/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 53 Solve the systems of linear equa... [FREE SOLUTION] | 91Ó°ÊÓ

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Solve the systems of linear equations over the indicated \(\mathbb{Z}_{p}.\) \(x+2 y=1\) over \(\mathbb{Z}_{3}\) \(x+y=2\)

Short Answer

Expert verified
The solution is \(x = 0, y = 2\) in \(\mathbb{Z}_3\).

Step by step solution

01

Understanding the Finite Field

We are solving the linear equations over \(\mathbb{Z}_3\), which means all arithmetic operations should be performed modulo 3. The set \(\mathbb{Z}_3\) includes the numbers \{0, 1, 2\}.
02

Write the System of Equations

We have the two equations: 1. \(x + 2y \equiv 1 \pmod{3}\)2. \(x + y \equiv 2 \pmod{3}\).
03

Isolate Variables

Subtract the second equation from the first: \((x + 2y) - (x + y) \equiv 1 - 2 \pmod{3}\).This simplifies to \(y \equiv -1 \equiv 2 \pmod{3}\).
04

Substitute Back to Find x

Substitute \(y = 2\) into the second equation: \(x + 2 \equiv 2 \pmod{3}\).This simplifies to \(x \equiv 0 \pmod{3}\).
05

Verify the Solution

Substitute \(x = 0\) and \(y = 2\) into both original equations to verify:- First equation: \(0 + 2(2) = 4 \equiv 1 \pmod{3}\).- Second equation: \(0 + 2 = 2 \equiv 2 \pmod{3}\).Both equations hold true.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Modular Arithmetic
Modular arithmetic is a system of arithmetic for integers that handles the concept of congruence. In simpler terms, when we perform operations like addition, subtraction, and multiplication in this system, we do so focusing on the remainder of these operations with respect to a certain number, called the modulus. For example, when we say we are working modulo 3, like in this problem, we are interested in the remainder of any number after it has been divided by 3. This means only three possible results exist in any modular computation based on the modulus: 0, 1, or 2.

In a modular arithmetic system such as \( \mathbb{Z}_3 \), the set \( \{0, 1, 2\} \) indicates these possible remainders. It's important to handle negative numbers carefully. For instance, \(-1 \equiv 2 \pmod{3}\) because when a division by 3 occurs, the remainder of \(-1+3\) is 2.
  • The operations wrap around upon reaching the modulus—this is often likened to a clock.
  • Calculations must respect the modulus, ensuring results stay within the set limits defined by it.
Linear Equations
A linear equation is an equation that makes a straight line when it is graphed. These equations involve variables with exponents of 1 and express relationships through addition, subtraction, or equality. In this context, the equations involve variables such as \(x\) and \(y\) and are solved in the finite context of \( \mathbb{Z}_3 \).

Understanding linear equations over finite fields requires adjusting typical concepts to ensure solutions reflect the modular constraints. For example, the problem gives two linear equations: \(x + 2y \equiv 1 \pmod{3}\) and \(x + y \equiv 2 \pmod{3}\). Both equations must be simultaneously satisfied within these constraints, which requires:
  • Appropriate transformations, like isolating variables when possible.
  • Ensuring calculations consistently apply the modulus.
Modulo Operation
The modulo operation finds the remainder after division of one number by another. It is fundamental to solving equations in finite fields and wraps answers to fit within a specified set of integers from 0 up to one less than the modulus. For example, when addressing "\(1 \mod 3\)", it evaluates to 1.

The modular arithmetic in \( \mathbb{Z}_3 \) always focuses on wrapping numbers around through this modulo operation, which becomes crucial in eliminating any answers that don't fit the correct remainder. In linear problems like these, the modulo operation ensures that both sides of each equation reflect numbers within the modular system. Operations might look different compared to regular arithmetic:
  • Subtracting two equations requires reapplying the modulo condition.
  • Adjusting results to fit within the confines of the modulus, for instance, remembering that \(-1 \equiv 2 \pmod{3}\).
Systems of Equations
Systems of equations involve solving multiple equations simultaneously. Each equation in the system shares common variables, and the solution is a set of values for each variable that makes all equations true at the same time. In this problem, we have:
  • \(x + 2y \equiv 1 \pmod{3}\)
  • \(x + y \equiv 2 \pmod{3}\)
Solving systems of equations over finite fields like \( \mathbb{Z}_3 \) requires finding a common solution that fits all equations under the modular constraint. Steps to solving:
  • Rearrange equations if needed to isolate a variable.
  • Substitute between equations to find one complete solution.
  • Verify the solution for both equations to ensure full compatibility within the modulus framework.
This process involves careful handling of each equation to balance between maintaining equality while respecting the modular structure.

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Most popular questions from this chapter

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