Let $A$ be a non-zero periodic matrix with period $4$ and $A^{12} + B = I$,where $I$ is the identity matrix and $B$ is any square matrix of the same order as $A$. The matrix product $AB$ is equal to:

Suppose the vectors $x_{1}, x_{2}$ and $x_{3}$ are the solutions of the system of linear equations $Ax = b$ when the vector $b$ on the right side is equal to $b_{1}, b_{2}$ and $b_{3}$ respectively. If $x_{1} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, x_{2} = \begin{bmatrix} 0 \\ 2 \\ 1 \end{bmatrix}, x_{3} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}, b_{1} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix}, b_{2} = \begin{bmatrix} 0 \\ 2 \\ 0 \end{bmatrix}$ and $b_{3} = \begin{bmatrix} 0 \\ 0 \\ 2 \end{bmatrix}$,then the determinant of $A$ is equal to

Let $A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$,then:

Let $A = \begin{bmatrix} 1 & 2 & 3 \\ 2 & 2 & -1 \\ 3 & 0 & k \end{bmatrix}$ and $f(x) = x^3 - 2x^2 - \alpha x + \beta = 0$. If $A$ satisfies $f(A) = 0$,then:

Let $P=\begin{bmatrix} 1 & 0 & 0 \\ 4 & 1 & 0 \\ 16 & 4 & 1 \end{bmatrix}$ and $I$ be the identity matrix of order $3$. If $Q=[q_{ij}]$ is a matrix such that $P^{50}-Q=I$,then $\frac{q_{31}+q_{32}}{q_{21}}$ equals

The value of $(1+\Delta)(1-\nabla)$ is