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

If $A = \begin{bmatrix} \cos^2 \theta & \sin \theta \cos \theta \\ \sin \theta \cos \theta & \sin^2 \theta \end{bmatrix}$,$B = \begin{bmatrix} \cos^2 \phi & \sin \phi \cos \phi \\ \sin \phi \cos \phi & \sin^2 \phi \end{bmatrix}$ and $\theta$ and $\phi$ differ by $\frac{\pi}{2}$,then $AB = $

Let $M = \begin{bmatrix} 0 & 1 & a \\ 1 & 2 & 3 \\ 3 & b & 1 \end{bmatrix}$ and $\operatorname{adj} M = \begin{bmatrix} -1 & 1 & -1 \\ 8 & -6 & 2 \\ -5 & 3 & -1 \end{bmatrix}$ where $a$ and $b$ are real numbers. Which of the following options is/are correct?
$(1)$ $a+b=3$
$(2)$ $\operatorname{det}(\operatorname{adj} M^2) = 81$
$(3)$ $(\operatorname{adj} M)^{-1} + \operatorname{adj} M^{-1} = -M$
$(4)$ If $M \begin{bmatrix} \alpha \\ \beta \\ \gamma \end{bmatrix} = \begin{bmatrix} 1 \\ 2 \\ 3 \end{bmatrix}$,then $\alpha - \beta + \gamma = 3$

If $D_1$ and $D_2$ are two $3 \times 3$ diagonal matrices,then

Find $x,$ if $[x \ -5 \ -1]\begin{bmatrix} 1 & 0 & 2 \\ 0 & 2 & 1 \\ 2 & 0 & 3 \end{bmatrix}\begin{bmatrix} x \\ 4 \\ 1 \end{bmatrix} = O$

If $A$ and $B$ are two square matrices of the same order and $(AB+BA)^{T}+(AB-BA)^{T}=2BA$, then: