1. Let $A=\left(\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right)$. Determine the set of matrices:

$$
M=\left\{X \in M_{2}(\mathbb{Z}) \mid A X=X^{2} A\right\}
$$

Prof.Burtea Marius, Alexandria
1. By passing to determinants, from the given relation, we obtain that $\operatorname{det}(X)=\operatorname{det}\left(X^{2}\right)$, from which $\operatorname{det}(X) \in\{0,1\}$. $2 \mathrm{p}$

Let $X=\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)$ and $u=\operatorname{Tr}(X)=a+d$.

From the Cayley-Hamilton relation, we obtain that $X^{2}-u X+\operatorname{det}(X) I_{2}=O_{2}$.

We distinguish the cases:

1) $\operatorname{det}(X)=0$ so $X^{2}=u X$.

We have that $A X=u X A$ or $\left(\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right)\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)=u\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)\left(\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right)$ and we obtain that $\left(\begin{array}{ll}c & d \\ a & b\end{array}\right)=u\left(\begin{array}{ll}b & a \\ d & c\end{array}\right)$. The equalities result in $\left\{\begin{array}{l}c=b u \\ d=a u \\ a=d u \\ b=c u\end{array}\right.$. Adding equations 2 and 3, we obtain that $u=u^{2}$ with solutions $u \in\{0,1\}$.

- For $u=0$ we obtain that $a=b=c=d=0$ and $X=O_{2}$.
- For $u=1$ we obtain that $\quad a=d=\frac{1}{2} \notin \mathbb{Z}$

2) $\operatorname{det}(X)=1$ so $X^{2}=u X-I_{2}=\left(\begin{array}{cc}u a-1 & u b \\ u c & u d-1\end{array}\right)$.

We have that $\left(\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right)\left(\begin{array}{ll}a & b \\ c & d\end{array}\right)=\left(\begin{array}{cc}u a-1 & u b \\ u c & u d-1\end{array}\right)\left(\begin{array}{ll}0 & 1 \\ 1 & 0\end{array}\right)$. The equalities result in $\left\{\begin{array}{l}c=b u \\ d=a u-1 \\ a=d u-1 \\ b=c u\end{array}\right.$.

Adding equations 2 and 3, we obtain that $u^{2}-u-2=0$ with solutions $u \in\{-1,2\}$

- For $u=2$ we obtain that $b=c=0, a=d=1$ and $X=I_{2}$.
- For $u=-1$ we obtain that $c=-b, d=-a-1$ and $X=\left(\begin{array}{cc}a & b \\ -b & -a-1\end{array}\right)$. From the condition $\operatorname{det} X=1$ it follows that $b^{2}=a^{2}+a+1$. By multiplying by 4, we obtain that

$$
\begin{aligned}
& 4 b^{2}-(2 a+1)^{2}=3 \text { or }(2 b-2 a-1)(2 b+2 a+1)=3 \text { with solutions } \\
& (a, b) \in\{(0,1),(-1,1),(-1,-1),(0,-1)\} \text {. We obtain the matrices in } M_{2}(\mathbb{Z}) \\
& x \in\left\{\left(\begin{array}{cc}
0 & 1 \\
-1 & -1
\end{array}\right),\left(\begin{array}{cc}
-1 & 1 \\
-1 & 0
\end{array}\right),\left(\begin{array}{cc}
-1 & -1 \\
1 & 0
\end{array}\right),\left(\begin{array}{cc}
0 & -1 \\
1 & -1
\end{array}\right)\right\} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots
\end{aligned}
$$

Therefore, $M=\left\{O_{2}, I_{2},\left(\begin{array}{cc}0 & 1 \\ -1 & -1\end{array}\right),\left(\begin{array}{cc}-1 & 1 \\ -1 & 0\end{array}\right),\left(\begin{array}{cc}-1 & -1 \\ 1 & 0\end{array}\right),\left(\begin{array}{ll}0 & -1 \\ 1 & -1\end{array}\right)\right\}$