Problem 18 - 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. $$ \left\\{\begin{aligned} 2 x-4 y+z &=-7 \\ x-2 y+2 z &=-2 \\ -x+4 y-2 z &=3 \end{aligned}\right. $$

Step-by-Step Solution

Answer

The solution is \((x, y, z) = (-3, \frac{1}{2}, 1)\). The system is consistent independent.

Step 1: Organize the Equations

Let's start by identifying the given system of equations: 1. \( 2x - 4y + z = -7 \) 2. \( x - 2y + 2z = -2 \) 3. \( -x + 4y - 2z = 3 \) Our goal is to convert this system into triangular form. This means we will aim to have leading coefficients of 1 in the first column and zeros below them.

Step 2: Eliminate x from Equations 2 and 3

First, we'll work to make the first pivot in the system (the 2 in the first equation) into a leading one and eliminate \(x\) from equation 2 and 3. - Let's keep equation 1 as is: \( 2x - 4y + z = -7 \)- Multiply the second equation by 2 and subtract from the first equation to eliminate \( x \): \( 2(x - 2y + 2z) - (2x - 4y + z) = 2(-2) - (-7) \) \( 0x + 0y + 3z = 3 \) (This simplifies to \( z = 1 \))- Add the first equation to the third equation to eliminate \( x \): \( -x + 4y - 2z + 2x -4y + z = 3 - 7 \) \( x - z = -4 \) (This simplifies to \( x - z = -4 \))

Step 3: Solve for z, then x

Now we substitute the value of \( z = 1 \) from the modified second equation into the expression for \( x \) from the modified third equation: - Substitute \( z = 1 \) into \( x - z = -4 \): \( x - 1 = -4 \) \( x = -3 \)

Step 4: Solve for y

Substitute \( z = 1 \) and \( x = -3 \) back into the first equation to solve for \( y \): - \( 2(-3) - 4y + 1 = -7 \) \( -6 - 4y + 1 = -7 \) \( -5 - 4y = -7 \) \( -4y = -2 \) \( y = \frac{1}{2} \)

Step 5: Classify the System

The system has a unique solution \( (x, y, z) = (-3, \frac{1}{2}, 1) \). Since it has one solution, this system is classified as consistent independent.

Key Concepts

Triangular Form

Triangular form is a crucial step in solving systems of linear equations. It simplifies the system into a staircase-like pattern, making it easier to solve. Imagine your original set of equations laid out like this:

• The first equation has variables all together, such as \(2x - 4y + z = -7\).
• The second equation will have one less variable, usually starting with zero in the first digit, such as \(0x + 0y + 3z = 3\).
• The third equation reduces further, ultimately focusing on the last variable or becoming a simpler form like \(x - z = -4\).

You achieve this form by strategically manipulating your equations through operations like swapping, multiplying, adding or subtracting entire equations. The result is a system where you can use back substitution to easily find the values of your variables systematically.

Consistent Independent Systems

In the world of linear equations, determining the type of solution you have is crucial. A consistent independent system is one where there exists a unique solution. This means the lines represented by the equations intersect at exactly one point.Think of it as each equation having enough distinct information to precisely determine one point of intersection.

• This intersection gives us a unique solution.
• Just like in the example, where the unique solution is \((x, y, z) = (-3, \frac{1}{2}, 1)\).
• In graph terms, this means the planes intersect at a single, distinct point.

These systems are predictable and reliable, as opposed to being infinite or nonexistent, making them straightforward once put into a triangular form.

Solving Linear Systems

Solving linear systems involves finding values for the variables that make all equations true. When tackling a system:

• First, aim for simplification using triangular form, allowing you to see clearer paths to solutions.
• Next, employ strategies like elimination or substitution, drawing closer to finding variable values.
• Look at each equation's modified version, like finding \(z = 1\) as a straightforward solution point.
• Work backward to deduce the values of remaining variables, keeping simplicity at the heart of the process.

Each step builds on the previous, turning complex, entangled systems into manageable pieces, ultimately yielding solutions that satisfy all given equations.

Substitution Method

The substitution method is an effective way to solve linear systems once they're broken down into simpler forms. It involves replacing one variable with an equivalent expression from another equation.Here's how it works:

• Identify a variable from one equation that can be expressed in terms of the others, such as finding \(z\) in \(z = 1\).
• Substitute this known variable back into the other equations, reducing the number of variables being solved at once.
• Repeat this, step-by-step, turning knowns into solved components, like finding \(x = -3\) after substituting \(z = 1\).
• This simplification hones in on effectively reducing complexity, solving each equation efficiently.

By methodically replacing variables, the substitution method thins out possibilities, leaving behind exact solutions for each variable.