Linear Algebra: Row Echelon Form and Applications

Keywords : Row Echelon Form, System of Linear Equations,
Gauss elimination method, Linear Independent, and Linearly Dependent.

Row Echelon Form (REF)

The Row Echelon Form (REF) of a matrix is a type of triangular matrix that is particularly useful in solving systems of linear equations. A matrix is in row echelon form if it satisfies the following conditions:

• All nonzero rows are above any rows of all zeros.
• Each leading entry (also known as a pivot) of a nonzero row is in a column to the right of the leading entry of the row above it.
• The leading entry in any nonzero row is 1.

Formally, a matrix \( A \) is in row echelon form if:

\[ A = \begin{bmatrix} 1 & * & * & \cdots & * \\ 0 & 1 & * & \cdots & * \\ 0 & 0 & 1 & \cdots & * \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 1 \\ 0 & 0 & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & 0 \\ \end{bmatrix} \]

Properties of Row Echelon Form:

• Uniqueness: The row echelon form of a matrix is not unique; different sequences of row operations can lead to different REF matrices.
• Existence: Every matrix can be transformed into row echelon form using elementary row operations.
• Simplification: REF simplifies the process of solving linear systems by enabling back-substitution.

Gauss Elimination Method

The Gauss Elimination Method is an algorithm for solving systems of linear equations. It systematically performs row operations to reduce the augmented matrix of the system to row echelon form (REF) or reduced row echelon form (RREF), from which the solutions can be easily obtained.

Steps of Gauss Elimination:

1. Form the Augmented Matrix: Represent the system of equations as an augmented matrix.

\[ \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n} & | & b_1 \\ a_{21} & a_{22} & \cdots & a_{2n} & | & b_2 \\ \vdots & \vdots & \ddots & \vdots & | & \vdots \\ a_{m1} & a_{m2} & \cdots & a_{mn} & | & b_m \\ \end{bmatrix} \]

2. Forward Elimination: Use elementary row operations to create zeros below the pivot positions, transforming the matrix into REF.
3. Back Substitution: Once in REF, solve for the variables starting from the last equation and moving upwards.

Elementary Row Operations:

• Swap Rows: Exchange two rows of the matrix.
• Multiply a Row by a Nonzero Scalar: Scale a row by a nonzero constant.
• Add a Multiple of One Row to Another Row: Replace one row with the sum of itself and a multiple of another row.

Example:

Consider the system:

\[ \begin{cases} 2x + 3y – z = 5 \\ 4x + y + z = 6 \\ -2x + 5y – 3z = 2 \\ \end{cases} \]

The augmented matrix is:

\[ \begin{bmatrix} 2 & 3 & -1 & | & 5 \\ 4 & 1 & 1 & | & 6 \\ -2 & 5 & -3 & | & 2 \\ \end{bmatrix} \]

Applying Gauss elimination, we transform this matrix to REF and then perform back substitution to find the values of \( x \), \( y \), and \( z \).

Linear Independence and Dependence

Understanding whether a set of vectors is linearly independent or linearly dependent is fundamental in linear algebra, especially when dealing with solutions to linear systems and vector spaces.

Linear Independence

A set of vectors \( \{ \mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k \} \) in a vector space \( V \) is said to be linearly independent if the only scalars \( c_1, c_2, \dots, c_k \) that satisfy the equation

\[ c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \dots + c_k \mathbf{v}_k = \mathbf{0} \]

are \( c_1 = c_2 = \dots = c_k = 0 \).

Geometric Interpretation:

• In \( \mathbb{R}^2 \), two vectors are linearly independent if they are not collinear.
• In \( \mathbb{R}^3 \), three vectors are linearly independent if they do not lie on the same plane.

Linear Dependence

A set of vectors \( \{ \mathbf{v}_1, \mathbf{v}_2, \dots, \mathbf{v}_k \} \) is linearly dependent if there exists a non-trivial combination of these vectors that equals the zero vector. In other words, there exist scalars \( c_1, c_2, \dots, c_k \), not all zero, such that

\[ c_1 \mathbf{v}_1 + c_2 \mathbf{v}_2 + \dots + c_k \mathbf{v}_k = \mathbf{0} \]

Implications of Linear Dependence:

• At least one vector in the set can be expressed as a linear combination of the others.
• The set does not span the vector space as effectively as a linearly independent set.

Testing for Linear Independence

To determine if a set of vectors is linearly independent, you can:

1. Form a Matrix: Arrange the vectors as columns of a matrix.
2. Row Reduce: Apply Gaussian elimination to bring the matrix to REF or RREF.
3. Check Pivots: If each column has a leading 1 (pivot), the vectors are linearly independent. Otherwise, they are dependent.

Example:

Consider the vectors in \( \mathbb{R}^3 \):

\[ \mathbf{v}_1 = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}, \quad \mathbf{v}_2 = \begin{bmatrix} 4 \\ 5 \\ 6 \end{bmatrix}, \quad \mathbf{v}_3 = \begin{bmatrix} 7 \\ 8 \\ 9 \end{bmatrix} \]

Form the matrix:

\[ A = \begin{bmatrix} 1 & 4 & 7 \\ 2 & 5 & 8 \\ 3 & 6 & 9 \\ \end{bmatrix} \]

Row reducing \( A \), we find that the third column is a linear combination of the first two, indicating that the vectors are linearly dependent.

Application to Systems of Linear Equations

Row echelon form and Gaussian elimination are powerful tools for solving systems of linear equations. By transforming the augmented matrix of the system to REF or RREF, we can easily identify the solutions, whether they are unique, infinite, or nonexistent.

Types of Solutions:

• Unique Solution: Occurs when the system is consistent and has exactly one solution.
• Infinite Solutions: Occurs when the system is consistent but has free variables, leading to infinitely many solutions.
• No Solution: Occurs when the system is inconsistent.

Determining the Nature of Solutions:

• Consistent System: At least one solution exists.
• Inconsistent System: No solutions exist.

By examining the REF or RREF of the augmented matrix:

• Unique Solution: Each variable has a pivot column.
• Infinite Solutions: At least one free variable exists (a column without a pivot).
• No Solution: A row of the form \( [0 \ 0 \ \cdots \ 0 \ | \ b] \) with \( b \neq 0 \) appears.

Example:

Solve the system:

\[ \begin{cases} x + 2y + 3z = 9 \\ 2x + 3y + 4z = 13 \\ 3x + 4y + 5z = 17 \\ \end{cases} \]

Form the augmented matrix and apply Gaussian elimination:

\[ \begin{bmatrix} 1 & 2 & 3 & | & 9 \\ 2 & 3 & 4 & | & 13 \\ 3 & 4 & 5 & | & 17 \\ \end{bmatrix} \]

Row reducing leads to:

\[ \begin{bmatrix} 1 & 2 & 3 & | & 9 \\ 0 & -1 & -2 & | & -5 \\ 0 & 0 & 0 & | & 0 \\ \end{bmatrix} \]

This indicates that \( z \) is a free variable, and the system has infinitely many solutions.

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