If $P = \begin{bmatrix} \frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2} \end{bmatrix}$,$A = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$ and $Q = PAP^T$,then $P^T(Q^{2005})P$ is equal to

If $A = \int_{1}^{\sin \theta} \frac{t}{1+t^2} dt$ and $B = \int_{1}^{\operatorname{cosec} \theta} \frac{1}{t(1+t^2)} dt$,then the value of $\left| \begin{array}{ccc} A & A^2 & B \\ e^{A+B} & B^2 & -1 \\ 1 & A^2+B^2 & -1 \end{array} \right| = $

The number of solutions of the matrix equation $X^2 = \begin{bmatrix} 1 & 1 \\ 2 & 3 \end{bmatrix}$ is

$A$ determinant is chosen at random from the set of all determinants of order $2$ with elements $0$ or $1$ only. The probability that the value of the chosen determinant is positive 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

For $M=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]$ and for any $n \in N$,the matrix $M^{n+1}-M^n=$