3. Let $A, B \in M_{2}(\mathbb{Q})$ such that $A B = B A$, $\operatorname{det}(A) = -3$, and $\operatorname{det}(A + \sqrt{3} B) = 0$. Calculate $\operatorname{det}\left(A^{2} + B^{2} - A B\right)$.

Local Olympiad, Constanța 2014
## Subject 3.

| Details of Solution | Associated Scoring |
| :---: | :---: |
| ![](https://cdn.mathpix.com/cropped/2024_06_07_6bf3897e7ddbf36cc1b7g-3.jpg?height=74&width=1567&top_left_y=485&top_left_x=190) | $1 \mathrm{p}$ |
| ![](https://cdn.mathpix.com/cropped/2024_06_07_6bf3897e7ddbf36cc1b7g-3.jpg?height=74&width=1567&top_left_y=558&top_left_x=190) | $1 \mathrm{p}$ |
| $P(\sqrt{3})=3 \operatorname{det}(B)+\alpha \sqrt{3}-3=0(1) \Rightarrow \alpha \sqrt{3}=3-3 \operatorname{det}(B)$ and since $\alpha, \operatorname{det}(B) \in \mathbb{Q}$ it follows that $\alpha=0$......................................................................................................... | $1 \mathrm{p}$ |
| ![](https://cdn.mathpix.com/cropped/2024_06_07_6bf3897e7ddbf36cc1b7g-3.jpg?height=92&width=1567&top_left_y=767&top_left_x=190) | $1 \mathrm{p}$ |
| Let $\varepsilon=-\frac{1}{2}+i \frac{\sqrt{3}}{2} \Rightarrow \varepsilon^{3}=1 \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots$ | $1 \mathrm{p}$ |
| ![](https://cdn.mathpix.com/cropped/2024_06_07_6bf3897e7ddbf36cc1b7g-3.jpg?height=74&width=1567&top_left_y=995&top_left_x=190) | $1 \mathrm{p}$ |
| $\operatorname{det}\left(A^{2}+B^{2}-A B\right)=P(\varepsilon) \cdot P\left(\varepsilon^{2}\right)=\left(\varepsilon^{2}-3\right) \cdot(\varepsilon-3)=13$ | $1 \mathrm{p}$ |