\( 5^8 \)
\( 3^8 \)
The determinant of A is:
$|A| = \begin{vmatrix} K & 4 \\ 4 & K \end{vmatrix} = K \cdot K - 4 \cdot 4 = K^2 - 16.$
Using the property of determinants:
$|A|^3 = (|A|)^3 = 729.$
Take the cube root:
$|A| = \sqrt[3]{729} = 9.$
Thus:
$K^2 - 16 = 9 \implies K^2 = 25.$
Therefore:
$K = \pm 5.$
The value of $K^8$ is:
$K^8 = (K^2)^4 = 25^4.$
Calculate:
$25^4 = (25^2)^2 = 625^2 = 390625.$
Final Answer:
$5^8$
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: