If $A = \begin{bmatrix} 1 & -1 \\ 2 & -1 \end{bmatrix}$,$B = \begin{bmatrix} a & 1 \\ b & -1 \end{bmatrix}$ and $(A + B)^2 = A^2 + B^2$,then the values of $a$ and $b$ are:

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

Let $A=\begin{bmatrix} \frac{1}{\sqrt{2}} & -2 \\ 0 & 1 \end{bmatrix}$ and $P=\begin{bmatrix} \cos \theta & -\sin \theta \\ \sin \theta & \cos \theta \end{bmatrix}, \theta > 0$. If $B=P A P^T$,$C=P^T B^{10} P$ and the sum of the diagonal elements of $C$ is $\frac{m}{n}$,where $\operatorname{gcd}(m, n)=1$,then $m+n$ is:

If $A = \begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$ and $I = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$,then which one of the following holds for all $n \ge 1$ (by the principle of mathematical induction)?

If a $2^{nd}$ order determinant with elements $0$ or $1$ is chosen from the set of all such determinants,find the probability that the determinant chosen is non-zero.

Matrix $A_r = \begin{bmatrix} r & r-1 \\ r-1 & r \end{bmatrix}$ for $r = 1, 2, 3, \dots$. If $\sum_{r=1}^{109} |A_r| = (\sqrt{10})^k$,then $k = $ . . . . . . . Where $|A_r| = \det(A_r)$.