Questão 7 - 1ª Fase - ITA 2026

$$\begin{gathered} \mathrm{A}=\left[\begin{array}{cc}\cos 30º& -\mathrm{sen}30º\\ \mathrm{sen}30º& \cos 30º\end{array}\right]\Rightarrow {\mathrm{A}}^2=\left[\begin{array}{cc}\cos 60º& -\mathrm{sen}60º\\ \mathrm{sen}60º& \cos 60º\end{array}\right]\Rightarrow \\ \Rightarrow {\mathrm{A}}^3=\left[\begin{array}{cc}\cos 90º& -\mathrm{sen}90º\\ \mathrm{sen}90º& \cos 90º\end{array}\right]\Rightarrow {\mathrm{A}}^6=\left[\begin{array}{cc}\cos 180º& -\mathrm{sen}180º\\ \mathrm{sen}180º& \cos 180º\end{array}\right]=-\mathrm{I}\Rightarrow \\ \Rightarrow {\mathrm{A}}^{12}=\left[\begin{array}{cc}\cos 360º& -\mathrm{sen}360º\\ \mathrm{sen}360º& \cos 360º\end{array}\right]=\mathrm{I}\Rightarrow {\mathrm{A}}^{24}=\mathrm{I} \Rightarrow {\mathrm{A}}^{30}=-\mathrm{I}\Rightarrow \\ \Rightarrow {\mathrm{A}}^{31}=-\mathrm{A} \end{gathered}$$

Logo:

${\mathrm{A}}^{31}· \mathrm{B}=-\mathrm{A} · \mathrm{B}=-\left[\begin{array}{cc}\cos 30º& -\mathrm{sen}30º\\ \mathrm{sen}30º& \cos 30º\end{array}\right]·\left[\begin{array}{c}\frac{\sqrt{2}}{2}\\ \frac{\sqrt{2}}{2}\end{array}\right]=\boxed{\left[\begin{array}{c}-\frac{\sqrt{2}}{4}\left(\sqrt{3}-1\right)\\ -\frac{\sqrt{2}}{4}\left(\sqrt{3}+1\right)\end{array}\right]}$