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Question #40644

Difficulty: 8/10 | 🟢 Free

The Question

Let $A$ and $B$ be two $11 \times 11$ matrices with row and column indices ranging from $0$ to $10$ . The entries of $A$ and $B$ are given by: $A_{i,j} = \frac{10!}{i! j! (10 - i - j)!}$ $B_{i,j} = (-1)^{i-j} \binom{i}{j} 2^j$ (Note: $\frac{1}{k!} = 0$ and $\binom{n}{k} = 0$ for $k < 0$ or $k > n$ ). Let $C = AB$ and $P = \prod_{k=0}^{10} \binom{10}{k}$ . What is the value of $\det(C)$ ?

Options

A.
$2^{55} P$
B.
$-2^{55} P$
C.
$2^{11} P$
D.
$-2^{11} P$

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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

Multinomial Theorem|Inversion and Determinant Product|Row/Column Exchange Parity

Why This Question Is Tricky

Common Mistakes

This question tests the properties of determinants for structured combinatorial matrices, specifically focusing on the parity trap of anti-triangular matrices.

Students often correctly identify the anti-diagonal elements and compute their product $P$ , but fail to account for the sign change induced by the column/row exchanges required to transform the matrix into a standard triangular form.

For an $11 \times 11$ matrix, reversing the columns requires $\lfloor 11/2 \rfloor = 5$ swaps, introducing a factor of $(-1)^5 = -1$ .

Furthermore, it tests the determinant of a lower triangular matrix $B$ and the multiplicative property $\det(AB) = \det(A)\det(B)$ , ensuring a comprehensive evaluation of matrix operations and combinatorial identities.

Quick Facts About This Question

Question Information

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