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 $A = \begin{bmatrix} \cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix}$ where $\theta = \frac{2 \pi}{19}$,then $A^{2017} = $

If $A = \begin{bmatrix} \cos^2 \alpha & \sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha & \sin^2 \alpha \end{bmatrix}$ and $B = \begin{bmatrix} \cos^2 \beta & \sin \beta \cos \beta \\ \sin \beta \cos \beta & \sin^2 \beta \end{bmatrix}$ are such that $AB$ is a null matrix,then which of the following must be an odd integral multiple of $\frac{\pi}{2}$?

Let $\alpha$ be a root of the equation $(a-c)x^2 + (b-a)x + (c-b) = 0$,where $a, b, c$ are distinct real numbers such that the matrix $\begin{bmatrix} \alpha^2 & \alpha & 1 \\ 1 & 1 & 1 \\ a & b & c \end{bmatrix}$ is singular. Then the value of $\frac{(a-c)^2}{(b-a)(c-b)} + \frac{(b-a)^2}{(a-c)(c-b)} + \frac{(c-b)^2}{(a-c)(b-a)}$ is:

The number of elements in the set $\{A=\begin{bmatrix} a & b \\ 0 & d \end{bmatrix} : a, b, d \in \{-1, 0, 1\} \text{ and } (I-A)^3 = I-A^3 \}$,where $I$ is the $2 \times 2$ identity matrix,is:

Let $M = \begin{bmatrix} 0 & 1 & a \\ 1 & 2 & 3 \\ 3 & b & 1 \end{bmatrix}$ and $\operatorname{adj} M = \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix}$ where $a$ and $b$ are real numbers. Which of the following options is/are correct?
$(1)$ $a+b=3$
$(2)$ $\operatorname{det}(\operatorname{adj} M^2) = 81$
$(3)$ $(\operatorname{adj} M)^{-1} + \operatorname{adj} M^{-1} = -M$
$(4)$ If $M \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$,then $\alpha - \beta + \gamma = 3$