How to resolve the algorithm Reduced row echelon form step by step in the C# programming language

Published on 12 May 2024 09:40 PM
#C_Sharp

How to resolve the algorithm Reduced row echelon form step by step in the C# programming language

Table of Contents

Problem Statement

Show how to compute the reduced row echelon form (a.k.a. row canonical form) of a matrix. The matrix can be stored in any datatype that is convenient (for most languages, this will probably be a two-dimensional array). Built-in functions or this pseudocode (from Wikipedia) may be used: For testing purposes, the RREF of this matrix: is:

Let's start with the solution:

Step by Step solution about How to resolve the algorithm Reduced row echelon form step by step in the C# programming language

Explanation of the C# Program:

This C# program performs row reduction (also known as echelon form reduction) on a given matrix. Row reduction is a mathematical operation used to simplify matrices into a form where each row contains a single leading coefficient (non-zero value) and zeros in all other columns.

Code Overview:

1. Input Matrix:

  • The program defines a 3x4 matrix with integer values.

2. rref Function:

  • This function performs row reduction on the input matrix and returns the reduced matrix.

3. Row Reduction Process:

  • The function iterates through the matrix's rows and columns, performing the following steps:
    • Find a Leading Coefficient:
      • It searches for a row where the element in the current column is non-zero. If none is found, it skips the column and moves to the next one.
    • Swap Rows:
      • If the leading coefficient is not in the current row, it swaps the current row with the row containing the leading coefficient.
    • Normalize the Leading Coefficient:
      • It divides the entire row by the value of the leading coefficient to make it 1.
    • Subtract Multiples:
      • For each other row in the matrix, it subtracts a multiple of the reduced row to eliminate non-zero elements in the current column.
    • Advance Leading Column:
      • Once the current column is in reduced form, it moves to the next column and repeats the process.

4. Output:

  • The reduced matrix is returned as the result of the rref function.

Example:

For the input matrix provided in the program:

1 2 -1 -4
2 3 -1 -11
-2 0 -3 22

The output after row reduction will be:

1 0 4 7
0 1 -1 2
0 0 1 -6

Source code in the csharp programming language

using System;

namespace rref
{
    class Program
    {
        static void Main(string[] args)
        {
            int[,] matrix = new int[3, 4]{
                {  1, 2, -1,  -4 },
                {  2, 3, -1, -11 },
                { -2, 0, -3,  22 }
            };
            matrix = rref(matrix);   
        }

        private static int[,] rref(int[,] matrix)
        {            
            int lead = 0, rowCount = matrix.GetLength(0), columnCount = matrix.GetLength(1);
            for (int r = 0; r < rowCount; r++)
            {
                if (columnCount <= lead) break;
                int i = r;
                while (matrix[i, lead] == 0)
                {
                    i++;
                    if (i == rowCount)
                    {
                        i = r;
                        lead++;
                        if (columnCount == lead)
                        {
                        lead--;
                        break;
                        }
                    }
                }
                for (int j = 0; j < columnCount; j++)
                {
                    int temp = matrix[r, j];
                    matrix[r, j] = matrix[i, j];
                    matrix[i, j] = temp;
                }
                int div = matrix[r, lead];
                if(div != 0)
                    for (int j = 0; j < columnCount; j++) matrix[r, j] /= div;                
                for (int j = 0; j < rowCount; j++)
                {
                    if (j != r)
                    {
                        int sub = matrix[j, lead];
                        for (int k = 0; k < columnCount; k++) matrix[j, k] -= (sub * matrix[r, k]);
                    }
                }
                lead++;
            }
            return matrix;
        }
    }
}

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