\((i)\begin{bmatrix}a&b\\-b&a\end{bmatrix}+\begin{bmatrix}a&b\\b&a\end{bmatrix}\)\(=\begin{bmatrix}a+a& b+b\\ -b+b& a+a\end{bmatrix}\)\(=\begin{bmatrix}2a& 2b\\ 0& 2a\end{bmatrix}\)
\((ii)\begin{bmatrix}a^2+b^2& b^2+c^2\\ a^2+c^2& a^2+b^2\end{bmatrix}+\begin{bmatrix}2ab& 2bc \\-2ac& -2ab\end{bmatrix}\)
\(=\begin{bmatrix}a^2+b^2+2ab& b^2+c^2+2bc\\ a^2+c^2-2ac& a^2+b^2-2ab\end{bmatrix}\)
\(=\begin{bmatrix}(a+b)^2& (b+c)^2\\ (a-c)^2& (a-b)^2\end{bmatrix}\)
\((iii)\begin{bmatrix}-1&4&-6\\ 8&5&16\\ 2&8&5\end{bmatrix}+\begin{bmatrix}12&7&6 \\8&0&5\\ 3&2&4\end{bmatrix}\)
\(=\begin{bmatrix}-1+12& 4+7& -6+6\\ 8+8& 5+0& 16+5\\ 2+3& 8+2& 5+4\end{bmatrix}\)
\(=\begin{bmatrix}11& 11& 0\\ 16& 5& 21\\ 5& 10& 9\end{bmatrix}\)
\((iv)\begin{bmatrix}cos^2x& sin^2x\\ sin^2x& cos^2x\end{bmatrix}+\begin{bmatrix}sin^2x& cos^2x\\ cos^2x& sin2x\end{bmatrix}\)
\(=\begin{bmatrix}cos^2x+sin^2x& sin^2x+cos^2x\\ sin^2x+cos^2x& cos^2x+sin^2x\end{bmatrix}\)
\(=\begin{bmatrix}1& 1\\ 1& 1\end{bmatrix}\)\((\because sin^2x+cos^2x=1)\)
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: