Solving Systems of Linear Equations using the Row Reduction Algorithm: A Step-by-Step Guide

The row reduction algorithm applies only to augmented matrices for a linear system

The row reduction algorithm, also known as Gaussian elimination, is a method used to solve systems of linear equations and manipulate matrices

The row reduction algorithm, also known as Gaussian elimination, is a method used to solve systems of linear equations and manipulate matrices. It can be applied to augmented matrices, which combine the coefficient matrix and the constant vector of a linear system.

An augmented matrix is a representation of a linear system in matrix form, where the coefficients of the variables are written in a matrix and the constants are appended as an additional column. For example, consider the following system of equations:

2x + 3y = 9
4x – y = 2

To write this system in augmented matrix form, we create a matrix with the coefficients of x and y as the first two columns, and the constants as the last column:

[ 2 3 | 9 ]
[ 4 -1 | 2 ]

The row reduction algorithm allows us to perform a series of elementary row operations on the augmented matrix to transform it into a simpler form, typically an upper triangular or row echelon form. These operations include:

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

The goal of these operations is to create zeros below the leading entries (non-zero values) in each row, resulting in an upper triangular form. Once the augmented matrix is in this form, we can easily solve the linear system by back-substitution.

For example, let’s apply the row reduction algorithm to the augmented matrix from above:

[ 2 3 | 9 ]
[ 4 -1 | 2 ]

We can start by dividing the first row by 2 to make the leading entry 1:

[ 1 3/2 | 9/2 ]
[ 4 -1 | 2 ]

Next, subtract 4 times the first row from the second row to make the second entry in the second row zero:

[ 1 3/2 | 9/2 ]
[ 0 -7 | -16 ]

Our augmented matrix is now in row echelon form. We can interpret this as the simplified system of equations:

x + (3/2)y = 9/2
-7y = -16

From the second equation, we can solve for y to find y = 16/7. Substituting this value back into the first equation, we can solve for x:

x + (3/2)(16/7) = 9/2
x + 24/7 = 9/2
x = 9/2 – 24/7
x = (63 – 48)/14
x = 15/14

Therefore, the solution to the original linear system is x = 15/14 and y = 16/7.

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