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Question #67478

Difficulty: 8/10 | 🟢 Free

The Question

Let $A$ be an $n \times n$ matrix with rows $r_1, r_2, \dots, r_n$ and $\det(A) = \Delta \neq 0$ . A new matrix $B$ is formed such that its $i$ -th row $R_i$ is given by $R_i = x \cdot r_{n-i+1} + y \sum_{\substack{j=1 \\\\ j \neq n-i+1}}^{n} r_j$ for all $1 \le i \le n$ . Which of the following represents $\det(B)$ ?

Options

A.
$(-1)^{\lfloor \frac{n}{2} \rfloor} (x + (n-1)y) (x-y)^{n-1} \Delta$
B.
$(x + (n-1)y) (x-y)^{n-1} \Delta$
C.
$(-1)^{\lfloor \frac{n}{2} \rfloor} (x - y)^{n} \Delta$
D.
$(-1)^{n-1} (x + (n-1)y) (x-y)^{n-1} \Delta$

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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

Elementary Row Operations|Properties of Transpose Determinant|Selection vs Arrangement

Why This Question Is Tricky

Examiner Levers

Variable Manipulation

Algebraic solutions only (No numbers); tests pure symbolic logic.

Difficulty Boost : +4

Common Mistakes

This question tests the 'Selection vs Arrangement' trap in the context of matrix transformations.

Students often correctly identify the transformation matrix $T$ as having elements $x$ on the anti-diagonal and $y$ elsewhere, and may even correctly find the eigenvalues of the related matrix $(x-y)I + yU$ .

However, they frequently assume that since the same set of vectors is selected, the determinant remains unchanged, confusing the transformation $T = J M$ with a similarity transformation $P^{-1} M P$ .

They fail to account for the parity of the row permutation induced by the exchange matrix $J$ , missing the crucial $(-1)^{\lfloor n/2 \rfloor}$ factor that arises from reversing the arrangements of the rows.

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Matrices - Class 12
Concept	Elementary Row Operations
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free