Problem 13 - Algebra and Trigonometry

Question

\(13-16\) . Perform an operation on the given system that eliminates the indicated variable. Write the new equivalent system. $$ \begin{array}{l}{\left\\{\begin{array}{l}{x-2 y-z=4} \\ {x-y+3 z=0} \\ {2 x+y+z=0}\end{array}\right.} \\ {\text { Eliminate the } x \text { -term }} \\\ {\text { rom the second equation. }}\end{array} $$

Step-by-Step Solution

Answer

New system: \[\begin{array}{l} x - 2y - z = 4 \\ y + 4z = -4 \\ 2x + y + z = 0 \end{array}\]

Step 1: Identify the Second Equation

We start by identifying the second equation from the system, which is given as: \[ x - y + 3z = 0 \] We need to eliminate the \(x\)-term from this equation.

Step 2: Find a Multiple to Cancel x-Term

To eliminate the \(x\)-term from the second equation, we'll need to subtract the first equation \(x - 2y - z = 4\) or the third equation \(2x + y + z = 0\) from it, after aligning the coefficients of \(x\). In this case, since both the coefficient of \(x\) in the third equation is \(2\), and we need to eliminate \(x\) in its own set of equations especially having coefficients different in zero, we can easily subtract twice the first equation from the second one.

Step 3: Multiply First Equation and Subtract

Multiply the first equation by 1 (since both equations are equal in the coefficient of \(x\)): \[ x - 2y - z = 4 \] Then subtract this from the second equation:\[ (x - y + 3z) - (x - 2y - z) = 0 - 4 \]

Step 4: Simplify the Resulting Equation

The subtraction results in:- The \(x\) terms cancel each other out.- Combine the \(y\) and \(z\) terms:\[ -y + 3z + 2y + z = -4 \] This simplifies to the new equation:\[ y + 4z = -4 \]

Step 5: Write the New System

Now, we write the new equivalent system excluding \(x\) from the second equation:\[\begin{array}{l} x - 2y - z = 4 \ y + 4z = -4 \ 2x + y + z = 0\end{array}\] This gives us the new system where the \(x\)-term has been eliminated from the second equation.

Key Concepts

System of Equations

A system of equations is a collection of two or more equations with the same set of variables. Solving these systems involves finding values for the variables that satisfy all equations simultaneously. In linear algebra, systems of equations are often represented in matrix form, which allows for efficient methods of solution through matrix operations.

Systems of equations can be classified based on the number of solutions they possess:

• Consistent Systems: These have at least one solution. Consistent systems can be further categorized as independent (one unique solution) or dependent (infinitely many solutions).
• Inconsistent Systems: These have no solution because the equations represent parallel lines, which never intersect.

Working with systems of equations is a crucial aspect of solving real-world problems where multiple conditions must be met simultaneously. Understanding how to manipulate and solve these systems provides a strong foundation in algebra and is essential in applied mathematics.

Variable Elimination

Variable elimination is a method used to solve a system of equations by removing one or more variables. This technique is often called the elimination method because it involves cancelling out a variable to simplify the system.

In the given exercise, we focused on eliminating the x-term from the second equation. To do this:

• Align the coefficients of the variable you seek to eliminate (in this case, x).
• Subtract one equation from another to cancel the variable out.
• Simplify the resulting equation to find an expression in terms of the remaining variables.

This method is particularly useful when substitution becomes cumbersome. Eliminating variables step-by-step allows for a systematic approach to home in on simple equations that can be more easily solved. It also demonstrates the interconnected nature of variables within linear equations.

Equivalent System

An equivalent system is achieved when you make operations on a system of equations that don't change the solution set. Techniques such as addition, subtraction, multiplication, or division of entire equations ensure that the new system has the same solutions as the original.

In the exercise, we derived a new equivalent system by subtracting one multiple of an equation from another. This manipulation didn't affect the overall solution, preserving all original intersect points in variable space.

Equivalent systems are a powerful concept because:

• They allow flexibility—enabling transformations that simplify complex systems without altering the solution.
• They are foundational in the principle of row operations in matrix algebra.

Understanding equivalent systems provides insight into how equations interact and the validity of their solutions.

Linear Algebra

Linear algebra is a branch of mathematics concerning linear equations and their representations through matrices and vector spaces. It serves as the backbone for many mathematical applications in engineering, physics, computer science, and economics.

Key concepts within linear algebra include:

• Vector Spaces: Collections of vectors that can be added together and multiplied by scalars.
• Matrices: Rectangular arrays of numbers representing a system of linear equations. They allow compact, efficient handling of multiple equations.
• Determinants and Eigenvalues: Attributes of matrices that provide insight into their properties and the systems they represent.

Employing linear algebra techniques simplifies the solving of systems of equations. In the exercise, the elimination method derives directly from these concepts, illustrating the practical use of linear algebra in reducing systems to simpler forms for straightforward analysis.