3. 

(a) Let the matrix $A=\left(\begin{array}{ccc}1 & 1 & 1 \\ 1 & \omega & \omega^{2} \\ 1 & \omega^{2} & \omega\end{array}\right) \in \mathcal{M}_{3}(\mathbb{C})$, where $\omega$ is a third root of unity, different from 1. Determine $A^{2015}$.

(b) Show that the matrix equation $X^{2}=\left(\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right)$ has no solutions in the set $\mathcal{M}_{3}(\mathbb{R})$, but admits solutions in the set $\mathcal{M}_{3}(\mathbb{C})$.
Solution.

(a) Using the relations $\omega^{3}=1$ and $\omega^{2}+\omega+1=0$, we obtain

$$
A^{2}=3\left(\begin{array}{lll}
1 & 0 & 0 \\
0 & 0 & 1 \\
0 & 1 & 0
\end{array}\right), A^{3}=3\left(\begin{array}{ccc}
1 & 1 & 1 \\
1 & \omega^{2} & \omega \\
1 & \omega & \omega^{2}
\end{array}\right) \text { and } A^{4}=9 I_{3}
$$

Then

$$
A^{2015}=\left(A^{4}\right)^{503} \cdot A^{3}=\left(3^{2} I_{3}\right)^{503} \cdot A^{3}=3^{1007}\left(\begin{array}{ccc}
1 & 1 & 1 \\
1 & \omega^{2} & \omega \\
1 & \omega & \omega^{2}
\end{array}\right)
$$

(b) Let $J=\left(\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right)$. We have $\operatorname{det}(J)=-1 . \quad$ But $\operatorname{det}\left(X^{2}\right)=(\operatorname{det}(X))^{2} \geqslant 0, \forall X \in \mathcal{M}_{3}(\mathbb{R})$.

It follows that the equation $X^{2}=J$ has no solutions in the set $\mathcal{M}_{3}(\mathbb{R}) .(\mathbf{1 p})$

Let $B=\frac{1}{\sqrt{3}}\left(\begin{array}{ccc}\omega & \omega^{2} & 1 \\ \omega^{2} & \omega & 1 \\ 1 & 1 & 1\end{array}\right) \in \mathcal{M}_{3}(\mathbb{C})$. We have $B^{2}=(-B)^{2}=J$, so the equation $X^{2}=J$ has solutions in the set $\mathcal{M}_{3}(\mathbb{C})$. (3p)