The value of $\theta$ lying between $\theta = 0$ and $\theta = \pi / 2$ and satisfying the equation : $\left| \begin{array}{ccc} 1 + \sin^2 \theta & \cos^2 \theta & 4 \sin 4 \theta \\ \sin^2 \theta & 1 + \cos^2 \theta & 4 \sin 4 \theta \\ \sin^2 \theta & \cos^2 \theta & 1 + 4 \sin 4 \theta \end{array} \right| = 0$ are :

Let $A = \begin{bmatrix} 1 & 2 & 7 \\ 4 & -2 & 8 \\ 3 & 8 & -7 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 0 & 0 \\ 0 & -5\alpha & 0 \\ 0 & 4\alpha & -2\alpha \end{bmatrix} + \text{adj}(A)$. If $\det(B) = 66$, then $\det(\text{adj}(A))$ equals:

If $A$ and $B$ are square matrices of the same order such that $AB = A$ and $BA = B$,then $(A + I)^5$ is equal to (where $I$ is the identity matrix).

Let $A$ and $B$ be two symmetric matrices of order $3$.
Statement $-1$: $A(BA)$ and $(AB)A$ are symmetric matrices.
Statement $-2$: $AB$ is a symmetric matrix if the matrix multiplication of $A$ with $B$ is commutative.

Let $A = \begin{bmatrix} 1 & -1 \\ 2 & \alpha \end{bmatrix}$ and $B = \begin{bmatrix} \beta & 1 \\ 1 & 0 \end{bmatrix}$,where $\alpha, \beta \in \mathbb{R}$. Let $\alpha_{1}$ be the value of $\alpha$ which satisfies $(A + B)^{2} = A^{2} + \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}$ and $\alpha_{2}$ be the value of $\alpha$ which satisfies $(A + B)^{2} = B^{2}$. Then $|\alpha_{1} - \alpha_{2}|$ 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