$A=\left[\begin{array}{ll}\omega & 0 \\ 0 & \omega\end{array}\right]$
Then, $A^2=\left[\begin{array}{ll}\omega & 0 \\ 0 & \omega\end{array}\right]\left[\begin{array}{ll}\omega & 0 \\ 0 & \omega\end{array}\right]=\left[\begin{array}{cc}\omega^2 & 0 \\ 0 & \omega^2\end{array}\right]$ and $A^3=A^2 \cdot A$
$$
\begin{aligned}
A^3 & =\left[\begin{array}{cc}
\omega^2 & 0 \\
0 & \omega^2
\end{array}\right]\left[\begin{array}{cc}
\omega & 0 \\
0 & \omega
\end{array}\right] \\
& =\left[\begin{array}{cc}
\omega^3 & 0 \\
0 & \omega^3
\end{array}\right]
\end{aligned}
$$
$A^3=\left[\begin{array}{ll}1 & 0 \\ 0 & 1\end{array}\right]\left[\because \omega\right.$ is cube root of unity $\left.\Rightarrow \omega^3=1\right]$
Now, $A^{50}=A^{48} \cdot A^2$
$$
\begin{aligned}
& =\left(A^3\right)^{16} \cdot A^2=(I)^{16} \cdot A^2 \\
& =I A^2=A^2 \\
& =\left[\begin{array}{cc}
\omega^2 & 0 \\
0 & \omega^2
\end{array}\right]=\omega\left[\begin{array}{ll}
\omega & 0 \\
0 & \omega
\end{array}\right] \\
A^{50} & =\omega A
\end{aligned}
$$