Gaussian Elimination to Solve Linear Equations

The article focuses on using an algorithm for solving a system of linear equations. We will deal with the matrix of coefficients. Gaussian Elimination does not work on singular matrices (they lead to division by zero).

Input: For N unknowns, input is an augmented 
       matrix of size N x (N+1). One extra 
       column is for Right Hand Side (RHS)
  mat[N][N+1] = {{3.0, 2.0,-4.0, 3.0},
                {2.0, 3.0, 3.0, 15.0},
                {5.0, -3, 1.0, 14.0}
               };
Output: Solution to equations is:
        3.000000
        1.000000
        2.000000
Explanation:
Given matrix represents following equations
3.0X1 + 2.0X2 - 4.0X3 =  3.0
2.0X1 + 3.0X2 + 3.0X3 = 15.0
5.0X1 - 3.0X2 +    X3 = 14.0
There is a unique solution for given equations, 
solutions is, X1 = 3.0, X2 = 1.0, X3 = 2.0, 

Row echelon form: Matrix is said to be in r.e.f. if the following conditions hold: 

1. The first non-zero element in each row, called the leading coefficient, is 1.
2. Each leading coefficient is in a column to the right of the previous row leading coefficient.
3. Rows with all zeros are below rows with at least one non-zero element.

Reduced row echelon form: Matrix is said to be in r.r.e.f. if the following conditions hold – 

1. All the conditions for r.e.f.
2. The leading coefficient in each row is the only non-zero entry in its column.

The algorithm is majorly about performing a sequence of operations on the rows of the matrix. What we would like to keep in mind while performing these operations is that we want to convert the matrix into an upper triangular matrix in row echelon form. The operations can be: 

1. Swapping two rows
2. Multiplying a row by a non-zero scalar
3. Adding to one row a multiple of another

The process:  

1. Forward elimination: reduction to row echelon form. Using it one can tell whether there are no solutions, or unique solution, or infinitely many solutions.
2. Back substitution: further reduction to reduced row echelon form.

Algorithm:  

1. Partial pivoting: Find the kth pivot by swapping rows, to move the entry with the largest absolute value to the pivot position. This imparts computational stability to the algorithm.
2. For each row below the pivot, calculate the factor f which makes the kth entry zero, and for every element in the row subtract the fth multiple of the corresponding element in the kth row.
3. Repeat above steps for each unknown. We will be left with a partial r.e.f. matrix.

Below is the implementation of the above algorithm.  

Solution for the system:
3.000000
1.000000
2.000000

Illustration: 
 

Time Complexity: Since for each pivot we traverse the part to its right for each row below it, O(n)*(O(n)*O(n)) = O(n³).
We can also apply Gaussian Elimination for calculating:

1. Rank of a matrix
2. Determinant of a matrix
3. Inverse of an invertible square matrix

Approach:

Another approach for implementing Gaussian Elimination is using Partial Pivoting. In partial pivoting, at each iteration, the algorithm searches for the row with the largest pivot element (absolute value) and swaps that row with the current row. This ensures that the largest absolute value is used as the pivot element, reducing the rounding errors that can occur when dividing by a small pivot element. The steps involved in Partial Pivoting Gaussian Elimination are:

1. Start with the given augmented matrix.
2. Find the row with the largest absolute value in the first column and swap that row with the first row.
3. Divide the first row by the pivot element to make it equal to 1.
4. Use the first row to eliminate the first column in all other rows by subtracting the appropriate multiple of the first row from each row.
5. Repeat steps 2-4 for the remaining columns, using the row with the largest absolute value as the pivot row for each column.
6. Back-substitute to obtain the values of the unknowns.

implementation of above approach:

Solution for the system:
3
1
2

complexity

The time complexity of Partial Pivoting Gaussian Elimination is O(N^3) for an N x N matrix.

This article is contributed by Yash Varyani. Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.