println("Scalar Product Matrix is : ");
for (int i = 0; i < N; i++)
{
for (int j = 0; j < N; j++)
System.out.print(mat[i][j] + " ");
System.out.println();
}
}
}
|
Python 3
N = 3
def scalarProductMat( mat, k):
for i in range( N):
for j in range( N):
mat[i][j] = mat[i][j] * k
if __name__ == "__main__":
mat = [[ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ]]
k = 4
scalarProductMat(mat, k)
print("Scalar Product Matrix is : ")
for i in range(N):
for j in range(N):
print(mat[i][j], end = " ")
print()
|
C#
using System;
class GFG{
static int N = 3;
static void scalarProductMat(int[,] mat,
int k)
{
for (int i = 0; i < N; i++)
for (int j = 0; j < N; j++)
mat[i,j] = mat[i, j] * k;
}
static public void Main ()
{
int[,] mat = {{1, 2, 3},
{4, 5, 6},
{7, 8, 9}};
int k = 4;
scalarProductMat(mat, k);
Console.WriteLine("Scalar Product Matrix is : ");
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++)
Console.Write(mat[i, j] + " ");
Console.WriteLine();
}
}
}
|
PHP
<?php
function scalarProductMat($mat,
$k)
{
$N = 3;
for ( $i = 0; $i < $N; $i++)
for ($j = 0; $j < $N; $j++)
$mat[$i][$j] = $mat[$i][$j] * $k;
return $mat;
}
$N = 3;
$mat = array(array(1, 2, 3 ),
array( 4, 5, 6 ),
array(7, 8, 9 ));
$k = 4;
$mat1 = scalarProductMat($mat, $k);
echo("Scalar Product Matrix is : " . "\n");
for ($i = 0; $i < $N; $i++)
{
for ($j = 0; $j < $N; $j++)
echo($mat1[$i][$j] . " ");
echo "\n";
}
|
Javascript
<script>
N = 3
function scalarProductMat(mat, k)
{
for (var i = 0; i < N; i++)
for (var j = 0; j < N; j++)
mat[i][j] = mat[i][j] * k;
}
var mat = [ [ 1, 2, 3 ],
[ 4, 5, 6 ],
[ 7, 8, 9 ] ];
var k = 4;
scalarProductMat(mat, k);
document.write("Scalar Product Matrix is : <br>");
for (var i = 0; i < N; i++)
{
for (var j = 0; j < N; j++)
document.write(mat[i][j]+" ");
document.write("<br>");
}
</script>
|
Output:
Scalar Product Matrix is :
4 8 12
16 20 24
28 32 36
Time Complexity: O(n2),
Auxiliary Space: O(1), since no extra space has been taken.
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