Determinants Matrices Question 20
Question: Let $ f(x)= \begin{vmatrix} n & n+1 & n+2 \\ ^{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. The $ f(x) $ is divisible by
Options:
A) $ n^{2}+n+1 $
B) $ (n+1)! $
C) $ (2n+1)! $
D) None of the above
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] $ \because f(x)= \begin{vmatrix} n & n+1 & n+2 \\ ^{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} $
$ = \begin{vmatrix} n & n+1 & n+2 \\ n! & (n+1)! & (n+2)! \\ 1 & 1 & 1 \\ \end{vmatrix} $
$ (\because {{}^{n}}P _{n}=n!,{{}^{n}}C _{n}=1) $
Applying $ C_2\to C_2-C_1 $ and $ C_3\to C_3-C_1 $ Then, $ f(x)= \begin{vmatrix} n & 1 & 2 \\ n! & n.n! & (n^{2}+3n+1)n! \\ 1 & 0 & 0 \\ \end{vmatrix} $
$ = \begin{vmatrix} 1 & 2 \\ n.n! & (n^{2}+3n+1)n! \\ \end{vmatrix}=n!(n^{2}+n+1) $
BETA
