For a square matrix $B$ of order $3$, if $B^T=B^{-1}$ and $|B|=1$, then $|B-I|=$

If $A = \begin{bmatrix} 0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0 \end{bmatrix}$ and $I$ is the identity matrix of order $2$,show that $I+A = (I-A) \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$.

Let $A$ be a $3 \times 3$ matrix of non-negative real elements such that $A\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} = 3\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}$. Then the maximum value of $\operatorname{det}(A)$ is:

Let $R = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}$ be a non-zero $3 \times 3$ matrix,where $x \sin \theta = y \sin \left(\theta + \frac{2 \pi}{3}\right) = z \sin \left(\theta + \frac{4 \pi}{3}\right) \neq 0$,$\theta \in (0, 2 \pi)$. For a square matrix $M$,let $\text{trace}(M)$ denote the sum of all the diagonal entries of $M$. Then,among the statements:
$(I) \text{ Trace}(R) = 0$
$(II) \text{ If trace}(\text{adj}(\text{adj}(R))) = 0, \text{ then } R \text{ has exactly one non-zero entry.}$

Let $A$ be a non-singular matrix of order $3$. If $\operatorname{det}(\operatorname{adj}(2 \operatorname{adj}((\operatorname{det} A) A))) = 3^{-13} \cdot 2^{-10}$ and $\operatorname{det}(\operatorname{adj}(2A)) = 2^m \cdot 3^n$,then $|3m + 2n|$ is equal to:

Let $A = \begin{bmatrix} 2 & 1 & 2 \\ 6 & 2 & 11 \\ 3 & 3 & 2 \end{bmatrix}$ and $P = \begin{bmatrix} 1 & 2 & 0 \\ 5 & 0 & 2 \\ 7 & 1 & 5 \end{bmatrix}$. The sum of the prime factors of $|P^{-1}AP - 2I|$ is equal to