Let $P$ be an $m \times m$ matrix such that $P^2=P$. Then,$(I+P)^n$ equals

Let $P=\begin{bmatrix} -30 & 20 & 56 \\ 90 & 140 & 112 \\ 120 & 60 & 14 \end{bmatrix}$ and $A=\begin{bmatrix} 2 & 7 & \omega^{2} \\ -1 & -\omega & 1 \\ 0 & -\omega & -\omega+1 \end{bmatrix}$,where $\omega=\frac{-1+ i \sqrt{3}}{2}$,and $I_{3}$ is the identity matrix of order $3$. If the determinant of the matrix $(P^{-1}AP - I_{3})^{2}$ is $\alpha \omega^{2}$,then the value of $\alpha$ is equal to:

If $A$ and $B$ are two square matrices such that $B = -A^{-1}BA$,then $(A + B)^2 = $

Let $M$ be a $2 \times 2$ symmetric matrix with integer entries. Then $M$ is invertible if:

$\left|\begin{array}{ll}2 & 1 \\ 3 & 1\end{array}\right|+\left|\begin{array}{cc}1 & 1/3 \\ 3 & 1\end{array}\right|+\left|\begin{array}{cc}1/2 & 1/9 \\ 3 & 1\end{array}\right|+\left|\begin{array}{cc}1/4 & 1/27 \\ 3 & 1\end{array}\right|+\ldots \infty=$

Let $A = \begin{bmatrix} 1 & 2 \\ 1 & \alpha \end{bmatrix}$ and $B = \begin{bmatrix} 3 & 3 \\ \beta & 2 \end{bmatrix}$. If $A^2 - 4A + I = O$ and $B^2 - 5B - 6I = O$, then among the two statements:
(S1): $[(B - A)(B + A)]^T = \begin{bmatrix} 13 & 15 \\ 7 & 10 \end{bmatrix}$
and
(S2): $\det(\text{adj}(A + B)) = -5$.