Problem 25 - 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. $$ \left\\{\begin{aligned} x_{1}-x_{3} &=-2 \\ 2 x_{2}-x_{4} &=0 \\ x_{1}-2 x_{2}+x_{3} &=0 \\ -x_{3}+x_{4} &=1 \end{aligned}\right. $$

Step-by-Step Solution

Answer

Consistent and independent with solution: \(x_1 = 1\), \(x_2 = 2\), \(x_3 = 3\), \(x_4 = 4\).

Step 1: Initial Setup

Start by writing out the given system of equations as they are presented:\[\begin{align*}x_1 - x_3 &= -2 \2x_2 - x_4 &= 0 \x_1 - 2x_2 + x_3 &= 0 \-x_3 + x_4 &= 1 \end{align*}\]

Step 2: Substitution for Independent Solution

The goal is to express one variable in terms of others to reduce this system to triangular form. Let's start by isolating variables where possible. From Equation 1, express \(x_1\) in terms of \(x_3\):\[x_1 = x_3 - 2\]

Step 3: Simplify Other Equations

Substitute \(x_1 = x_3 - 2\) in Equation 3 and simplify:\[(x_3 - 2) - 2x_2 + x_3 = 0 \]This simplifies to:\[2x_3 - 2x_2 = 2 \]Divide through by 2 to simplify:\[ x_3 - x_2 = 1 \]

Step 4: Organize into Triangular Form

Now organize the system in a triangular form by expressing the three derived equations:\[\begin{align*}x_1 &= x_3 - 2 \x_3 - x_2 &= 1 \2x_2 - x_4 &= 0 \-x_3 + x_4 &= 1\end{align*}\]

Step 5: Solve System of Equations

Start solving the system by using back substitution:1. From Equation 3: \(x_4 = 2x_2\).2. Substitute \(x_4 = 2x_2\) into Equation 4: \[-x_3 + 2x_2 = 1\] Use Equation 2 \(x_3 - x_2 = 1\) to express \(x_3\) and solve for \(x_2\). Solve for \(x_3 = x_2 + 1\).3. Substitute into 4: \[-(x_2 + 1) + 2x_2 = 1\], yielding \[x_2 = 2\]. 4. Hence, \(x_3 = 3\) and \(x_1 = 1\), compute \(x_4 = 4\).

Step 6: Classify the System

This is a consistent independent system as all variables hold distinct values that satisfy all equations.

Key Concepts

Triangular Form

The triangular form of a system of linear equations is an arrangement of the equations such that they resemble a staircase or upper triangular matrix structure. This form simplifies the process of solving the system, making it easier to apply methods like back substitution.

The goal is to organize the equations so that each successive equation focuses on one additional variable. For example, the simplest case might involve:

• The first equation containing the first variable.
• The second equation containing the first and second variables.
• The third equation containing the first two variables, and so on.

Converting a system into triangular form often involves re-arranging and combining equations to systematically isolate variables in a step-by-step manner.

Consistent Independent System

A consistent independent system is one where there is exactly one solution for the variables. This occurs when all the equations intersect at only one point in space.

In the context of linear equations, being consistent means that the equations do not contradict each other. Being independent means that none of the equations are duplicates or multiples of one another.

When we classify a system as consistent independent, we are saying:

• All equations have a unique intersection point.
• There is no redundancy among the equations.
• Each variable has a distinct solution within the system.

This characteristic ensures that the system has a solvable, unique answer for all variables involved.

Solving Linear Systems

Solving linear systems involves finding the values of the variables that satisfy all equations in the system simultaneously.

There are multiple methods to solve these systems, including substitution, elimination, and matrix operations. However, transforming the equations into a specific form often simplifies the solving process.

For systems transformed into triangular form, back substitution is commonly used. Steps generally include:

• Expressing one variable explicitly in terms of the others.
• Reorganizing remaining equations to continue isolating variables.
• Using known values to find remaining variables progressively.

This methodical approach helps in unraveling the relationships between variables step-by-step.

Back Substitution

Back substitution is a process used to solve systems of equations once they are in triangular form. It is most effective when the system is re-arranged such that each equation builds on the previous one.

Starting with the last equation, you solve for one variable and substitute this back into prior equations to find other variables. For example:

• Solve the last equation for one variable.
• Substitute this known value into the penultimate equation to find another variable.
• Continue this process until all variables are known.

This structured approach allows you to efficiently unravel the entire system by working backwards from what you already know, ensuring that each step is verified and consistent with the equations given.