Problem 45 - Algebra and Trigonometry

Question

35–46 Solve the system of linear equations. $$\left\\{\begin{aligned} x-y \qquad +w=0 \\ 3 x \qquad -z+2 w=0 \\ x-4 y+z+2 w =0 \end{aligned}\right.$$

Step-by-Step Solution

Answer

The solution involves parameters: \( (x, y, z, w) = (t-s, t, 3t-s, s) \).

Step 1: Rewrite the System of Equations

We start by rewriting the given system of equations:1. \( x - y + w = 0 \)2. \( 3x - z + 2w = 0 \)3. \( x - 4y + z + 2w = 0 \)

Step 2: Solve for One Variable from the First Equation

From the first equation, \( x - y + w = 0 \), we can express \( x \) in terms of \( y \) and \( w \):\[ x = y - w \]

Step 3: Substitute x in the Second Equation

Substitute \( x = y - w \) into the second equation:\( 3(y - w) - z + 2w = 0 \)Simplify:\( 3y - 3w - z + 2w = 0 \)Combine like terms:\( 3y - z - w = 0 \)

Step 4: Solve for z from the Modified Second Equation

Rearrange the equation \( 3y - z - w = 0 \) to express \( z \) in terms of \( y \) and \( w \):\[ z = 3y - w \]

Step 5: Substitute x and z in the Third Equation

Substitute \( x = y - w \) and \( z = 3y - w \) into the third equation:\( (y - w) - 4y + (3y - w) + 2w = 0 \)Simplify:\( y - w - 4y + 3y - w + 2w = 0 \)Combine like terms:\( 0 = 0 \)

Step 6: Conclude the Solution

Since the simplified third equation is always true (\( 0 = 0 \)), it does not provide additional constraints, indicating the system has dependent equations. The system has a free variable, so solutions exist for any chosen \( y \) and \( w \), with \( x = y - w \) and \( z = 3y - w \).

Step 7: Parameterize the Solution

To describe the solution set, choose free variables. Let \( y = t \) and \( w = s \), where \( t \) and \( s \) are parameters. Thus:- \( x = t - s \)- \( z = 3t - s \)- \( y = t \)- \( w = s \)The general solution is:\[(x, y, z, w) = (t - s, t, 3t - s, s) \]

Key Concepts

Free Variable

In a system of linear equations, a free variable is one that can take any value from the set of real numbers. While other variables are determined in terms of the free variable, the free variable itself has no specific value.
In our exercise, the equations simplify to express other variables in terms of two main variables: let's name them \( y \) and \( w \). Here, \( y \) and \( w \) act as free variables. These variables can be assigned any value, and they define the possible solutions for the entire system.
Choosing free variables is a common technique in solving systems of equations, particularly when there are infinitely many solutions. These free variables allow us to generalize the solution as they can change freely, determining the values of the dependent variables.

Dependent Equations

Dependent equations occur when one or multiple equations in the system do not contribute additional unique information. Instead, they are derivable from others by combining or rescaling the equations.

In the exercise given, after substitutions and simplifications, the third equation becomes \( 0 = 0 \). This identity indicates dependency, as it doesn't constrain the values of the variables any further.
The presence of dependent equations often signals that the system does not have a unique solution, but instead, an entire family of solutions exists. Understanding dependent equations helps to determine why and when free variables occur, further allowing us to understand the nature of the solution set.

Parameterization

Parameterization is a method to express the solutions of a system with free variables using parameters. These parameters are free to vary over the set of real numbers, and they provide a way to describe the entire solution set efficiently.

In the system from the exercise, by letting \( y = t \) and \( w = s \), where \( t \) and \( s \) are parameters, we can express \( x \) and \( z \) in terms of these parameters:

• \( x = t - s \)
• \( z = 3t - s \)

The process of parameterization allows us to clearly represent all possible solutions as a function of \( t \) and \( s \), demonstrating flexibility and clarity in dealing with solutions of systems of equations.

Solution Set

The solution set is a complete collection of all possible solutions that satisfy the given system of linear equations.

For our system, the solution set can be represented in parameterized form, showcasing the role of the free variables. Using parameters \( t \) and \( s \), it is expressed as \((x, y, z, w) = (t - s, t, 3t - s, s)\). This representation captures all combinations of \( x, y, z, \) and \( w \) that satisfy every equation in the system.
Understanding the solution set is crucial because it not only shows us what solutions exist, but also how they relate to each other through the parameters. It confirms that infinitely many solutions exist, each consistent with the equations provided.