Let \( A = \begin{bmatrix} \alpha & -1 \\6 & \beta \end{bmatrix},\ \alpha > 0 \), such that \( \det(A) = 0 \) and \( \alpha + \beta = 1 \). If \( I \) denotes the \( 2 \times 2 \) identity matrix, then the matrix \( (1 + A)^5 \) is:
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:
For the reaction \( A \rightarrow \) products,
The reaction was started with 2.5 mol L\(^{-1}\) of A.
Consider the following electrochemical cell at standard condition. \[ \text{Au(s) | QH}_2\text{ | QH}_X(0.01 M) \, \text{| Ag(1M) | Ag(s) } \, E_{\text{cell}} = +0.4V \] The couple QH/Q represents quinhydrone electrode, the half cell reaction is given below: \[ \text{QH}_2 \rightarrow \text{Q} + 2e^- + 2H^+ \, E^\circ_{\text{QH}/\text{Q}} = +0.7V \]
Consider the following compound (X):
The most stable and least stable carbon radicals, respectively, produced by homolytic cleavage of corresponding C - H bond are: