If $A = \begin{vmatrix} -1 & 2 & 4 \\ 3 & 1 & 0 \\ -2 & 4 & 2 \end{vmatrix}$ and $B = \begin{vmatrix} -2 & 4 & 2 \\ 6 & 2 & 0 \\ -2 & 4 & 8 \end{vmatrix}$,then $B$ is given by

If $P$ is a square matrix with $P^2=P$ and if $I$ is the unit matrix of the same order as of $P$,then $(P+I)^4=$

Let $A$ be a $3 \times 3$ matrix such that $A \begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 5 \\ 2 \\ 2 \end{bmatrix}$ and $A \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 3 \\ 1 \\ 1 \end{bmatrix}$. If $\det(A) = 1$ and the matrix $A$ satisfies $A \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 1 \end{bmatrix}$,then $\det(\text{adj}(A^2 + A))$ is equal to:

If the minimum and the maximum values of the function $f : \left[\frac{\pi}{4}, \frac{\pi}{2}\right] \rightarrow \mathbb{R}$,defined by $f(\theta) = \left|\begin{array}{ccc} -\sin^2 \theta & -1-\sin^2 \theta & 1 \\ -\cos^2 \theta & -1-\cos^2 \theta & 1 \\ 12 & 10 & -2 \end{array}\right|$ are $m$ and $M$ respectively,then the ordered pair $(m, M)$ is equal to:

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 $M$ denote the set of all real matrices of order $3 \times 3$ and let $S=\{-3,-2,-1,1,2\}$. Let $S_1=\{A=[a_{ij}] \in M: A=A^{T} \text{ and } a_{ij} \in S, \forall i, j\}$,$S_2=\{A=[a_{ij}] \in M: A=-A^{T} \text{ and } a_{ij} \in S, \forall i, j\}$,and $S_3=\{A=[a_{ij}] \in M: a_{11}+a_{22}+a_{33}=0 \text{ and } a_{ij} \in S, \forall i, j\}$. If $n(S_1 \cup S_2 \cup S_3)=125 \alpha$,then $\alpha$ equals.