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:

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Complex Numbers

Let

α

be a root of the equation

x^{2} + x + 1 = 0

and the matrix

A = \frac{1}{3} 111 1 α α^{2} 1 α^{2} α^{4}

, then the matrix

A^{31}

is equal to:

Answer: (A)

Since

α

satisfies

x^{2} + x + 1 = 0

, it is a primitive cube root of unity, so

α^{3} = 1

and

α^{4} = α

, and

A = \frac{1}{3} 111 1 ω ω^{2} 1 ω^{2} ω

. Direct multiplication gives

A^{2} = 100001010

, a permutation whose square is the identity, so

A^{4} = I

. Therefore

A^{31} = (A^{4})^{7} A^{3} = A^{3}

JEE Main · 2020 · 07 Jan 2020 · Shift 1 · Single correct option