Problem 81 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

The set of equations : \(\lambda x-y+(\cos \theta) z=0 ; 3 x+y+2 z\) \(=0 ;(\cos \theta) x+y+2 z=0,0 \leq \theta<2 \pi\), has non-trivial solutions. (A) for no values of \(\lambda\) and \(\theta\) (B) for all values of \(\lambda\) and \(\theta\) (C) for all values of \(\lambda\) and only two values of \(\theta\) (D) for only one value of \(\lambda\) and all values of \(\theta\)

Step-by-Step Solution

Answer

No non-trivial solutions exist for any values of \(\lambda\) and \(\theta\) (option A).

Step 1: Set up the Equations as a Matrix

Convert the system of equations into a matrix form:\[\begin{bmatrix}\lambda & -1 & \cos \theta \3 & 1 & 2 \\cos \theta & 1 & 2\end{bmatrix}\begin{bmatrix}x \y \z\end{bmatrix}= \begin{bmatrix}0 \0 \0\end{bmatrix}\]

Step 2: Find the Determinant for Non-trivial Solutions

The system has non-trivial solutions if the determinant of the matrix is zero. Calculate the determinant of the 3x3 matrix:\[\text{det} = \lambda \cdot \left(1 \times 2 - 1 \times 2\right) + 1 \cdot \left(3 \cdot 2 - 2 \cdot \cos \theta\right) + \cos \theta \cdot \left(3 \cdot 1 - 1 \cdot \cos \theta\right)\]

Step 3: Simplify the Determinant Expression

Simplifying the expression, we get:\[\text{det} = \lambda \cdot (0) + 6 - 2 \cos \theta + 3 \cos \theta - (\cos^2 \theta) = 6 + \cos \theta - \cos^2 \theta\]

Step 4: Set the Determinant to Zero

Set the determinant equal to zero for non-trivial solutions:\[6 + \cos \theta - \cos^2 \theta = 0\]Rearrange it into a quadratic form in terms of \(\cos \theta\):\[\cos^2 \theta - \cos \theta - 6 = 0\]

Step 5: Solve the Quadratic Equation

Solve the quadratic equation \(\cos^2 \theta - \cos \theta - 6 = 0\). Using the quadratic formula:\[\cos \theta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]where \(a=1\), \(b=-1\), and \(c=-6\):\[\cos \theta = \frac{1 \pm \sqrt{1 + 24}}{2} = \frac{1 \pm 5}{2}\]This results in \(\cos \theta = 3\) or \(\cos \theta = -2\), neither of which are possible values for \(\cos \theta\), which must be between -1 and 1.

Key Concepts

Systems of Equations

The concept of systems of equations involves solving for multiple unknowns by using more than one equation. In this exercise, we deal with a system comprised of three equations:

• \( \lambda x - y + \cos \theta \, z = 0 \)
• \( 3x + y + 2z = 0 \)
• \( \cos \theta \, x + y + 2z = 0 \)

These equations are interlinked, meaning any changes in one variable could affect the others. The main goal is to find values for \(x\), \(y\), and \(z\) that satisfy all three equations simultaneously.
This real-world application often requires transforming equations into systems to simplify the solution process. Equations can represent various scenarios such as supply-demand problems, or other relationships where multiple variables are interconnected.

Determinants

Determinants play a significant role in solving systems of equations. They help us determine whether a system has a unique solution, no solution, or infinitely many solutions. In this exercise, we need to find non-trivial solutions, which means solutions other than all zero values.
For the matrix derived from our system, the determinant is calculated to check if it equals zero:

• If the determinant is non-zero, the system has only the trivial solution (all variables equal zero).
• If the determinant is zero, there are non-trivial solutions.

Here, the determinant is initially expressed as \(\text{det} = 6 + \cos \theta - \cos^2 \theta\).
Simplifying the expression helps us understand the behavior and feasibility of the system's solutions. It involves mathematical operations that need careful handling to avoid errors.

Quadratic Equation

A quadratic equation is an algebraic equation of the second degree, usually in the form of \(ax^2 + bx + c = 0\). To solve these equations, we often use the quadratic formula:

• \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

This formula provides the solutions to the equation based on the coefficients \(a\), \(b\), and \(c\).
In the exercise, we derived a quadratic equation in terms of \(\cos \theta\):\[\cos^2 \theta - \cos \theta - 6 = 0\]
Applying the quadratic formula gives potential solutions for \(\cos \theta\), which were computed as \(\cos \theta = 3\) and \(\cos \theta = -2\).
However, these values are outside the permissible range for \(\cos \theta\), which indicates there are certain restrictions on our original parameters.

Matrix Representation

Matrix representation is a powerful method for organizing and solving systems of equations. By converting equations into a matrix form, we can utilize linear algebra techniques to find solutions effectively. In this exercise, the system is represented as the matrix:

• \[\begin{bmatrix}\lambda & -1 & \cos \theta \3 & 1 & 2 \\cos \theta & 1 & 2\end{bmatrix}\begin{bmatrix}x \y \z\end{bmatrix}= \begin{bmatrix}0 \0 \0\end{bmatrix}\]
  

This form allows us to perform operations such as finding the determinant to explore solution conditions. Matrices facilitate a clear and structured approach, enabling us to leverage properties like linear transformations and matrix equivalence.
Understanding matrices provides a foundational tool in linear algebra, aiding in tasks ranging from theoretical proofs to computational simulations.