Determinants Matrices Question 46
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}=0,\forall x\in R $
$ \Rightarrow a+1=n+2 $ or $ a=n+1 $ (as it will make first and third row is identical)
BETA
