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

Let $J_{n, m}=\int_{0}^{\frac{1}{2}} \frac{x^{n}}{x^{m}-1} d x, \quad \forall n>m$ and $n, m \in N$. Consider a matrix $A=\left[a_{i j}\right]_{3 \times 3}$ where $a_{i j}=J_{6+i, 3}-J_{i+3,3}$ for $i \leq j$ and $a_{i j}=0$ for $i>j$. Then $\left|\operatorname{adj} A^{-1}\right|$ is:

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

Let $A$ be a matrix of order $2 \times 2$,whose entries are from the set $\{0, 1, 2, 3, 4, 5\}$. If the sum of all the entries of $A$ is a prime number $p$,where $2 < p < 8$,then the number of such matrices $A$ is:

Let $\quad P_1=I=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{array}\right], \quad P_2=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 0 & 1 \\ 0 & 1 & 0\end{array}\right], \quad P_3=\left[\begin{array}{lll}0 & 1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\end{array}\right], \quad P_4=\left[\begin{array}{lll}0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0\end{array}\right], \quad P_5=\left[\begin{array}{lll}0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{array}\right], \quad P_6=\left[\begin{array}{lll}0 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 0\end{array}\right]$ and $X=\sum_{k=1}^6 P_k \left[\begin{array}{lll}2 & 1 & 3 \\ 1 & 0 & 2 \\ 3 & 2 & 1\end{array}\right] P_k^{\top}$ where $P_k^{\top}$ denotes the transpose of the matrix $P_k$. Then which of the following options is/are correct?
$(1)$ $X - 30I$ is an invertible matrix
$(2)$ The sum of diagonal entries of $X$ is $18$
$(3)$ If $X \left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]=\alpha\left[\begin{array}{l}1 \\ 1 \\ 1\end{array}\right]$,then $\alpha=30$
$(4)$ $X$ is a symmetric matrix

$A$ and $B$ are two square matrices such that $A^2B = BA$. If $(AB)^{10} = A^K B^{10}$,then $k$ is: