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Question #40628

Difficulty: 8/10 | 🟢 Free

The Question

Let $M$ be a $7 \times 7$ matrix whose $(i,j)$ -th entry is given by the coefficient of $t^5$ in the polynomial expansion of $(1+t)^i (1+xt)^j$ , for $1 \le i, j \le 7$ . Let $f(x) = \det(M^3)$ . Find the coefficient of $x^4$ in the polynomial $f(x)$ .

Options

A.
0
B.
1
C.
\binom{7}{4}
D.
\binom{12}{5}

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

Coefficient of x^r in Expansion|Determinant Product Rule|Nilpotent Matrix Logic

Why This Question Is Tricky

Common Mistakes

This question tests the student's ability to recognize rank-nullity conflicts hidden within matrices formed by polynomial coefficients.

The trap is set by asking for the coefficient of a specific power of $x$ in $\det(M^3)$ , which strongly suggests a non-zero polynomial determinant that requires manual expansion, Vandermonde identities, or complex row operations.

Students often fall for this and attempt intractable calculations.

The elegant solution requires invoking the Determinant Product Rule (Cauchy-Binet) to decompose the $n \times n$ matrix into a product of $n \times (k+1)$ and $(k+1) \times n$ rectangular matrices, where $k$ is the target power in the expansion.

Since the matrix dimension $n=7$ is strictly greater than $k+1=6$ , the matrix is structurally singular (rank $\le 6$ ), making the determinant identically zero regardless of $x$ .

Quick Facts About This Question

Question Information

Subject	Mathematics
Chapter	Binomial Theorem - Class 11
Concept	Coefficient of x^r in Expansion
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free