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:

Let $A$ be a $2 \times 2$ matrix with real entries. Let $I$ be the $2 \times 2$ identity matrix. $\operatorname{Tr}(A)$ denotes the sum of diagonal entries of $A$. Assume that $A^2=I$.
Statement $I$: If $A \neq I$ and $A \neq -I$,then $\operatorname{det}(A) = -1$.
Statement $II$: If $A \neq I$ and $A \neq -I$,then $\operatorname{Tr}(A) \neq 0$.

If $p, q, r, s$ are in $A.P.$ and $f(x) = \left| \begin{array}{ccc} p + \sin x & q + \sin x & p - r + \sin x \\ q + \sin x & r + \sin x & -1 + \sin x \\ r + \sin x & s + \sin x & s - q + \sin x \end{array} \right|$ such that $\int_{0}^{\pi} f(x) dx = -4$,then the common difference of the $A.P.$ can be:

In a square matrix $A$ of order $3$,$a_{ii} = i + m_i$,where $i = 1, 2, 3$ and $m_i$ are the slopes (in increasing order of their absolute value) of the $3$ normals concurrent at the point $(9, -6)$ to the parabola $y^2 = 4x$. All other entries of the matrix are $1$. The value of $\det(A)$ is equal to

If $A=\left[\begin{array}{ll}1 & 0 \\ 2 & 1\end{array}\right]$ and $B=\left[\begin{array}{ll}1 & 3 \\ 0 & 1\end{array}\right]$,then $\operatorname{det}\left(A^6+B^6\right)=$

Let $M$ and $N$ be two $3 \times 3$ matrices such that $MN = NM$. Further,if $M \neq N^2$ and $M^2 = N^4$,then:
$(A)$ determinant of $(M^2 + MN^2)$ is $0$
$(B)$ there is a $3 \times 3$ non-zero matrix $U$ such that $(M^2 + MN^2)U$ is the zero matrix
$(C)$ determinant of $(M^2 + MN^2) \geq 1$
$(D)$ for a $3 \times 3$ matrix $U$,if $(M^2 + MN^2)U$ equals the zero matrix then $U$ is the zero matrix