Problem 19 - Intermediate Algebra

Question

Solve each of the following systems. If the solution set is \(\varnothing\) or if it contains infinitely many solutions, then so indicate. $$ \left(\begin{array}{r} x+y+z=1 \\ 2 x-3 y+6 z=1 \\ -x+y+z=0 \end{array}\right) $$

Step-by-Step Solution

Answer

The system has infinitely many solutions.

Step 1: Write the System of Equations

We begin by writing down the system of equations as given. 1. \( x + y + z = 1 \) 2. \( 2x - 3y + 6z = 1 \) 3. \( -x + y + z = 0 \)

Step 2: Solve the First Two Equations

Eliminate one variable from the first two equations. Add Equation 1 and Equation 3: \[ (x + y + z) + (-x + y + z) = 1 + 0 \] This simplifies to: \[ 2y + 2z = 1 \] Dividing by 2 gives: \[ y + z = \frac{1}{2} \] (Equation 4).

Step 3: Substitute Equation 4 into Equation 2

Substitute \( y + z = \frac{1}{2} \) from Equation 4 into Equation 2:\[ 2x - 3(y + z) + 6z = 1 \] \[ 2x - 3\left(\frac{1}{2}\right) + 6z = 1 \] Simplify to get: \[ 2x - \frac{3}{2} + 6z = 1 \] This can be rearranged to: \[ 2x + 6z = \frac{5}{2} \] Multiply through by 2 to clear the fraction: \[ 4x + 12z = 5 \] (Equation 5).

Step 4: Solve the Reduced System

Now we have:1. \( y + z = \frac{1}{2} \) 2. \( 4x + 12z = 5 \) Solve for \( y \) from Equation 4: \[ y = \frac{1}{2} - z \] Substitute values from Equation 5 back into the original structure.

Step 5: Interpret the Results

There is a linear dependency among the equations suggesting a consistent modified form which results in multiple solutions along said parameters: \( x = 0, y = \frac{1}{2} - z, z = z \). Thus, varying \( z \) freely manipulates complimentary \( y \) solutions, revealing the system's infinite dimensional linear solutions.

Key Concepts

Gaussian elimination

Gaussian elimination is a method used to solve systems of linear equations. It involves a sequence of operations that transforms the coefficient matrix of the system into a simpler form, typically row-echelon form. The main steps involved in Gaussian elimination are:

• Swapping rows: If needed, rearrange the equations by swapping the rows in the matrix so the process can proceed smoothly.
• Eliminating variables: One variable is eliminated systematically from consecutive rows, leading to simpler equations.
• Back substitution: Once the equations are in row-echelon form, solve them from the last equation upwards to find the values of the variables.

In this exercise, Gaussian elimination was used to simplify the equations to a system where some variables could be expressed in terms of others. This helped uncover the relationships between variables and determine consistency in the system. Our solution led to an expression for one variable in terms of another, allowing for more straightforward interpretations of the results.

dependent system

A dependent system of linear equations is one where the equations are not independent but rather express the same relationship in different forms. In such systems, one equation can be derived from another by multiplying by a scalar or adding the equations.

In the given problem, after performing Gaussian elimination, we discovered that the system was dependent. This is identified by the fact that rearranging and simplifying the equations led to expressions such as \( y + z = \frac{1}{2} \). This equation depends on \( z \), indicating that several solutions can be formed based on different values for \( z \). A dependent system means the solutions are interlinked, and one or more equations don't add new information about the values of the variables. As a result, such systems usually lead to infinite solutions.

infinite solutions

When a system of linear equations has infinite solutions, it means that there are endless combinations of values for the variables that satisfy all the equations simultaneously. This typically occurs in dependent systems, where the number of unique equations is less than the number of variables.

In our step-by-step solution, after simplifying the system, we found that the solutions were expressed as \( x = 0, y = \frac{1}{2} - z, z = z \). The free choice of values for \( z \) means that every possible value of \( z \) yields corresponding values for \( x \) and \( y \) that satisfy all the equations in the system. Infinite solutions come from the freedom to adjust certain variables without restriction, reflecting a whole line or plane of possibilities rather than a single point solution.