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 real roots for the equation
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 }^{3}} $ , $ {{\omega }^{3}} $ , $ {{\omega }^{2}} $ , $ {{\omega }^{2}} $ .
BETA
