Problem 102 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

In the following questions an Assertion (A) is given followed by a Reason. (R). Mark your responses from the following options: (A) Assertion(A) is True and Reason(R) is True; Reason(R) is a correct explanation for Assertion(A) (B) Assertion(A) is True, Reason(R) is True; Reason(R) is not a correct explanation for Assertion(A) (C) Assertion(A) is True, Reason( \(\mathrm{R}\) ) is False (D) Assertion(A) is False, Reason(R) is True Assertion: Let \(\lambda\) and \(\alpha\) be real. The set of all values of \(\lambda\) for which the system of linear equations \(\lambda x+(\sin \alpha) y+(\cos \alpha) z=0\) \(x+(\cos \alpha) y+(\sin \alpha) z=0\) \(-x+(\sin \alpha) y-(\cos \alpha) z=0\) has a non-trivial solution, is \([-\sqrt{2}, \sqrt{2}]\) Reason: The equations \(a_{1} x+b_{1} y+c_{1} z=0, a_{2} x\) \(+b_{2} y+c_{2} z=0, a_{3} x+n_{3} y+c_{3} z=0\) have a non-trivial solution if $$ \left|\begin{array}{lll} a_{1} & b_{1} & c_{1} \\ a_{2} & b_{2} & c_{2} \\ a_{3} & b_{3} & c_{3} \end{array}\right|=0 $$

Step-by-Step Solution

Answer

Option (C) Assertion is True, Reason is False.

Step 1: Determine the system of equations

The given system of linear equations is as follows: \( \lambda x + (\sin \alpha) y + (\cos \alpha) z = 0 \), \( x + (\cos \alpha) y + (\sin \alpha) z = 0 \), \( -x + (\sin \alpha) y - (\cos \alpha) z = 0 \). The task is to find values of \( \lambda \) for which there is a non-trivial solution.

Step 2: Evaluate condition for non-trivial solution

For a system of equations to have a non-trivial solution, the determinant of the coefficient matrix must be zero. This is given by \[ \left| \begin{array}{ccc} \lambda & \sin \alpha & \cos \alpha \ 1 & \cos \alpha & \sin \alpha \ -1 & \sin \alpha & -\cos \alpha \end{array} \right| = 0 \].

Step 3: Calculate the determinant

Calculate the determinant using the matrix: \[ \left| \begin{array}{ccc} \lambda & \sin \alpha & \cos \alpha \ 1 & \cos \alpha & \sin \alpha \ -1 & \sin \alpha & -\cos \alpha \end{array} \right| \]. The expansion of this gives \( \lambda (-\cos^2 \alpha - \sin^2 \alpha) - (\sin \alpha \cdot 0 - \cos \alpha \cdot 0) + (\cos \alpha \cdot 0 - \sin \alpha \cdot 0) = 0 \). Since \( \cos^2 \alpha + \sin^2 \alpha = 1 \), the simplified determinant is \( -\lambda = 0 \). Therefore, \( \lambda = 0 \).

Step 4: Verify and conclude values of λ

The calculation indicates that \( \lambda = 0 \) is the only value satisfying the determinant condition derived. However, the assertion requires the set \([-\sqrt{2}, \sqrt{2}] \), which arises from misunderstanding. The conclusion that \( \lambda = 0 \) is the real solution conflicts with the given assertion.

Step 5: Evaluation of Assertion and Reason

The Assertion states a specific range \([-\sqrt{2}, \sqrt{2}]\) as the set of solutions, but the correct solution requires \( \lambda = 0 \). The Reason provided is a general condition for non-trivial solutions of systems of equations, which although valid, does not hold under the assertion conditions.

Key Concepts

Systems of Equations

A system of equations is a set of two or more equations that you need to solve together. Each equation has the same set of variables. In our case, the given problem involves three equations:

• \( \lambda x + (\sin \alpha) y + (\cos \alpha) z = 0 \)
• \( x + (\cos \alpha) y + (\sin \alpha) z = 0 \)
• \( -x + (\sin \alpha) y - (\cos \alpha) z = 0 \)

Systems of equations are often solved using various methods such as substitution, elimination, and matrix approaches.
In this scenario, solving the system means finding values for \(x\), \(y\), and \(z\) that satisfy all three equations simultaneously.

Determinants

Determinants are special numbers calculated from the elements of a square matrix. They are vital in determining the properties of matrices, such as whether a matrix is invertible. In the context of solving systems of equations, determinants help us check the solvability of the system.
A non-zero determinant means a unique solution; a zero determinant indicates either no solution or infinitely many solutions. For our system to have a non-trivial solution (a solution where not all of \(x\), \(y\), and \(z\) are zero), the determinant of the coefficient matrix must be zero: \[\left| \begin{array}{ccc}\lambda & \sin \alpha & \cos \alpha ewline1 & \cos \alpha & \sin \alpha ewline-1 & \sin \alpha & -\cos \alpha ewline\end{array} \right| = 0\]
This determinant checks the dependency of the equations, which is essential for the existence of non-trivial solutions.

Non-trivial Solutions

In linear algebra, a non-trivial solution of a homogeneous system of equations is a solution other than the trivial one (where all variables are zero).
A non-trivial solution implies that the equations are dependent. In other words, they essentially describe the same geometric object or line, leading to solutions that are not unique or that have a parametric form.
For our study, the condition for non-trivial solutions is that the determinant of the system's coefficient matrix equals zero. This indicates the presence of at least one free variable, and thus, non-zero solutions for the variables \(x\), \(y\), and \(z\). Non-trivially is powerful as it often reflects deeper symmetries or functionalities within mathematical systems.

Matrix Theory

Matrix theory underpins much of linear algebra, offering a structured way to handle systems of equations and transformations. A matrix organizes coefficients of equations into rows and columns, enabling systematic operations and simplifications.
The matrix in our problem describes the coefficients of our system of linear equations. By analyzing this matrix through operations like determinants and inversions, we gain insights into the system's behavior such as solvability and solution types.
Understanding matrix theory is crucial for interpreting results correctly, such as identifying the actual solution set (e.g., discovering \( \lambda = 0 \) versus a broader range of values \([-\sqrt{2}, \sqrt{2}]\)). It empowers students to manipulate complex algebraic concepts concisely and consistently.