Problem 23 - Precalculus

Question

In Exercises \(9-26\), put each system of linear equations into triangular form and solve the system if poesible. Classify each system as consistent independent, consistent dependent, or inconsistent. $$ \left\\{\begin{aligned} 2 x-3 y+z &=-1 \\ 4 x-4 y+4 z &=-13 \\ 6 x-5 y+7 z &=-25 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The system is consistent dependent, with infinitely many solutions.

Step 1: Write the System

We start with the system of equations given:\[\begin{align*} 2x - 3y + z &= -1 \ 4x - 4y + 4z &= -13 \ 6x - 5y + 7z &= -25 \end{align*}\] We will transform this system into an upper triangular form.

Step 2: Eliminate Variables from Row 2

To eliminate the \( x \) variable from the second equation, subtract twice the first equation from the second:\[ 4x - 4y + 4z - 2(2x - 3y + z) = -13 - 2(-1) \]Simplifying the equation, we obtain:\[0x + 2y + 2z = -11\] So the new second equation becomes:\[2y + 2z = -11\]

Step 3: Eliminate Variables from Row 3

For the third equation, eliminate the \( x \) variable by subtracting three times the first equation:\[ 6x - 5y + 7z - 3(2x - 3y + z) = -25 - 3(-1) \]Simplify this to get:\[0x + 4y + 4z = -22\] The new third equation becomes:\[4y + 4z = -22\]

Step 4: Simplify the System Further

We simplify the second and third equations:- Divide the second equation by 2: \[ y + z = -\frac{11}{2} \]- Divide the third equation by 4: \[ y + z = -\frac{11}{2} \]Now, the second and third modified equations are identical.

Step 5: Classify the System

Since the second and third equations are identical but consistent with the first equation, this indicates the system is consistent and dependent, meaning it has infinitely many solutions.

Key Concepts

Upper Triangular Form

The upper triangular form of a system of linear equations is a transformation where each equation has fewer variables than the one before it. This helps to simplify the system of equations, making it more manageable to solve. It's like climbing a staircase – each step leads you to the solution.
In practice, this means rewriting the system so that all terms below the main diagonal (a diagonal from the upper-left to the lower-right part of a matrix) are zero.
For our system:

• We initially have three equations with three variables: \(x, y,\) and \(z\).
• Our goal is to eliminate the \(x\) terms from the second and third equations.
• By manipulating these equations using addition or subtraction, we transform the second and third equations to leave us with two-variable equations or repeat constants like in this exercise.

This triangular transformation is crucial in simplifying the solution process and understanding the system's nature.

Consistent Dependent System

A consistent dependent system is a system of equations where all the equations are dependent on each other. This means they share the same solution set, resulting in infinitely many solutions. It's like having multiple maps that lead you to the same destination.
For example, if we end up with two identical equations after elimination, like:

• \( y + z = -\frac{11}{2} \)
• \( y + z = -\frac{11}{2} \)

This repetition confirms the equations are dependent. In other words, both express the same line or plane in a graphical representation.
A consistent dependent system never contradicts itself and will always confirm the existence of solutions fitting all its equations.

Infinitely Many Solutions

When a system has infinitely many solutions, it means there are countless values of the variables that satisfy all the equations simultaneously. Imagine a line extending infinitely in all directions, representing countless points that fit the equations.
This situation arises when:

• The equations are not unique and fit together perfectly.
• They are essentially different forms of the same expression.

For our exercise, after transforming our system to upper triangular form, we discovered identical equations. This means any combination of \(x, y,\) and \(z\) that satisfies \( y + z = -\frac{11}{2} \) also satisfies the entire system.
Systems with infinitely many solutions often signal redundancy in their original equations – they don't add new information beyond what's already given.

Elimination Method

The elimination method is a systematic approach to remove variables from equations in a system. It helps us transform the system into a simpler form, often revealing solutions more clearly.
To perform elimination:

• Align similar variables and constants from your equations.
• Decide which variable you want to remove first – often the variable in the leading position of your equations.
• Add or subtract equations so selected terms effectively cancel out.

In our system:

• We chose to eliminate \(x\) from the second and third equations using the first equation.
• This required subtracting multiples of the first equation from the others to zero out \(x\) coefficients.

The elimination method, also known as Gauss-Jordan elimination in matrix form, is an efficient way to tackle many linear equation systems, providing a clear path to their solutions.