If F(x)= \(\begin{bmatrix}\cos x&\sin x&0\\\sin x&cos x&0\\0&0&1\end{bmatrix}\)and F(y)=\(\begin{bmatrix}\cos y&-\sin y&0\\\sin y&cos y&0\\0&0&1\end{bmatrix}\),show that F(x)+F(y)=F(x+y)
F(x)=\(\begin{bmatrix}\cos x&-\sin x&0\\\sin x&cos x&0\\0&0&1\end{bmatrix}\),F(y)=\(\begin{bmatrix}\cos y&-\sin y&0\\\sin y&cos y&0\\0&0&1\end{bmatrix}\)
F (x+y)=\(\begin{bmatrix}\cos (x+y)&-\sin (x+y)&0\\\sin (x+y)&cos (x+y)&0\\0&0&1\end{bmatrix}\)
F(x)F(y)=\(\begin{bmatrix}\cos x&-\sin x&0\\\sin x&cos x&0\\0&0&1\end{bmatrix}\)\(\begin{bmatrix}\cos y&-\sin y&0\\\sin y&cos y&0\\0&0&1\end{bmatrix}\)
=\(\begin{bmatrix}\cos x\cos y-\sin x\sin y+0&-\cos x\sin y-\sin x\cos y+0&0\\\sin x\cos y+\cos x\sin y&-\sin x\sin y+\cos x\cos y+0&0\\0&0&0\end{bmatrix}\)
=\(\begin{bmatrix}\cos (x+y)&-\sin(x+y)&0\\\sin (x+y)&\cos(x+y)&0\\0&0&1\end{bmatrix}\)
=F(x+y)
\(\therefore\) F(x)+F(y)=F(x+y)
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: