If $\omega$ is an imaginary cube root of unity,then the value of $\left[\begin{array}{ccc}1 & \omega^{2} & 1-\omega^{4} \\ \omega & 1 & 1+\omega^{5} \\ 1 & \omega & \omega^{2}\end{array}\right]$ is

If $a, b, c, d, e, f$ are in $G.P.$,then the value of $\left| \begin{array}{ccc} a^2 & d^2 & x \\ b^2 & e^2 & y \\ c^2 & f^2 & z \end{array} \right|$ depends on

If $A = \begin{bmatrix} a & b \\ c & d \end{bmatrix}$ (where $bc \neq 0$) satisfies the equation $x^2 + k = 0$,then:

$A$ determinant is chosen at random from the set of all determinants of order $2$ with elements $0$ or $1$ only. The probability that the value of the chosen determinant is positive 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 $A = \begin{bmatrix} \cos \theta & 0 & -\sin \theta \\ 0 & 1 & 0 \\ \sin \theta & 0 & \cos \theta \end{bmatrix}$. If for some $\theta \in (0, \pi)$,$A^2 = A^T$,then the sum of the diagonal elements of the matrix $(A + I)^3 + (A - I)^3 - 6A$ is equal to . . . . . . .