If $\omega$ is a complex cube root of unity and $A=\left[\begin{array}{cc}\omega & 0 \ 0 & \omega\end{array} ight]$, then $A^{-1}=$

Question

If $\omega$ is a complex cube root of unity and $A=\left[\begin{array}{cc}\omega & 0 \ 0 & \omega\end{array}ight]$, then $A^{-1}=$

Aicharya · Accepted Answer

Given $A=\left[\begin{array}{ll}\omega & 0 \ 0 & \omega\end{array}ight] \Rightarrow|A|=\left(\omega^{2}-0ight)=\omega^{2}$ and adj $A=\left[\begin{array}{cc}\omega & 0 \ 0 & \omega\end{array}ight]$ $	herefore A^{-1}=\frac{1}{\omega^{2}}\left[\begin{array}{ll}\omega & 0 \ 0 & \omega\end{array}ight]=\left[\begin{array}{cc}\frac{1}{\omega} & 0 \ 0 & \frac{1}{\omega}\end{array}ight]=\left[\begin{array}{cc}\omega^{2} & 0 \ 0 & \omega^{2}\end{array}ight] \quad \ldots\left[\because \omega^{3}=1ight]$ $	herefore A^{2}=\left[\begin{array}{ll}\omega & 0 \ 0 & \omega\end{array}ight]\left[\begin{array}{ll}\omega & 0 \ 0 & \omega\end{array}ight]=\left[\begin{array}{cc}\omega^{2} & 0 \ 0 & \omega^{2}\end{array}ight]$

If $\omega$ is a complex cube root of unity and $A=\left[\begin{array}{cc}\omega & 0 \\ 0 &…

Solution