If $A$ and $B$ are both $3 \times 3$ matrices,then which of the following statements are true?
$(i)$ $AB=0 \Rightarrow A=0$ or $B=0$
(ii) $AB=I_3 \Rightarrow A^{-1}=B$
(iii) $(A-B)^2=A^2-2AB+B^2$

Let $A$ be a $3 \times 3$ invertible matrix. If $|\operatorname{adj}(24A)| = |\operatorname{adj}(3 \operatorname{adj}(2A))|$,then $|A^2|$ is equal to

Let $d \in \mathbb{R}$,and $A = \begin{bmatrix} -2 & 4+d & \sin \theta - 2 \\ 1 & \sin \theta + 2 & d \\ 5 & 2\sin \theta - d & -\sin \theta + 2 + 2d \end{bmatrix}$,where $\theta \in [0, 2\pi]$. If the minimum value of $\det(A)$ is $8$,then a value of $d$ 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 $A = \begin{bmatrix} 1 & -1 \\ 2 & \alpha \end{bmatrix}$ and $B = \begin{bmatrix} \beta & 1 \\ 1 & 0 \end{bmatrix}$,where $\alpha, \beta \in \mathbb{R}$. Let $\alpha_{1}$ be the value of $\alpha$ which satisfies $(A + B)^{2} = A^{2} + \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}$ and $\alpha_{2}$ be the value of $\alpha$ which satisfies $(A + B)^{2} = B^{2}$. Then $|\alpha_{1} - \alpha_{2}|$ is equal to:

The determinant $\left| \begin{array}{ccc} a^2 & a^2 - (b - c)^2 & bc \\ b^2 & b^2 - (c - a)^2 & ca \\ c^2 & c^2 - (a - b)^2 & ab \end{array} \right|$ is divisible by :