$8 I$
$5 I$
Given \(A^2 = I\), the identity matrix, we analyze \((I + A)^4\):
\[(I + A)^2 = I^2 + 2IA + A^2 = I + 2A + I = 2I + 2A\]
\[(I + A)^4 = ((I + A)^2)^2 = (2I + 2A)^2\]
Expanding \((2I + 2A)^2\):
\[(2I + 2A)^2 = 4I^2 + 8IA + 4A^2\]
Since \(A^2 = I\), substitute \(4A^2 = 4I\):
\[(2I + 2A)^2 = 4I + 8A + 4I = 8I + 8A\]
Now compute \((I + A)^4 - 8A\):
\[(I + A)^4 - 8A = (8I + 8A) - 8A = 8I\]
Thus, the result is: \[8I\]
Let $A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{bmatrix}$. If for some $\theta \in (0, \pi)$, $A^2 = A^T$, then the sum of the diagonal elements of the matrix $(A + I)^3 + (A - I)^3 - 6A$ is equal to
Let $ A $ be a $ 3 \times 3 $ matrix such that $ | \text{adj} (\text{adj} A) | = 81.
$ If $ S = \left\{ n \in \mathbb{Z}: \left| \text{adj} (\text{adj} A) \right|^{\frac{(n - 1)^2}{2}} = |A|^{(3n^2 - 5n - 4)} \right\}, $ then the value of $ \sum_{n \in S} |A| (n^2 + n) $ is: