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Question #67779

Difficulty: 8/10 | 🟢 Free

The Question

Let $A = \frac{1}{6} \begin{pmatrix} 5 & -1 & \alpha \\ -1 & 5 & \alpha \\ \alpha & \alpha & 2 \end{pmatrix}$ be an idempotent matrix with $\alpha > 0$ , and let $k = \text{tr}(A)$ . Consider the column vector $B = \begin{pmatrix} 1 \\ 1 \\ -k \end{pmatrix}$ . Let $S_1$ be the number of solutions to the system of equations $A^{2024} X = B$ . Let $Y_0$ be the unique solution to the system of equations $(I - k A)^{2025} Y = B$ . Find the value of $S_1 + \|Y_0\|^2$ .

Options

A.
4
B.
6
C.
8
D.
10

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

Idempotent and Involutory Properties|Injective Mapping (One-One)|Extraneous Solution Rejection

Why This Question Is Tricky

Examiner Levers

Recursive Logic

The answer to part A is required to solve part B (Step-wise failure test).

Difficulty Boost : +5

Common Mistakes

This question tests the properties of idempotent and involutory matrices while embedding an 'Extraneous Solution' trap.

Students often assume matrix equations like $AX = B$ are consistent or try to invert $A$ .

By defining $A$ as an idempotent matrix ( $A^2=A$ ), the mapping $T(X)=AX$ is non-injective and non-surjective.

The vector $B$ is deliberately placed in the kernel (null space) of $A$ .

If a student fails to verify if $B$ is in the range of $A$ , they might falsely assume solutions exist.

Multiplying by $A$ reveals the extraneous condition $B=O$ , proving 0 solutions.

Furthermore, it tests the involutory property $(I-2A)^2=I$ to bypass massive matrix exponentiation.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Matrices - Class 12
Concept	Idempotent and Involutory Properties
Difficulty	8
Cognitive Load	7
Target Solve Time	170 sec
Access	Free