Problem 16
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. $$ \left\\{\begin{aligned} 2 x-y+z &=-1 \\ 4 x+3 y+5 z &=1 \\ 5 y+3 z &=4 \end{aligned}\right. $$
Step-by-Step Solution
Verified Answer
The system is inconsistent, meaning it has no solutions.
1Step 1: Write the System of Equations
First, list the given system of linear equations: 1. \( 2x - y + z = -1 \) 2. \( 4x + 3y + 5z = 1 \) 3. \( 5y + 3z = 4 \). Our aim is to express this system in triangular form to facilitate easy solving.
2Step 2: Eliminate x from Equation 2
Multiply Equation 1 by 2: \( 4x - 2y + 2z = -2 \). Subtract this result from Equation 2: \((4x + 3y + 5z) - (4x - 2y + 2z) = 1 - (-2) \). This simplifies to \( 5y + 3z = 3 \).
3Step 3: Compare New Equations
Notice that the new Equation 2 obtained, \( 5y + 3z = 3 \), is distinct from Equation 3, \( 5y + 3z = 4 \). Since these two equations are inconsistent (same coefficients for \( y \) and \( z \) but different constants), the system does not have a solution.
4Step 4: Classify the System
Given the inconsistency between the two equations for \( y \) and \( z \) resulting from step 3, the system of equations is classified as inconsistent. An inconsistent system does not have any solutions.
Key Concepts
Triangular FormConsistent vs Inconsistent SystemsSolving Linear SystemsLinear Equations Classification
Triangular Form
To solve a system of linear equations efficiently, we often aim to arrange it into a "triangular form". This format is akin to a triangular matrix in linear algebra. Here's what this means: the system is organized such that each equation has fewer variables than the one before it.
- The first equation should involve all variables. In our case, it was made up of terms like \(2x - y + z = -1\).
- The second equation should have one less variable, often eliminating the first variable. We achieve this by eliminating \(x\) through steps like multiplication and subtraction.
- The third, or last equation, involves the fewest variables, for example, just \(y\) and \(z\) in this exercise.
Consistent vs Inconsistent Systems
In linear algebra, a system of equations is either consistent or inconsistent. A consistent system has at least one solution, while an inconsistent system doesn’t have any.
To determine consistency, examine the resulting equations after applying methods like substitution or elimination:
To determine consistency, examine the resulting equations after applying methods like substitution or elimination:
- If any simplification leads to contradictory equations, such as \(5y + 3z = 4\) and \(5y + 3z = 3\), the system is inconsistent.
- A consistent system either has a unique solution \((consistent independent)\) or infinitely many solutions \((consistent dependent)\).
Solving Linear Systems
Solving linear systems involves finding values for variables that satisfy all given equations concurrently. Here's a step-by-step strategy:
1. Convert the system to triangular form for easier management.
2. Apply elimination methods: use operations like addition, subtraction, or multiplication to zero out variables in successive equations.
1. Convert the system to triangular form for easier management.
2. Apply elimination methods: use operations like addition, subtraction, or multiplication to zero out variables in successive equations.
- This example involved eliminating \(x\) from the second equation, forming a simpler two-variable equation.
- Compare and contrast these transformed equations to identify inconsistencies, if any.
Linear Equations Classification
When dealing with linear systems, classification is vital to predict the nature of solutions effectively. Linear systems fall into three distinct categories:
- Consistent Independent: A system with exactly one solution set, characterized by distinct parallel equations.
- Consistent Dependent: Infinitely many solutions where all equations describe the same line or plane.
- Inconsistent: No solution due to contradictory equations, as identified through operations like eliminating variables and comparison.
Other exercises in this chapter
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