Determinants Matrices Question 32
Question: If $ f(x)=a+bx+cx^{2} $ and $ \alpha ,\beta ,\lambda $ are roots of the equation $ x^{3}=1, $ then $ \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{vmatrix} $ is equal to
Options:
A) $ f(\alpha )+f(\beta )+f(\lambda ) $
B) $ f(\alpha )f(\beta )+f(\beta )f(\lambda )+f(\gamma )+f(\alpha ) $
C) $ f(\alpha )f(\beta )f(\gamma ) $
D) $ -f(\alpha )f(\beta )f(\gamma ) $
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] $ \begin{vmatrix} a & b & c \\ b & c & a \\ c & a & b \\ \end{vmatrix}=-(a^{3}+b^{3}+c^{3}-3abc) $
$ =-(a+b+c)(a+b{{\omega }^{2}}+c\omega )(a+b\omega +c{{\omega }^{2}}) $
$ =-f(\alpha )f(\beta )f(\lambda )[\because \alpha =1,\beta =\omega ,={{\omega }^{2}}] $
BETA
