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

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-

Let $A = \begin{bmatrix} \frac{1}{6} & \frac{-1}{3} & \frac{-1}{6} \\ \frac{-1}{3} & \frac{2}{3} & \frac{1}{3} \\ \frac{-1}{6} & \frac{1}{3} & \frac{1}{6} \end{bmatrix}$. If $A^{2016l} + A^{2017m} + A^{2018n} = \frac{1}{\alpha} A$, for every $l, m, n \in N$, then the value of $\alpha$ is

Let $A$ be a square matrix such that $AA^T = I$. Then $\frac{1}{2} A[(A+A^T)^2 + (A-A^T)^2]$ is equal to

If $x, y$ are any two non-zero real numbers, $a_{i j} = xi + yj$, $A = \{a_{i j}\}_{n \times n}$ and $P, Q$ are two $n \times n$ matrices such that $A = xP + yQ$, then

The number of $3 \times 3$ non-singular matrices,with four entries as $1$ and all other entries as $0$,is