MCQ
If $\omega$ is a complex cube root of unity and $A=\left[\begin{array}{lll}\omega & 0 & 0 \\ 0 & \omega^2 & 0 \\ 0 & 0 & 1\end{array}\right]$, then $A^{-1}$ is equal to
  • A
    $\left[\begin{array}{llr}1 & 0 & 0 \\ 0 & \omega^2 & 0 \\ 0 & 0 & \omega\end{array}\right]$
  • B
    $\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right]$
  • $\left[\begin{array}{ccc}\omega^2 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & 1\end{array}\right]$
  • D
    $\left[\begin{array}{lll}\omega & 0 & 0 \\ 0 & \omega^2 & 0 \\ 0 & 0 & 1\end{array}\right]$

Answer

Correct option: C.
$\left[\begin{array}{ccc}\omega^2 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & 1\end{array}\right]$
(C) $A=\left[\begin{array}{ccc}\omega & 0 & 0 \\ 0 & \omega^2 & 0 \\ 0 & 0 & 1\end{array}\right]$
$|A|=\omega^3=1 \neq 0$
$\operatorname{adj} A =\left[\begin{array}{ccc}\omega^2 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & \omega^3\end{array}\right]^{ T }$
$=\left[\begin{array}{ccr}\omega^2 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & \omega^3\end{array}\right]$
$=\left[\begin{array}{ccc}\omega^2 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & 1\end{array}\right]$
$A^{-1}=\left[\begin{array}{ccc}\omega^2 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & 1\end{array}\right]$

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