Let $A$ be a matrix of order $2 \times 2$,whose entries are from the set $\{0, 1, 2, 3, 4, 5\}$. If the sum of all the entries of $A$ is a prime number $p$,where $2 < p < 8$,then the number of such matrices $A$ is:

If $\left| \begin{array}{ccc} a & a^2 & 1 + a^3 \\ b & b^2 & 1 + b^3 \\ c & c^2 & 1 + c^3 \end{array} \right| = 0$ and the vectors $\vec{a} = (1, a, a^2)$,$\vec{b} = (1, b, b^2)$,and $\vec{c} = (1, c, c^2)$ are non-coplanar,then $abc$ is equal to

If $A = \begin{bmatrix} 1 & 3 \\ 2 & 1 \end{bmatrix}$,then the determinant of $A^2 - 2A$ is

If $A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}$,$B = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix}$ and $(A + B)^2 = A^2 + B^2$,then the values of $a$ and $b$ are:

Let $z = \frac{-1 + \sqrt{3}i}{2}$, where $i = \sqrt{-1}$, and $r, s \in \{1, 2, 3\}$. Let $P = \begin{bmatrix} (-z)^r & z^{2s} \\ z^{2s} & z^r \end{bmatrix}$ and $I$ be the identity matrix of order $2$. Then the total number of ordered pairs $(r, s)$ for which $P^2 = -I$ is

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