We have \(\begin{bmatrix} 1 & 2 & 1 \end{bmatrix}\)\(\begin{bmatrix} 1 & 2 & 0\\ 2 & 0 & 1 \\1&0&2 \end{bmatrix}\)\(\begin{bmatrix} 0 \\2\\x\end{bmatrix}\)\(=O\)
⇒ \(\begin{bmatrix} 1+4+1 & 2+0+0 & 0+2+2 \end{bmatrix}\) \(\begin{bmatrix} 0 \\2\\x\end{bmatrix}\)\(=O\)
⇒ \(\begin{bmatrix} 6 & 2 & 4 \end{bmatrix}\) \(\begin{bmatrix} 0 \\2\\x\end{bmatrix}\)\(=O\)
⇒\(\begin{bmatrix} 6(0) +2(2)+4(x) \end{bmatrix}=O\)
⇒[4+4x]=[0]
\(\therefore 4+4x=0\)
⇒x=-1
Thus, the required value of x is −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: