Problem 22 - A Complete Resource Book in Mathematics for JEE Main

Question

Let \(A\) be a \(2 \times 2\) matrix with real entries. Let \(I\) be the \(2 \times 2\) identity matrix. Denote by \(\operatorname{tr}(A)\), the sum of diagonal entries of \(A\). Assume that \(A^{2}=I\) Statement 1: If \(A \neq I\) and \(A \neq-I\), then \(\operatorname{det} A=-1\). Statement 2: If \(A \neq I\) and \(A \neq-I\), then \(\operatorname{tr}(A) \neq 0\) (A) Statement 1 is false, Statement 2 is true (B) Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 1 (C) Statement 1 is true, Statement 2 is true; Statement 2 is not a correct explanation for Statement 1 (D) Statement 1 is true, Statement 2 is false

Step-by-Step Solution

Answer

(D) Statement 1 is true, Statement 2 is false.

Step 1: Understanding Matrix Properties

The given matrix property is \(A^2 = I\), where \(I\) is the identity matrix. This means that \(A\) is an involutory matrix, satisfying \(A^2 = I\). Such matrices have eigenvalues \(\pm 1\) and can be either the identity matrix, the negative identity matrix, or have both eigenvalues as \(1\) and \(-1\).

Step 2: Analyze Statement 1

We need to analyze if \(\det(A) = -1\) when \(A eq I\) and \(A eq -I\). Since \(A^2 = I\), the eigenvalues of \(A\) are \(\pm 1\). The determinant, being the product of the eigenvalues, is \(\det(A) = \lambda_1 \cdot \lambda_2\). If \(\lambda_1 = 1\) and \(\lambda_2 = -1\), then \(\det(A) = -1\). Hence, Statement 1 is true if \(A eq I\) and \(A eq -I\).

Step 3: Analysis of eigenvalues for Statement 2

Assume \(A\) has eigenvalues \(\lambda_1 = 1\) and \(\lambda_2 = -1\) due to the constraints \(A eq I\) and \(A eq -I\). The trace is given by \(\operatorname{tr}(A) = \lambda_1 + \lambda_2 = 1 + (-1) = 0\). Statement 2 asserts \(\operatorname{tr}(A) eq 0\), therefore it is false as the trace in this case is 0.

Step 4: Conclusion about the Correct Answer

From analyzing Statements 1 and 2, we conclude that Statement 1 is true, and Statement 2 is false. Statement 2 is not a correct explanation for Statement 1 as it is false.

Key Concepts

Eigenvalues and Eigenvectors

In matrix theory, eigenvalues and eigenvectors are fundamental concepts that relate to linear transformations. When we have a square matrix like our matrix \( A \), an eigenvalue \( \lambda \) is a special number such that, for some non-zero vector \( \mathbf{v} \), the equation \( A\mathbf{v} = \lambda\mathbf{v} \) holds true. This vector \( \mathbf{v} \), which corresponds to the eigenvalue, is known as an eigenvector.
Eigenvalues capture significant information about a matrix’s properties. In our exercise, since the matrix \( A^2 = I \) where \( I \) is the identity matrix, it implies the eigenvalues of \( A \) are \( +1 \) and \( -1 \). This feature of \( A \) being involutory (meaning it squares to the identity matrix) highlights a key aspect – it flips some vectors while maintaining others.
Here’s how eigenvalues work in practice:

• They are found as roots of the characteristic polynomial of the matrix \( A \).
• In our specific scenario, an eigenvalue could be \( 1 \) which means the transformation keeps the eigenvector’s direction unchanged.
• If an eigenvalue is \( -1 \), the vector is flipped through the origin, reversing its direction.

These insights help us determine other properties, such as the determinant and trace, as the exercise explores.

Determinant of a Matrix

The determinant of a matrix is a scalar value that can indicate if a matrix is invertible and carries vital geometrical and algebraic information. For a \(2 \times 2\) matrix \( A \) having entries \( a, b, c, \) and \( d \), the determinant \( \det(A) \) is calculated as \( ad - bc \).
In our exercise context, it is established that \( \det(A) = -1 \) when \( A eq I \) and \( A eq -I \). This result arises because the eigenvalues of \( A \) are \( +1 \) and \( -1 \), and the determinant equals the product of these eigenvalues. Therefore,\(det(A) = 1 \times (-1) = -1 \).
This tells us two things:

• The matrix \( A \) is not singular; it is invertible because its determinant is not zero.
• Being determinant \( -1 \), matrix \( A \) enacts a transformation that involves reflection over a subspace.

The determinant is particularly useful in understanding how the transformation scales volumes in transforms and provides insights into matrix behavior under multiplication.

Trace of a Matrix

The trace of a matrix is the sum of its diagonal elements, and it reflects some of the intrinsic properties of the matrix's linear transformation. For our \(2 \times 2\) matrix \( A \), the trace is symbolized as \( \operatorname{tr}(A)\).
In terms of eigenvalues, the trace of a matrix is also the sum of its eigenvalues. With the given condition that \( A eq I \) and \( A eq -I \), and the eigenvalues being \( +1 \) and \( -1 \), we calculate \( \operatorname{tr}(A) = 1 + (-1) = 0 \).
This calculation shows:

• The trace thus reflects the net rotating effect of the matrix, being zero when the eigenvalues cancel each other out.
• Since the trace in this scenario is zero and the statement in the exercise was \( \operatorname{tr}(A) eq 0\), we confirm that part of the statement was false.

Understanding a matrix through its trace helps in identifying fixed points and the nature of transformation without needing a detailed decomposition.