Let $B=\begin{bmatrix} 1 & 3 & \alpha \\ 1 & 2 & 3 \\ \alpha & \alpha & 4 \end{bmatrix}, \alpha > 2$ be the adjoint of a matrix $A$ and $|A|=2$. Then the value of $\begin{bmatrix} \alpha & -2\alpha & \alpha \end{bmatrix} B \begin{bmatrix} \alpha \\ -2\alpha \\ \alpha \end{bmatrix}$ is equal to:

If $A = \begin{bmatrix} 3 & 4 \\ 5 & 6 \end{bmatrix}$ and $B = \begin{bmatrix} x & 0 \\ 0 & y \end{bmatrix}$,where $x, y \in \mathbb{N}$,then:

Read the following mathematical statements carefully:
$I$. There can exist two triangles such that the sides of one triangle are all less than $1 \text{ cm}$ while the sides of the other triangle are all bigger than $10 \text{ m}$,but the area of the first triangle is larger than the area of the second triangle.
$II$. If $x, y, z$ are all different real numbers,then $\frac{1}{(x - y)^2} + \frac{1}{(y - z)^2} + \frac{1}{(z - x)^2} = \left( \frac{1}{x - y} + \frac{1}{y - z} + \frac{1}{z - x} \right)^2$.
$III$. $\log_3 x \cdot \log_4 x \cdot \log_5 x = (\log_3 x \cdot \log_4 x) + (\log_4 x \cdot \log_5 x) + (\log_5 x \cdot \log_3 x)$ is true for exactly one real value of $x$.
$IV$. $A$ matrix has $12$ elements. The number of possible orders it can have is $6$. Now indicate the correct alternative.

Let $\theta = \frac{\pi}{5}$ and $A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$. If $B = A + A^4$,then $\det(B)$

Let $\alpha$ and $\beta$ be real numbers. Consider a $3 \times 3$ matrix $A$ such that $A^2 = 3A + \alpha I$. If $A^4 = 21A + \beta I$,then:

If $A$ is a square matrix of $3 \times 3$ order,and $|A| = 2$,then $|(A-A^T)^6| + |(A^T-A)^7|$ is equal to (where $A^T$ denotes the transpose of matrix $A$).