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JEE Main 20207 JanuaryMorning Shift
Exam
JEE Main
Year
2020
Date
7 January
Shift
Morning Shift
Subject
Mathematics
Chapter
Matrices

JEE Main 2020 · 7 January · Morning Shift · Mathematics · Matrices

Let α be a root of the equation x ^ 2 + x + 1 = 0 and the matrix A...

Let α\alpha be a root of the equation x2+x+1=0{x}^{2} + x + 1 = 0 and the matrix A=13A = \dfrac{1}{\sqrt{3}}\left[{1111αα21α2α4\begin{cases}1 & 1 & 1 \\ 1 & \alpha & {\alpha}^{2} \\ 1 & {\alpha}^{2} & {\alpha}^{4}\end{cases}\right],thenthematrixthen the matrix{A}^{31} is equal to

Options

  1. AA3{A}^{3}
  2. BI3{I}_{3}
  3. CA2{A}^{2}
  4. DAA

Correct answer

A. A3{A}^{3}

Step-by-step solution

Here, α=ω,ω2\alpha = \omega, {\omega}^{2} where ω\omega is complex cube root of unity.

{A}^{2} = \dfrac{1}{3}\left[\begin{cases}1 & 1 & 1 \\ 1 & \omega & {\omega}^{2} \\ 1 & {\omega}^{2} & \omega\end{cases}\right]\left[\begin{cases}1 & 1 & 1 \\ 1 & \omega & {\omega}^{2} \\ 1 & {\omega}^{2} & \omega\end{cases}\right] = \left[\begin{cases}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{cases}\right]

⇒ {A}^{4} = \left[\begin{cases}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{cases}\right]\left[\begin{cases}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{cases}\right] = \left[\begin{cases}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{cases}\right]

⇒A31=A28×A3=I×A3=A3⇒ {A}^{31} = {A}^{28} \times {A}^{3} = I \times {A}^{3} = {A}^{3}

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