JEE PYQ: Quadratic Equation Question 25
Question 25 - 2020 (07 Jan Shift 1)
Let $\alpha$ be a root of the equation $x^2 + x + 1 = 0$ and the matrix $A = \frac{1}{\sqrt{3}}\begin{bmatrix} 1 & 1 & 1 \ 1 & \alpha & \alpha^2 \ 1 & \alpha^2 & \alpha^4 \end{bmatrix}$, then the matrix $A^{31}$ is equal to:
(1) $A$
(2) $I_3$
(3) $A^2$
(4) $A^3$
Show Answer
Answer: (4)
Solution
Solution of $x^2 + x + 1 = 0$ is $\omega, \omega^2$
So, $\alpha = \omega$ and $\alpha^3 = \omega^3, \alpha = \omega$
$A^2 = \frac{1}{3}\begin{bmatrix} 1 & 1 & 1 \ 1 & \omega & \omega^2 \ 1 & \omega^2 & \omega \end{bmatrix}\begin{bmatrix} 1 & 1 & 1 \ 1 & \omega & \omega^2 \ 1 & \omega^2 & \omega \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 0 & 0 & 1 \ 0 & 1 & 0 \end{bmatrix}$
$\Rightarrow A^4 = I$
$\Rightarrow A^{30} = A^{28} \times A^3 = A^3$
BETA
