For $A = \begin{bmatrix} 0 & 0 & -2 \\ 0 & -2 & 0 \\ -2 & 0 & 0 \end{bmatrix}$,which of the following is true?

Let $f(x)=\left|\begin{array}{ccc}1+\sin ^2 x & \cos ^2 x & \sin 2 x \\ \sin ^2 x & 1+\cos ^2 x & \sin 2 x \\ \sin ^2 x & \cos ^2 x & 1+\sin 2 x\end{array}\right|$,where $x \in\left[\frac{\pi}{6}, \frac{\pi}{3}\right]$. If $\alpha$ and $\beta$ are the maximum and minimum values of $f(x)$ respectively,then:

If $A$ is a $2 \times 2$ matrix such that $\operatorname{det} A = -21$ and $\operatorname{trace}(A^3) = 2024$, then the trace of $A$ is

$\det \left[ \begin{array}{ccc} \frac{a^2+b^2}{c} & c & c \\ a & \frac{b^2+c^2}{a} & a \\ b & b & \frac{c^2+a^2}{b} \end{array} \right] = $

If a $2^{nd}$ order determinant with elements $0$ or $1$ is chosen from the set of all such determinants,find the probability that the determinant chosen is non-zero.

$A$ and $B$ are two $3 \times 3$ non-singular matrices such that $\operatorname{adj} A = |A| B$. If $\operatorname{tr}(X)$ denotes the trace of a square matrix $X$ and $C = \begin{bmatrix} 4 & 4 & 7 \\ 3 & -2 & 5 \\ -2 & 3 & 6 \end{bmatrix}$, then $\sum_{k=1}^{\infty} \operatorname{tr}\left(\frac{1}{3^k}(A B)^k C\right)$ is equal to