Simplify \(\cos\theta\) \(\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}\)+\(\sin\theta\) \(\begin{bmatrix}\sin\theta&-\cos\theta\\\cos\theta&\sin\theta\end{bmatrix}\)
\(\cos\theta\)\(\begin{bmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{bmatrix}\)+\(\sin\theta\)\(\begin{bmatrix}\sin\theta&-\cos\theta\\\cos\theta&\sin\theta\end{bmatrix}\)
=\(\begin{bmatrix}\cos^2\theta&\cos\theta\sin\theta\\-\sin\theta\cos\theta&\cos^2\theta\end{bmatrix}\)+\(\begin{bmatrix}\sin^2\theta&-\sin\theta\cos\theta\\\sin\theta\cos\theta&\sin^2\theta\end{bmatrix}\)
=\(\begin{bmatrix}\cos^2\theta+\sin^2\theta&\cos\theta\sin\theta-\sin\theta\cos\theta\\-\sin\theta\cos\theta+\sin\theta\cos\theta&\cos^2\theta+\sin^2\theta\end{bmatrix}\)
=\(\begin{bmatrix}1&0\\0&1\end{bmatrix}\)
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:
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is: