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Question #40625

Difficulty: 8/10 | 🟢 Free

The Question

Let $k$ be the term independent of $x$ in the binomial expansion of $\left( \frac{x}{\sqrt{2}} + \frac{\alpha}{x^2} \right)^9$ , where $\alpha \in \mathbb{R}$ . Let $A = \begin{pmatrix} k-2 & 2 & 2 \\ 2 & k-2 & 2 \\ 2 & 2 & k-2 \end{pmatrix}$ Let $S$ be the set of all real values of $\alpha$ for which $\text{adj}(A)$ is the null matrix. Then the value of $\sum_{\alpha \in S} 21\alpha^3$ is equal to:

Options

A.
8
B.
4
C.
-4
D.
12

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Bridge

Practice problems that bridge the gap between foundational theory and advanced practical application.

Concepts Behind This Question

Term Independent of x|Adjoint-Determinant Relationship|Singular vs Non-Singular Matrices

Why This Question Is Tricky

Common Mistakes

This question tests the intersection of the Binomial Theorem and Matrix properties, specifically targeting a common misconception regarding singular matrices.

The primary trap lies in the condition $\text{adj}(A) = O$ .

Many students incorrectly equate this to $|A| = 0$ .

While $\text{adj}(A) = O \implies |A| = 0$ , the converse is false.

For a $3 \times 3$ matrix, $|A| = 0$ only implies $\text{rank}(A) \le 2$ , whereas $\text{adj}(A) = O$ strictly requires $\text{rank}(A) \le 1$ .

Solving $|A| = 0$ yields an extraneous root ( $k=-2$ ) which leads to an incorrect sum of 4.

Recognizing the rank condition isolates the true root ( $k=4$ ), rewarding deep conceptual clarity.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Binomial Theorem - Class 11
Concept	Term Independent of x
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free