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Question #41845

Difficulty: 8/10 | 🟢 Free

The Question

Let $A = \begin{pmatrix} \sin \theta & \tan \theta & \sec \theta \\ 0 & \cos \theta & \csc \theta \\ 0 & 0 & 1 \end{pmatrix}$ for some $\theta \in \left(0, \frac{\pi}{2}\right)$ . Let $\lambda_1, \lambda_2, \lambda_3$ be the eigenvalues of $A$ . A discrete random variable $X$ takes values $\lambda_1, \lambda_2, \lambda_3$ with probability $\frac{1}{3}$ each. If the variance of $X$ is $\frac{3 - 2\sqrt{2}}{9}$ , then the value of $\text{Tr}(A A^T)$ is:

Options

A.
$7$
B.
$\frac{23}{3}$
C.
$\frac{31}{3}$
D.
$5$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Specialist

Targeted practice for niche, high-difficulty concepts that frequently appear in JEE Advanced exams.

Concepts Behind This Question

Variance of Discrete Variable|Matrix Trace Properties|Triangular/Diagonal Matrix Shortcut

Why This Question Is Tricky

Common Mistakes

The question tests the intersection of matrix properties and probability.

The primary trap lies in the distinction between $\text{Tr}(A^2)$ and $\text{Tr}(A A^T)$ .

For an upper triangular matrix, the eigenvalues are strictly the diagonal elements, meaning the sum of their squares (needed for $E[X^2]$ ) is simply the sum of the squares of the diagonal entries, which equals $\text{Tr}(A^2)$ .

Students frequently memorize that $\text{Tr}(A A^T)$ relates to squared terms and mistakenly use it to compute the variance, inadvertently including the 'ghost' off-diagonal elements ( $\tan \theta, \sec \theta$ , etc.).

By asking for $\text{Tr}(A A^T)$ at the end, the problem rewards students who correctly isolated the diagonal for the variance calculation and then correctly incorporated the entire matrix for the Frobenius norm squared.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 12
Concept	Variance of Discrete Variable
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free