If $\omega$ is a cube root of unity and $\Delta = \begin{vmatrix} 1 & 2\omega \\ \omega & \omega^2 \end{vmatrix}$,then $\Delta^2$ is equal to

If $5$ is one root of the equation $\left| \begin{array}{ccc} x & 3 & 7 \\ 2 & x & -2 \\ 7 & 8 & x \end{array} \right| = 0$,then the other two roots of the equation are:

For real numbers $x, y$ and $z$,if $x \neq y \neq z$,$\left|\begin{array}{ccc}x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3\end{array}\right|=0$ and $\left|\begin{array}{ccc}1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2\end{array}\right| \neq 0$,then $xyz = $ . . . . . . .

If $A = \begin{bmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{bmatrix}$,then for all $\theta \in \left( \frac{3\pi}{4}, \frac{5\pi}{4} \right)$,$\det(A)$ lies in the interval:

If $A \neq O$ and $B \neq O$ are $n \times n$ matrices such that $AB = O$,then

If $A=\left[\begin{array}{lll}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{array}\right]$,then show that $|3 A|=27|A|$.