Determinants Matrices Question 47
Question: If $ \begin{vmatrix} x^{n} & {x^{n+2}} & x^{2n} \\ 1 & x^{a} & a \\ {x^{n+5}} & {x^{a+6}} & {x^{2n+5}} \\ \end{vmatrix}=0\forall x\in R, $ where $ n\in N $ then value of ‘a’ is
Options:
A) $ n $
B) $ n-1 $
C) $ n+1 $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] Taking $ x^{5} $ common from last row, we get, $ x^{5} \begin{vmatrix} x^{n} & {x^{n+2}} & x^{2n} \\ 1 & x^{a} & a \\ x^{n} & {x^{a+1}} & x^{2n} \\ \end{vmatrix}=0x\in R, $
$ \Rightarrow a+1=n+2\Rightarrow a=n+1 $ (as it will make first and third row identical)
BETA
