Problem 30 - Algebra and Trigonometry

Question

25–34 Determine whether the system of linear equations is inconsistent or dependent. If it is dependent, find the complete solution. $$\left\\{\begin{array}{rr}{-2 x+6 y-2 z=} & {-12} \\ {x-3 y+2 z=} & {10} \\\ {-x+3 y+2 z=} & {6}\end{array}\right.$$

Step-by-Step Solution

Answer

The system is dependent. Solution: \( x = 3y + 2, y = y, z = 4 \).

Step 1: Set up the augmented matrix

We write the system of equations into an augmented matrix:\[\begin{bmatrix}-2 & 6 & -2 & | & -12 \1 & -3 & 2 & | & 10 \-1 & 3 & 2 & | & 6 \\end{bmatrix}\]

Step 2: Row reduce the matrix

Perform row operations to make the matrix simpler to understand. We aim to convert it into row-echelon form.- Swap Row 1 and Row 2: \[ \begin{bmatrix} 1 & -3 & 2 & | & 10 \ -2 & 6 & -2 & | & -12 \ -1 & 3 & 2 & | & 6 \ \end{bmatrix} \]- Add 2 times Row 1 to Row 2: \[ \begin{bmatrix} 1 & -3 & 2 & | & 10 \ 0 & 0 & 2 & | & 8 \ -1 & 3 & 2 & | & 6 \ \end{bmatrix} \]- Add Row 1 to Row 3: \[ \begin{bmatrix} 1 & -3 & 2 & | & 10 \ 0 & 0 & 2 & | & 8 \ 0 & 0 & 4 & | & 16 \ \end{bmatrix} \]

Step 3: Analyze the row-reduced form

In the current row-echelon form, neither the second nor the third row suggest an inconsistency. Instead, they indicate redundancy in the system. To continue:- Row 2 can be simplified by dividing by 2: \[ \begin{bmatrix} 1 & -3 & 2 & | & 10 \ 0 & 0 & 1 & | & 4 \ 0 & 0 & 4 & | & 16 \ \end{bmatrix} \]Since Row 3 is 4 times Row 2, it can be eliminated. This confirms dependency rather than inconsistency.

Step 4: Determine the complete solution

The pivot elements indicate that:\x and z are leading variables, and y can be expressed in terms of z.From Row 2: \( z = 4 \)From Row 1: \( x - 3y + 2(4) = 10 \)\( x - 3y + 8 = 10 \)\( x = 3y + 2 \)Thus, the solution in terms of y is:\[\begin{aligned}x &= 3y + 2 \y &= y \z &= 4 \\end{aligned}\]

Key Concepts

Augmented Matrix

An augmented matrix is a valuable tool when working with systems of linear equations. It combines the coefficients of the variables and the constants from each equation into a single matrix, facilitating more efficient manipulation. The augmented matrix provides a compact representation of the system that can be used to perform row operations to simplify the equations. For the given system of equations:

• First equation: \-2x + 6y - 2z = -12\
• Second equation: \x - 3y + 2z = 10\
• Third equation: \-x + 3y + 2z = 6\

The corresponding augmented matrix is:\[ \begin{bmatrix} -2 & 6 & -2 & | & -12 \ 1 & -3 & 2 & | & 10 \ -1 & 3 & 2 & | & 6 \end{bmatrix} \]This matrix aligns each row with the respective equation, allowing us to apply row operations to simplify and possibly solve the system.

Row-Reduction

Row-reduction is a series of operations designed to simplify the system of equations represented by an augmented matrix. The goal is to transform the matrix into a form that makes it easier to identify solutions. The three main operations used in row-reduction are:

• Swapping two rows
• Multiplying a row by a non-zero constant
• Adding or subtracting multiples of one row to another

For our system, row-reduction included steps such as swapping rows and adding multiples of rows to each other. Eventually, these operations resulted in:\[ \begin{bmatrix} 1 & -3 & 2 & | & 10 \ 0 & 0 & 1 & | & 4 \ 0 & 0 & 4 & | & 16 \end{bmatrix} \]This form makes it evident that the system can be further simplified, confirming the dependency and facilitating the finding of the solution.

Dependent Systems

A dependent system in the context of linear equations occurs when there is a redundancy in the equations. This means that at least one equation can be derived from others, leading to infinitely many solutions rather than a unique solution. In terms of the row-reduced matrix:

• Each non-zero row corresponds to a significant equation in the system
• If a zero row appears, it typically indicates dependence among the equations

In our case, once row-reduction was applied, Row 3 became a multiple of Row 2. This redundancy confirms the system is dependent. Consequently, rather than revealing an inconsistency or no solution, the dependency suggests multiple solutions exist, expressed in terms of free variables.

Solution Sets

When dealing with dependent systems, the solution set is generally expressed in terms of free variables, which allow for multiple solutions. In the reduced system, we observe:

• From Row 2, we have \( z = 4 \)
• From Row 1, simplifying gives \( x = 3y + 2 \)
• The variable \( y \) is free, meaning it can take any value

Thus, the solution set can be expressed parametrically as:\[ \begin{aligned} x &= 3y + 2 \ y &= y \ z &= 4 \end{aligned} \]Here, \( y \) is the parameter offering infinite potential solutions corresponding to different values that \( y \) might take. This formulation clearly presents the full set of solutions based on the dependent nature of the system.