If $A = \begin{bmatrix} 3 & -2 \\ 4 & 2 \end{bmatrix}$, then $A^2 - 5A + 14I = 0$. Which of the following is equivalent to $A^2$?

If both $\left( A - \frac{I}{2} \right)$ and $\left( A + \frac{I}{2} \right)$ are orthogonal matrices,then:

If $A$ is a non-singular matrix of order $3$ such that $(A-2I)(A-4I)=0$,then $\frac{1}{6}A + \frac{4}{3}A^{-1}$ is (where $I$ is a unit matrix of order $3$ and $0$ is a null matrix of order $3$).

Let $A$ and $B$ be two $3 \times 3$ non-singular matrices such that $\operatorname{det}(A^T B A) = 27$ and $\operatorname{det}(A B^{-1}) = 8$. Then $\operatorname{det}(B^T A^{-1} B) = $

Suppose $A$ is a $3 \times 3$ matrix consisting of integer entries that are chosen at random from the set $\{-1000, -999, \ldots, 999, 1000\}$. Let $P$ be the probability that either $A^2 = -I$ or $A$ is diagonal,where $I$ is the $3 \times 3$ identity matrix. Then,

Let $A = \begin{bmatrix} 0 & 1 & 2 \\ a & 0 & 3 \\ 1 & c & 0 \end{bmatrix}$,where $a, c \in \mathbb{R}$. If $A^3 = A$ and the positive value of $a$ belongs to the interval $(n-1, n]$,where $n \in \mathbb{N}$,then $n$ is equal to $..........$.