Determinants Matrices Question 130
Question: For the equation , $ \begin{vmatrix} 1 & x & x^{2} \\ x^{2} & 1 & x \\ x & x^{2} & 1 \\ \end{vmatrix} $= 0
Options:
A) There are exactly two distinct roots
B) There is one pair of equation real roots
C) There are three pairs of equal roots
D) Modulus of each root is 2
Show Answer
Answer:
Correct Answer: C
Solution:
- [c] $ \Delta =(1+x+x^{2}) \begin{vmatrix} 1 & 1 & 1 \\ x^{2} & 1 & x \\ x & x^{3} & 1 \\ \end{vmatrix}=(1+x+x^{2}){{(x-1)}^{2}} $ Therefore, $ \Delta =0 $ has roots 1, 1, $ {{\omega }^{{}}} $ , $ {{\omega }^{{}}} $ , $ {{\omega }^{2}} $ , $ {{\omega }^{2}} $ .
BETA
