Romanian Mathematical Olympiad

The number $c=\sqrt{b-a^2}$ is real. We have $$A^2-2aA+b{I}_2=(A-(a+ic)I_2)(A-(a-ic)I_2),$$ whence the characteristic polynomial of $A$ has as a root $a+ic$, $a-ic$, or both. Since the polynomial has real coefficients, both are roots, hence eigenvalues. Therefore the polynomial is $x^2-2ax+b$. It follows $A^2-2aA+b{I}_2=0$, and so $\mathrm{tr}(A)=2a$, $\det A=b$.

In conclusion, the matrices bearing the given property are $$\left(\begin{array}{cc}a+x& y\\ \frac{a^2-x^2-b}{y}& a-x\end{array}\right),$$ where $x\in \mathbb{R}$ and $y\in {\mathbb{R}}^{\ast }$.