Problem 136 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

SYSTEM OF LINEAR EQUATIONS WITH PARAMETERS. $$ \left\\{\begin{array}{l} x+a y-1=0 \\ a x-3 a y-(2 a+3)=0 \end{array}\right\\} $$

Step-by-Step Solution

Answer

For the given system of linear equations with parameters: \(x = 1 - a\frac{a - 2a - 3}{a^2 + 3a}\) \(y = \frac{a - 2a - 3}{a^2 + 3a}\) There are two cases to consider: 1. If \(a = 0\), there are infinitely many solutions as there's only one independent equation. 2. If \(a ≠ 0\), there is a unique solution in terms of the parameter a.

Step 1: Rewrite the given equations in a more familiar form.

For the system of equations to be solvable, we first need to rewrite them in the following form: \(x + ay - 1 = 0\) and \(ax -3ay - (2a + 3) = 0\).

Step 2: Solving for x or y in one equation

From the first equation, we can isolate x and solve for it: x = 1 - ay

Step 3: Substitute x in the second equation

Now, substitute x from our expression found in step 2 into the second equation: a(1 - ay) - 3ay - (2a + 3) = 0

Step 4: Simplify the expression

Simplify the equation derived in step 3: a - a^2y - 3ay - 2a - 3 = 0

Step 5: Analyze the simplified equation

The simplified equation is a quadratic equation in terms of y. The coefficients of the quadratic equation depend on the parameter a. Now, we need to analyze different cases depending on the value of the parameter a.

Step 6: Determine cases for the parameter a

Analyze \(a = 0\): If a = 0, then the second equation becomes irrelevant as it will be 0 = 0. The first equation will be x = 1. So in this case, we have infinitely many solutions as there is only one independent equation. If a ≠ 0, then the equation we derived in step 4 no longer degenerates. We can rewrite the equation as follows: \(y(a^2 + 3a) = a - 2a - 3\)

Step 7: Solve for y

Now, we can solve for y by dividing both sides by \((a^2 + 3a)\) provided that \(a^2 + 3a ≠ 0\): \(y = \frac{a - 2a - 3}{a^2 + 3a}\)

Step 8: Substitute y back into the expression for x

Replace y with the expression found in step 7 in the expression for x: x = 1 - a\(\frac{a - 2a - 3}{a^2 + 3a}\) Now, we have the general solution for the given system of linear equations with parameters in terms of x and y as: x = 1 - a\(\frac{a - 2a - 3}{a^2 + 3a}\) y = \(\frac{a - 2a - 3}{a^2 + 3a}\) Case 1: \(a = 0\) - Infinitely many solutions as there's only one independent equation. Case 2: \(a ≠ 0\) - Unique solution in terms of the parameter a.

Key Concepts

Parameters in Equations

In many algebraic problems, parameters play a crucial role in understanding and manipulating equations. Parameters are essentially constants that can take on various values, affecting the general behavior of a set of equations.
In the given problem, the parameter "a" is a variable which can modify both equations in the system, altering the solution set.
Parameters can be:

• Constant values that simplify equations
• Variables that allow for flexibility in solutions

By treating parameters differently from regular variables, we can analyze how equations behave under different conditions and see how solutions might vary.

Solving Equations with Parameters

When faced with equations involving parameters, the first step is often to isolate variables as much as possible. In our exercise, we manipulated the equations to express one variable in terms of the other and the parameter "a".
Solving with parameters might involve:

• Isolating one key variable
• Substituting back into other equations
• Simplifying expressions to find a general solution

After simplifying, it’s important to analyze different cases of the parameter to fully understand every potential solution.
This analysis is necessary because parameters can introduce conditions where usual algebraic operations (like division) might no longer be valid nor defined.

Linear Algebra Concepts

Linear algebra provides us with a toolkit for solving systems of equations, often involving more than one equation and several variables. Conceptually, each equation can be thought of as representing a geometric object, such as a line.
Key linear algebra concepts include:

• Coefficients: Numbers multiplying variables that dictate the line's slope.
• Variables: Represent unknown quantities we aim to find.

By rewriting the original equations and isolating variables, the step-by-step solution effectively uses these concepts to manipulate and solve a parameter-based system, ensuring our solutions meet all given equations.

Dependent and Independent Equations

Understanding whether equations are dependent or independent is vital when solving systems of equations. Independent equations each provide unique information, leading to a pinpoint solution when solved together.
In contrast, dependent equations share overlap, reducing the amount of independent information and potentially leading to multiple solutions.
The original problem illustrates this by showing how the value of the parameter "a" affects the independence of the equations:

• If "a = 0", we only have one valid equation left, leading to an infinite number of solutions.
• If "a ≠ 0", both equations uniquely contribute to a single synthesis, providing a unique solution.

Grasping these concepts helps predict outcomes and tailors the method for solving the system effectively.