Matrices & Determinants
Matrices and Determinants
star_batch_jee_advanced_2025
Grade None
Question:
Let $f(n) = \begin{vmatrix} ^n P_n & ^{n+1} P_{n+1} & ^{n+2} P_{n+2} \\ ^n C_n & ^{n+1} C_{n+1} & ^{n+2} C_{n+2} \end{vmatrix}$, where the symbols have their usual meanings. Then $f(n)$ is divisible by
$n^2 + n + 1$
$(n+1)!$
$n!$
None of these
Step-by-Step Solution
Key Concept: Expanding along a row with many zeros simplifies determinant computation significantly.
Compute the determinant of $f(x) = \begin{vmatrix}n & n+1 & n+2 \\ n! & n! n & (n+1)(n+1)! \\ 0 & 0 & 1\end{vmatrix}$ by expanding along the third row. This gives $(n+1)(n+1)! - n! = n![(n+1)^2 - n] = n![n^2 + 2n + 1 - n] = n![n^2 + n + 1]$.
Correct Answer: 1,3