If \(A = \begin{bmatrix} \frac{2}{3} & 1 & \frac 53 \\[0.3em] \frac{1}{3} & \frac 23 & \frac{4}{3} \\[0.3em] \frac 73 & 2 & \frac{2}{3} \end{bmatrix}\) and \(B = \begin{bmatrix} \frac{2}{5} & \frac 35 & 1 \\[0.3em] \frac{1}{5} & \frac 25 & \frac{4}{5} \\[0.3em] \frac 75 & \frac 65 & \frac{2}{5} \end{bmatrix}\) then compute 3A-5B.
3A - 5B = 3\(\begin{bmatrix} \frac{2}{3} & 1 & \frac 53 \\[0.3em] \frac{1}{3} & \frac 23 & \frac{4}{3} \\[0.3em] \frac 73 & 2 & \frac{2}{3} \end{bmatrix}\)- 5 \(\begin{bmatrix} \frac{2}{5} & \frac 35 & 1 \\[0.3em] \frac{1}{5} & \frac 25 & \frac{4}{5} \\[0.3em] \frac 75 & \frac 65 & \frac{2}{5} \end{bmatrix}\)
= \(\begin{bmatrix} 2 & 3 & 5 \\[0.3em] 1 & 2 & 4 \\[0.3em] 7 & 6 & 2 \end{bmatrix}\)- \(\begin{bmatrix} 2 & 3 & 5 \\[0.3em] 1 & 2 & 4 \\[0.3em] 7 & 6 & 2 \end{bmatrix}\)
= \(\begin{bmatrix} 0 & 0 & 0 \\[0.3em] 0 & 0 & 0 \\[0.3em] 0 & 0 & 0 \end{bmatrix}\)
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: