Problem 1. Consider $A \in \mathcal{M}_{2020}(\mathbb{C})$ such that

$$
\begin{aligned}
& A+A^{\times}=I_{2020} \\
& A \cdot A^{\times}=I_{2020}
\end{aligned}
$$

where $A^{\times}$is the adjugate matrix of $A$, i.e., the matrix whose elements are $a_{i j}=(-1)^{i+j} d_{j i}$, where $d_{j i}$ is the determinant obtained from $A$, eliminating the line $j$ and the column $i$.

Find the maximum number of matrices verifying (1) such that any two of them are not similar.
Solution. It is known that

$$
A \cdot A^{\times}=\operatorname{det} A \cdot I_{2020}
$$

hence, from the second relation we get $\operatorname{det} A=1$, so $A$ is invertible. Next, multiplying in the first relation by $A$, we get

$$
A^{2}-A+I_{2020}=\mathcal{O}_{2020}
$$

It follows that the minimal polynomial of $A$ divides

$$
X^{2}-X+1=(X-\omega)(X-\bar{\omega})
$$

where

$$
\omega=\frac{1}{2}+i \frac{\sqrt{3}}{2}=\cos \frac{\pi}{3}+i \sin \frac{\pi}{3}
$$

Because the factors of the minimal polynomial of $A$ are of degree 1 , it follows that $A$ is diagonalizable, so $A$ is similar to a matrix of the form

$$
A_{k}=\left(\begin{array}{cc}
\omega I_{k} & \mathcal{O}_{k, 2020-k} \\
\mathcal{O}_{2020-k, k} & \bar{\omega} I_{n-k}
\end{array}\right), \quad k \in\{0,1, \ldots, 2020\}
$$

But $\operatorname{det} A=1$, so we must have

$$
\begin{aligned}
\omega^{k} \bar{\omega}^{2020-k} & =1 \Leftrightarrow \omega^{2 k-2020}=1 \Leftrightarrow \cos \frac{(2 k-2020) \pi}{3}+i \sin \frac{(2 k-2020) \pi}{3}=1 \\
& \Leftrightarrow \cos \frac{(2 k+2) \pi}{3}+i \sin \frac{(2 k+2) \pi}{3}=1 \\
& \Leftrightarrow k=3 n+2 \in\{0, \ldots, 2020\} \Leftrightarrow k \in\{2,5,8, \ldots, 2018\}
\end{aligned}
$$

Two matrices that verify the given relations are not similar if and only if the numbers $k_{1}, k_{2}$ corresponding to those matrices are different, so the required maximum number of matrices is 673 .