91Ó°ÊÓ

A system of linear equations consists of two or more linear equations involving the same set of variables. The goal is to find values for these variables that satisfy all the equations in the system simultaneously. In the given problem, we have three equations:

• \[ x + y = 1 \]
• \[ y + z = 0 \]
• \[ x + z = 1 \]

Each equation is linear since the variables are only raised to the power of one and do not multiply each other. In practical terms, solving these means finding such values of \( x, y, \) and \( z \) that make all three equations true at the same time. Solving systems of linear equations often involves methods like substitution or elimination. However, in this exercise, we need to consider them within the constraints of modular arithmetic, which adds an interesting twist.

Modular arithmetic is a system of arithmetic for integers, where numbers "wrap around" after they reach a certain value called the modulus. In this scenario, we are working with \( \mathbb{Z}_2 \), which implies our modulus is 2. This means we only have two numbers: 0 and 1. Operations are performed with respect to these two numbers:

• Addition: If the sum is equally divisible by 2, it results in 0, otherwise 1. For example, \(1 + 1 = 2 \equiv 0 \pmod{2}\).
• Multiplication: Similarly, results stay within the same range, for instance, \(1 \times 1 = 1 \equiv 1 \pmod{2}\).

Understanding modular arithmetic is crucial for solving the problem, as it determines how we interpret operations like addition and multiplication. Applying these rules allows us to analyze each equation and find valid solutions within the limited set of numbers \(0\) and \(1\).

The concept of solving equations over \( \mathbb{Z}_2 \) requires a bit of a shift from regular arithmetic. In essence, the primary twist is that we interpret the addition and multiplication operations differently, as described by modular arithmetic. For the system of equations given:

• The possible solution sets we derive must also be subsets of \(0, 1\), the only elements in \( \mathbb{Z}_2 \).
• When solving \( x + y = 1 \), the combinations (0, 1) and (1, 0) for \((x, y)\) satisfy this equation.
• For \( y + z = 0 \), the valid solutions are (0, 0) and (1, 1) for \((y, z)\).
• The equation \( x + z = 1 \) permits solutions like (0, 1) or (1, 0) for \((x, z)\).

Combining these insights, the correct solution satisfying all conditions is \((x, y, z) = (1, 0, 1)\). Verifying by substituting back proves consistency across the system of equations, thus confirming it's a valid solution. Solutions over modular systems like \( \mathbb{Z}_2 \) can seem tricky at first but are simply a different view of constraints on arithmetic operations.