Given that \(AA^\top = I\), we can substitute this property in the expression.
\[ \frac{1}{2} A \left[(A + A^\top)^2 + (A - A^\top)^2\right] \]
Expanding \((A + A^\top)^2\) and \((A - A^\top)^2\), we get:
\[ \frac{1}{2} A \left[A^2 + (A^\top)^2 + 2AA^\top + A^2 + (A^\top)^2 - 2AA^\top\right] \]
\[ = A \left[A^2 + (A^\top)^2\right] \]
\[ = A^3 + A^\top \]
So, the correct answer is: \(A^3 + A^\top\)
Let $ P_n = \alpha^n + \beta^n $, $ n \in \mathbb{N} $. If $ P_{10} = 123,\ P_9 = 76,\ P_8 = 47 $ and $ P_1 = 1 $, then the quadratic equation having roots $ \alpha $ and $ \frac{1}{\beta} $ is: