Determining The Product Of Coefficients In A Linear Function

In the realm of mathematics, particularly within the study of functions, we often encounter situations where we need to determine the parameters of a function based on given data points. This article delves into such a problem, focusing on a linear function defined over integers. We are given a set of points that the function passes through, and our objective is to find the coefficients of the linear function and subsequently their product. This problem not only reinforces our understanding of linear functions but also highlights the importance of algebraic manipulation and problem-solving skills in mathematics. Understanding this concept is crucial for anyone delving into mathematical analysis, discrete mathematics, or computer science, as it lays the groundwork for more complex problem-solving strategies and algorithm design. Specifically, this article will guide you through the process of using a set of points to uniquely determine a linear function and then calculate a specific property of its coefficients, demonstrating a practical application of mathematical principles.

Problem Statement

Let P = {(1,1), (2,3), (0,-1)} be a function from Z to Z defined by P(x) = ax + b for some integers a and b. Our task is to determine the product of these integer coefficients, a.b. This problem requires us to find the values of 'a' and 'b' that define the linear function P(x) and then compute their product. The given set P represents points (x, y) that satisfy the equation P(x) = ax + b. We will use these points to create a system of linear equations, solve for 'a' and 'b', and finally calculate a.b. This type of problem is fundamental in understanding how to model relationships between variables using linear functions and is a common exercise in introductory algebra and discrete mathematics courses. Successfully solving it demonstrates a solid grasp of algebraic techniques and the ability to apply mathematical concepts to practical problems.

Solution

Step 1: Setting up the Equations

We are given that P(x) = ax + b and the function passes through the points (1,1), (2,3), and (0,-1). This gives us three equations:

1. P(1) = a(1) + b = 1
2. P(2) = a(2) + b = 3
3. P(0) = a(0) + b = -1

These equations form a system of linear equations that we can solve to find the values of 'a' and 'b'. The first two equations come from substituting x = 1 and x = 2 into the function P(x), respectively, and setting the results equal to the corresponding y-values (1 and 3). The third equation is derived similarly using the point (0, -1). This step is critical because it translates the problem from a functional description into a set of algebraic equations that are easier to manipulate and solve. Understanding how to set up these equations is fundamental in many areas of mathematics and science, particularly in modeling relationships between variables using linear models.

Step 2: Solving for b

The third equation, a(0) + b = -1, simplifies directly to b = -1. This provides us with the value of 'b' immediately, which is a significant step forward. This simplification occurs because any number multiplied by zero is zero, leaving us with a direct solution for 'b'. Recognizing such direct solutions is a valuable skill in problem-solving, as it can significantly reduce the complexity of the remaining steps. In this context, knowing 'b' allows us to reduce the number of unknowns in the other equations, making them easier to solve. This technique of identifying and utilizing direct solutions is applicable in a wide range of mathematical problems and is a key component of efficient problem-solving strategies.

Step 3: Solving for a

Now that we know b = -1, we can substitute this value into the first equation:

a(1) + (-1) = 1 a - 1 = 1 a = 2

Thus, we find that a = 2. Substituting the value of 'b' into the first equation allows us to isolate 'a' and solve for its value. This step demonstrates the principle of substitution, a fundamental technique in algebra for solving systems of equations. By substituting a known value into another equation, we reduce the number of unknowns and simplify the problem. This process is not only effective in solving linear equations but also applicable in more complex mathematical contexts. The ability to strategically substitute values is a key skill in mathematical problem-solving and is essential for mastering algebraic manipulations.

Step 4: Calculating a.b

Finally, we calculate the product of a and b:

a.b = 2 * (-1) = -2

Therefore, the product of the coefficients a and b is -2. This final step is a straightforward calculation that utilizes the values of 'a' and 'b' we previously determined. It highlights the importance of accurately solving for the individual components before combining them to find the final answer. The calculation itself is simple, but it underscores the overall problem-solving process: first, we established the equations, then we solved for the unknowns, and finally, we used those values to answer the original question. This structured approach is a valuable strategy for tackling complex mathematical problems and ensures that all steps are logically connected and contribute to the correct solution.

Conclusion

In conclusion, given the function P(x) = ax + b that passes through the points (1,1), (2,3), and (0,-1), we determined that a = 2 and b = -1. Therefore, the product a.b is -2. This problem demonstrates the application of basic algebraic principles to solve for unknown coefficients in a linear function. By setting up a system of equations and using substitution, we were able to efficiently find the values of 'a' and 'b'. This type of problem-solving is fundamental in various fields, including mathematics, physics, and engineering, where linear models are frequently used to represent relationships between variables. The ability to solve such problems is a key skill for students and professionals alike, as it provides a foundation for understanding more complex mathematical concepts and real-world applications. Moreover, the structured approach used in this solution—setting up equations, solving for unknowns, and then calculating the final result—is a valuable problem-solving strategy that can be applied to a wide range of problems.