Problem 2 - Precalculus: Mathematics for Calculus

Question

These exercises refer to the following system: $$\left\\{\begin{array}{rr}x-y+z= & 2 \\\\-x+2 y+z= & -3 \\\3 x+y-2 z= & 2\end{array}\right.$$ To eliminate \(x\) from the third equation, we add ____________ times the first equation to the third equation. The third equation becomes ___________ \(=\) ______.

Step-by-Step Solution

Answer

Add -3 times the first equation; the third becomes \(3y - 5z = -4\).

Step 1: Identify Coefficient Matching

We need to eliminate \(x\) from the third equation by adding a multiple of the first equation to it. To do this, we must ensure that the coefficients of \(x\) in both equations match in magnitude so they can cancel each other out.

Step 2: First Equation Coefficient

The coefficient of \(x\) in the first equation is 1. We will use this coefficient to determine how much we need to multiply the first equation to cancel out \(x\) in the third equation.

Step 3: Third Equation Coefficient

The coefficient of \(x\) in the third equation is 3. To eliminate \(x\), we need the coefficients to be opposites, i.e., add \(-3\) times the first equation to the third equation to cancel out \(x\).

Step 4: Multiply the First Equation

Multiply the entire first equation by \(-3\) to get the terms needed to cancel out \(x\) from the third equation:\[-3(x - y + z) = -3 \times 2\]which simplifies to:\[-3x + 3y - 3z = -6\]

Step 5: Add to the Third Equation

Add the scaled first equation to the third equation:\[(3x + y - 2z) + (-3x + 3y - 3z) = 2 + (-6)\]This simplification results in:\[3y - 5z = -4\]

Step 6: Conclusion: Resulting Equation

The third equation transformed after adding \(-3x + 3y - 3z\) (from scaling the first equation) to it becomes:\[3y - 5z = -4\]

Key Concepts

Matrix Operations

Matrix operations are fundamental in solving systems of linear equations. When dealing with multiple equations, matrices help us manage the data in an organized and simplified manner. To start, think of a matrix as a table of numbers with rows representing each equation and columns denoting the coefficients of the variables.

There are several matrix operations, but addition and multiplication are the most relevant when solving linear systems. In the process of eliminating variables, like in our exercise, we implicitly perform matrix operations, specifically row operations, which modify one row based on another:

• Adding or subtracting one row to or from another
• Multiplying a row by a scalar (a constant number)
• Swapping rows

Understanding these operations is crucial as they mirror algebraic operations, helping us manipulate equations to achieve a solution.
For example, multiplying the entire first equation by \(-3\) is akin to multiplying a row in a matrix—this action is crucial for matching coefficients and simplifying the system.

Linear Algebra

Linear algebra is the field of mathematics dealing with vectors, matrices, and systems of linear equations. It supplies the foundational tools and concepts needed to study these systems efficiently.

One core concept in linear algebra is the "solution of linear systems." Here, we aim to find the values for variables that satisfy all equations simultaneously. Each step, such as identifying coefficients or transforming equations, is undertaken with these solutions in mind.

In the context of the given exercise, linear algebra provides the framework for understanding how manipulating one equation affects the overall system. By leveraging principles from linear algebra, we can systematically reduce our system into simpler, equivalent forms. This particular exercise utilizes the elimination method, a technique derived from linear algebraic strategies, where we aim to reduce the system to a simpler form where solutions can be easily extracted.

Equation Manipulation

Equation manipulation involves altering the structure or form of algebraic equations to facilitate their solution. In our exercise, this is exemplified through the process of subtracting and adding equations, which helps to eliminate one variable at a time.

The main types of equation manipulation used in solving systems are:

• Multiplying an equation by a constant (scaling)
• Adding or subtracting equations
• Rearranging equations

This manipulation requires careful attention to coefficient matching, as seen when we multiply the first equation by \(-3\) to ensure the coefficients of \(x\) are opposites.

Effective manipulation allows us to simplify complex systems into manageable parts, gradually forming solutions. The goal is to isolate variables, which in our case, means transforming the third equation into a two-variable equation, \(3y - 5z = -4\), simplifying further steps in finding the solution.