Problem 45 - Precalculus: Mathematical for Calculus

Question

Solve the system of linear equations. $$\left\\{\begin{array}{r} x-y+w=0 \\ 3 x-z+2 w=0 \\ x-4 y+z+2 w=0 \end{array}\right.$$

Step-by-Step Solution

Answer

Solution is expressed in terms of parameters: \(x = s-t\), \(y = s\), \(z = 3s-t\), \(w = t\).

Step 1: Analyze the System

The system of linear equations has three equations and four variables: \(x\), \(y\), \(z\), and \(w\). This indicates that there may be infinitely many solutions, as we have more variables than equations.

Step 2: Express Variables in Terms of One Another

Let's aim to eliminate some variables. Start by expressing \(x\) from the first equation: \(x = y - w\).

Step 3: Substitute into Second Equation

Substitute \(x = y - w\) into the second equation: \(3(y-w) - z + 2w = 0\).Simplifying gives \(3y - 3w - z + 2w = 0\) or \(3y - z - w = 0\).

Step 4: Substitute into Third Equation

Substitute \(x = y - w\) into the third equation:\(y - w - 4y + z + 2w = 0\).Simplifying gives \(-3y + z + w = 0\).

Step 5: Solve for Relationship Between \(y\), \(z\), and \(w\)

We now have two new simpler equations: 1. \(3y - z - w = 0\) 2. \(-3y + z + w = 0\). Adding these yields \(0 = 0\), which is true for any \(y, z, w\), confirming the dependency between the variables.

Step 6: Parameterize the Solution

We need to choose a parameter for the solution, such as expressing \(y\) in terms of \(w\):From \(3y - z - w = 0\) and \(-3y + z + w = 0\), solve for \(y\), \(z\):From \(3y = z + w\): \( z = 3y - w\). From \( -3y = -z - w\), confirming \(z = 3y - w\).Let \(w = t\), a free parameter. Thus, \(y = s\) and \(z = 3s - t\).

Step 7: Write the Solution Set

The variables can be expressed in terms of parameters \(s\) and \(t\):- \(x = s - t\)- \(y = s\)- \(z = 3s - t\)- \(w = t\).This provides the general solution to the system.

Key Concepts

Linear Equations

Linear equations are equations where each term is either a constant or the product of a constant and a single variable. They are called 'linear' because they represent straight lines when graphed. In our example system of linear equations, we have:

• \(x - y + w = 0\)
• \(3x - z + 2w = 0\)
• \(x - 4y + z + 2w = 0\)

Each of these has variables raised only to the first power. The sum of these terms equals zero, meaning they form a linear combination. Solving systems of linear equations involves finding values for the variables that satisfy all equations simultaneously. In our system, we see four variables (\(x, y, z, w\)) and three equations, which typically implies the potential for infinitely many solutions. This is because there are more degrees of freedom (variables) than equations.

Parametric Solutions

When we solve systems with more variables than equations, it's common to express the solution in terms of parameters. A parametric solution assigns one or more variables to act as free parameters.In our problem, there is a need to parameterize because there are infinitely many solutions possible. We represent some of the variables in terms of others to simplify the equations:

• Start by expressing one variable in terms of another, as shown, replacing \(x\) with \(y - w\).
• Simplify the system so that the interdependencies between variables become clear.
• Assign free parameters to simplify further. This involves choosing one or more variables to set as parameters, here \(w = t\) and \(y = s\).

This method greatly simplifies solving by providing a framework to generate specific solutions for any selected parameter values.

Variables and Parameters

In any system of equations, variables are the unknowns we solve for, while parameters are constants that help express the variables.

• Variables in our system are \(x\), \(y\), \(z\), and \(w\). These are what we aim to solve.
• Parameters are chosen to express solutions flexibly. In this exercise, \(w = t\) and \(y = s\) are used as parameters.

This approach allows us to fully describe the potential solutions. By analyzing and manipulating equations, variables and their relationships evolve, enabling a wide range of outcomes.Understanding how to strategically choose and use these parameters is essential in systems with more variables than equations. It allows us to explore solutions beyond initial constraints, highlighting the depth and flexibility of linear systems.