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Question #41815

Difficulty: 8/10 | 🟢 Free

The Question

Let $B$ be a $3 \times 3$ matrix such that $B^2 = I$ , $B \neq I$ , and $B \neq -I$ . Let $A = \frac{1}{2}(I - B)$ . The probability of success of an event is given by $p = \det(A)$ . A trial is repeated $216$ times independently. Let $X$ be the number of successes, and let $\lambda = P(X \ge 1)$ . If $P(x) = \det(A - x I)$ and $Q(x) = \det(B - x I)$ , find the value of $P(\lambda) + (Q(\lambda))^2$ .

Options

A.
0
B.
1
C.
2
D.
4

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Monster

Ultra-high difficulty questions designed to test your limit, stamina, and deep conceptual understanding.

Concepts Behind This Question

Binomial Distribution Tail-Ends|Characteristic Equation|Idempotent and Involutory Properties

Why This Question Is Tricky

Common Mistakes

This question creates a 'Lethal Collision' between matrix transformations and probability.

The primary trap lies in the involutory-to-idempotent transformation $A = \frac{1}{2}(I - B)$ .

Students often attempt to calculate $\det(A)$ algebraically or assume a non-zero probability, leading them into a computationally impossible binomial tail-end summation for $n=216$ .

The conceptual bypass requires recognizing that since $B \neq -I$ , $A$ must have at least one eigenvalue of 0, making $p = \det(A) = 0$ .

Consequently, the binomial distribution collapses, yielding $\lambda = 0$ .

The final trap tests the property $\det(B)^2 = \det(B^2) = \det(I) = 1$ , rewarding spectral analysis over brute force.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Probability - Class 12
Concept	Binomial Distribution Tail-Ends
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free