Discussion Forum



1)

 if $\omega$  is a complex cube root of unity and 

$A=\begin{bmatrix}\omega & 0&0 \\0& \omega^{2} &0 \\ 0 & 0 &1\end{bmatrix}$  then A-1=


A) $\begin{bmatrix}\omega^{2} &0&0 \\0 & \omega^{} &0 \\ 0 & 0 &1\end{bmatrix}$

B) $\begin{bmatrix}1& 0&0 \\0& 1 &0 \\ 0 & 0 &1\end{bmatrix}$

C) $\begin{bmatrix}1 & 0&0 \\0& \omega^{2} &0 \\ 0 & 0 &\omega^{}\end{bmatrix}$

D) $\begin{bmatrix}0 & 0&\omega \\0& \omega^{2} &0 \\ 1 & 0 &0\end{bmatrix}$

Answer:

Option A

Explanation:

 Given , $\omega$ is a complex cube root of unity

 $\therefore$  $\omega^{3}$ =1

Given matrix A can be written  as A =lA

  $A=\begin{bmatrix}\omega & 0&0 \\0& \omega^{2} &0 \\ 0 & 0 &1\end{bmatrix}= \begin{bmatrix}1 & 0&0 \\0& 1&0 \\ 0 & 0 &1\end{bmatrix}A$

 On apply R1→$\omega ^{2}$ R1,   R2→$\omega ^{}$ R2 , we get 

 $\Rightarrow$ $\begin{bmatrix}\omega^{3} & 0&0 \\0& \omega^{3} &0 \\ 0 & 0 &1\end{bmatrix}= \begin{bmatrix}\omega^{2} & 0&0 \\0& \omega&0 \\ 0 & 0 &1\end{bmatrix}A$

$\Rightarrow$   $\begin{bmatrix}1 & 0&0 \\0& 1 &0 \\ 0 & 0 &1\end{bmatrix}= \begin{bmatrix}\omega^{2} & 0&0 \\0& \omega&0 \\ 0 & 0 &1\end{bmatrix}A$

    $A^{-1}= \begin{bmatrix}\omega^{2} & 0&0 \\0& \omega&0 \\ 0 & 0 &1\end{bmatrix}$