If $ω$ is a complex cube root of unity, and $A =\begin{bmatrix}ω&0\\0&ω\end{bmatrix}$, then $A^{100}$ is equal to
Answer & explanation
Correct answer: option 1
We have,
$A =ω\begin{bmatrix}1&0\\0&1\end{bmatrix}=ω\,I_2$
$∴A^{100}=ω^{100}(I_2)^{100}=ω^{100}I_2=ω\,I_2=A$.