Problem 9 - Precalculus

Question

The following matrices are in reduced row echelon form. Determine the solution of the corresponding system of linear equations or state that the system is inconsistent. $$ \left[\begin{array}{llll|r} 1 & 0 & 0 & 3 & 4 \\ 0 & 1 & 0 & 6 & -6 \\ 0 & 0 & 1 & 0 & 2 \end{array}\right] $$

Step-by-Step Solution

Answer

The system has infinite solutions, parameterized by \( x_4 \).

Step 1: Identify the Linear System

The given matrix is in the form \( \begin{bmatrix} A | B \end{bmatrix} \) where \( A \) is the coefficient matrix and \( B \) is the constants on the right-hand side of the equations. Given the matrix: \[\begin{bmatrix} 1 & 0 & 0 & 3 & | & 4 \ 0 & 1 & 0 & 6 & | & -6 \ 0 & 0 & 1 & 0 & | & 2 \end{bmatrix}\] can be translated into the system: \[\begin{align*} x_1 + 3x_4 &= 4 \ x_2 + 6x_4 &= -6 \ x_3 &= 2 \end{align*}\]

Step 2: Assign Variables

In a matrix of size \( m \times n \) in reduced row echelon form, the columns that have leading 1's correspond to basic variables. In this system, columns 1 to 3 correspond to the basic variables \( x_1, x_2, \) and \( x_3 \). The column corresponding to \( x_4 \), which is not a leading column, is free.

Step 3: Express Basic Variables

Use the system of equations to express \( x_1 \), \( x_2 \), and \( x_3 \) in terms of the free variable \( x_4 \). \[\begin{align*} x_1 &= 4 - 3x_4 \ x_2 &= -6 - 6x_4 \ x_3 &= 2 \end{align*}\]

Step 4: State the Solution

The solution to the system is given in terms of the free variable \( x_4 \). Thus, the solution set is: \[\{ (4 - 3x_4, -6 - 6x_4, 2, x_4) \mid x_4 \in \mathbb{R} \} \] This indicates that there are infinite solutions parameterized by \( x_4 \).

Key Concepts

System of Linear Equations

A system of linear equations is a collection of two or more equations with the same set of variables, often represented in the form:

• Linear equations consist of variables raised to the power of one.
• The goal is to find values for these variables that satisfy all equations simultaneously.

In our exercise, the matrix provides us with a system of linear equations:\[\begin{align*}x_1 + 3x_4 &= 4 \x_2 + 6x_4 &= -6 \x_3 &= 2\end{align*}\]Each equation is a line, and solving the system means finding the point(s) where these lines intersect.

Free Variable

A free variable in a system of linear equations is a variable that can take any value from the set of real numbers \( \mathbb{R} \).

• In reduced row echelon form, columns without leading 1's are identified with free variables.
• Free variables give flexibility and indicate multiple solutions.

In our given matrix, the variable \( x_4 \) is free because its column does not have a leading 1:\[ x_4 \text{ is free, allowing } x_1, x_2, \text{ and } x_3 \text{ to depend on } x_4 \].

Basic Variables

Basic variables correspond to columns with leading 1's in a matrix in reduced row echelon form.

• These variables depend on the free variables.
• In essence, they form the foundation of the solution.

For the given system:

• \( x_1 \), \( x_2 \), and \( x_3 \) are basic variables.
• They have specific equations derived from the matrix:
\[ \begin{align*}x_1 &= 4 - 3x_4 \x_2 &= -6 - 6x_4 \x_3 &= 2\end{align*} \]These equations show that basic variables are expressed in terms of the free variable \( x_4 \).

Infinite Solutions

When a system of linear equations has a free variable, it generally leads to an infinite number of solutions.

• This occurs because the free variable can take infinitely many values.
• Each valid value of the free variable represents a unique solution to the system.

For the exercise:

• The solution set is represented as \( \{ (4 - 3x_4, -6 - 6x_4, 2, x_4) \mid x_4 \in \mathbb{R} \} \).
• The presence of \( x_4 \) in the solution expresses this infinite nature.

Each possible value of \( x_4 \) provides a point on the line of solutions in the solution space.