If the determinant of a $3^{\text{rd}}$ order matrix $A$ is $K$,then the sum of the determinants of the matrices $(AA^T)$ and $(A-A^T)$ is

Consider the following relation $R$ on the set of real square matrices of order $3$. $R = \{(A,B) | A = P^{-1}BP \text{ for some invertible matrix } P\}$.
\textbf{Statement-$1$:} $R$ is an equivalence relation.
\textbf{Statement-$2$:} For any two invertible $3 \times 3$ matrices $M$ and $N$,$(MN)^{-1} = N^{-1}M^{-1}$.

Let the numbers $2, b, c$ be in an $A.P.$ and $A = \begin{bmatrix} 1 & 1 & 1 \\ 2 & b & c \\ 4 & b^2 & c^2 \end{bmatrix}$. If $\det(A) \in [2, 16]$,then $c$ lies in the interval:

Let $[A]_{3 \times 3}$ be a non-singular matrix such that $A^{-1}=\frac{1}{3}(A^2-5A+7I)$. Then $17A^8-85A^7+119A^6-51A^5-19A^4+95A^3-133A^2+58A+I=$

The value of $\theta$ lying between $0$ and $\pi / 2$ and satisfying the equation $\left| \begin{array}{ccc} 1 + \sin^2 \theta & \cos^2 \theta & 4 \sin 4 \theta \\ \sin^2 \theta & 1 + \cos^2 \theta & 4 \sin 4 \theta \\ \sin^2 \theta & \cos^2 \theta & 1 + 4 \sin 4 \theta \end{array} \right| = 0$ is:

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: