Find determinant of matrix generated by array rotation

Given an array of three elements. The task is to construct a matrix of order 3×3 by using all three rotations of the array as a row of the matrix and find the determinant of the resultant matrix.

Examples:  

Input : arr[] = {1, 2, 3}
Output : 18

Input : arr[] = {1, 1, 1}
Output : 0

Approach: As per the problem statement, construct a 3*3 matrix using the given array. If a1, a2, a3 are array elements then the corresponding matrix will be:  

{{a1, a2, a3}, 
 {a3, a1, a2},
 {a2, a3, a1}}

The task is to calculate the determinant of the above matrix.
The determinant can be calculated by using the proper method for this but on the other hand, if the resultant matrix is expanded for calculation, the result will be a1³ + a2³ + a3³ – 3*a1*a2*a3. Hence, instead of calculating determinants by proper expansion use the above-generated formula.
Therefore, the Determinant of the above Matrix will be:  

a13 + a23 + a33 - (3*a1*a2*a3)

Below is the implementation of the above approach:

C++

// C++ program for finding determinant of generated matrix

#include <bits/stdc++.h>
#define N 3
using namespace std;

// Function to calculate determinant
int calcDeterminant(int arr[])
{
    int determinant = 0;

    for (int i = 0; i < N; i++) {
        determinant += pow(arr[i], 3);
    }

    determinant -= 3 * arr[0] * arr[1] * arr[2];

    return determinant;
}

// Driver code
int main()
{
    int arr[] = { 4, 5, 3 };
    cout << calcDeterminant(arr);
    return 0;
}

Java

// Java program for finding determinant
// of generated matrix
import java.util.*;
import java.lang.*;

class GFG
{
static int N = 3;

// Function to calculate determinant
static double calcDeterminant(int arr[])
{
    double determinant = 0;

    for (int i = 0; i < N; i++)
    {
        determinant += Math.pow(arr[i], 3);
    }

    determinant -= 3 * arr[0] *
                    arr[1] * arr[2];

    return determinant;
}

// Driver code
static public void main (String args[])
{
    int []arr = { 4, 5, 3 };
    System.out.println(calcDeterminant(arr));
}
}

// This code is contributed
// by Akanksha Rai

Python3

# Python3 program for finding determinant of generated matrix

# Function to calculate determinant
def calcDeterminant(arr,n):
    determinant =0

    for i in range(n):
        determinant+= pow(arr[i],3)

    determinant -= 3*arr[0]*arr[1]*arr[2]

    return determinant

# Driver code
arr = [4,5,3]
n = len(arr)
print(calcDeterminant(arr,n))

# This code is contributed by Shrikant13

C#

// C# program for finding determinant
// of generated matrix
using System;

class GFG
{
static int N = 3;

// Function to calculate determinant
static double calcDeterminant(int []arr)
{
    double determinant = 0;

    for (int i = 0; i < N; i++)
    {
        determinant += Math.Pow(arr[i], 3);
    }

    determinant -= 3 * arr[0] *
                       arr[1] * arr[2];

    return determinant;
}

// Driver code
static public void Main ()
{
    int []arr = { 4, 5, 3 };
    Console.WriteLine(calcDeterminant(arr));
}
}

// This code is contributed by akt_mit

PHP

<?php
// PHP program for finding determinant
// of generated matrix

$N = 3;

// Function to calculate determinant
function calcDeterminant($arr)
{
    global $N;
    $determinant = 0;

    for ($i = 0; $i < $N; $i++)
    {
        $determinant += pow($arr[$i], 3);
    }

    $determinant -= 3 * $arr[0] *
                        $arr[1] * $arr[2];

    return $determinant;
}

// Driver code
$arr = array( 4, 5, 3 );
echo calcDeterminant($arr);

// This code is contributed by ajit
?>

Javascript

<script>
// Java script  program for finding determinant
// of generated matrix

let  N = 3;

// Function to calculate determinant
function calcDeterminant(arr)
{
    let determinant = 0;

    for (let i = 0; i < N; i++)
    {
        determinant += Math.pow(arr[i], 3);
    }

    determinant -= 3 * arr[0] *
                    arr[1] * arr[2];

    return determinant;
}

// Driver code

    let arr =[ 4, 5, 3 ];
    document.write(calcDeterminant(arr));


//contributed by bobby

</script>

Output:

36

Time Complexity : O(N) as only one traversal is requires on array.

Auxiliary Space : O(1), since no extra space has been taken.