If $A$ and $B$ are two invertible square matrices of the same order such that $(A + B)(A - B) = A^2 - B^2$,then $(A^2BA^{-1}B^{-1})^3$ is equal to-

If $\begin{bmatrix} a & b & c \\ d & e & f \\ g & h & i \end{bmatrix}$ is a skew-symmetric matrix and $b, c, f$ are non-zero real numbers,then $\frac{b}{c} = $

Let $A = [a_{ij}]$ be a square matrix of order $2$ with entries either $0$ or $1$. Let $E$ be the event that $A$ is an invertible matrix. Then the probability $P(E)$ is:

Let $\omega$ be the complex number $\cos \frac{2 \pi}{3} + i \sin \frac{2 \pi}{3}$. Then the number of distinct complex numbers $z$ satisfying $\left|\begin{array}{ccc} z+1 & \omega & \omega^2 \\ \omega & z+\omega^2 & 1 \\ \omega^2 & 1 & z+\omega \end{array}\right| = 0$ is equal to

Matrix $A$ is such that ${A^2} = 2A - I$,where $I$ is the identity matrix. Then for $n \ge 2$,${A^n} = $

Let $p$ be a non-singular matrix such that $I + p + p^2 + .... + p^n = O$ (where $O$ denotes the null matrix and $I$ denotes the identity matrix),then $p^{-1} = $