$A$ determinant is chosen at random from the set of all determinants of order $2 \times 2$ with elements $0$ or $1$ only. The probability that the determinant chosen is non-zero is

Let $\alpha \in(0, \infty)$ and $A=\begin{bmatrix} 1 & 2 & \alpha \\ 1 & 0 & 1 \\ 0 & 1 & 2 \end{bmatrix}$. If $\operatorname{det}(\operatorname{adj}(2A-A^{T}) \cdot \operatorname{adj}(A-2A^{T}))=2^8$,then $(\operatorname{det}(A))^2$ is equal to:

The number of positive integral solutions of the equation $\left| {\begin{array}{*{20}{c}}{{x^3} + 1}&{{x^2}y}&{{x^2}z}\\{x{y^2}}&{{y^3} + 1}&{{y^2}z}\\{x{z^2}}&{y{z^2}}&{{z^3} + 1}\end{array}} \right| = 11$ is

$A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 1 \\ 0 & 1 & 0\end{array}\right] \Rightarrow A^2-2 A=$

Let integers $a, b \in [-3, 3]$ be such that $a + b \neq 0$. Then the number of all possible ordered pairs $(a, b)$,for which $|\frac{z-a}{z+b}|=1$ and $\left|\begin{array}{ccc}z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega\end{array}\right|=1$ for some $z \in \mathbb{C}$,where $\omega$ and $\omega^2$ are the roots of $x^2+x+1=0$,is equal to . . . . . .