If $A$ is a square matrix of order $3$ and $A^2+A+2I=0$,then

If $A = \begin{bmatrix} 0 & -\tan \frac{\alpha}{2} \\ \tan \frac{\alpha}{2} & 0 \end{bmatrix}$ and $I$ is the identity matrix of order $2$,show that $I+A = (I-A) \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$.

Let $B=\begin{bmatrix} 1 & 3 \\ 1 & 5 \end{bmatrix}$ and $A$ be a $2 \times 2$ matrix such that $AB^{-1}=A^{-1}$. If $BCB^{-1}=A$ and $C^4+\alpha C^2+\beta I=O$,then $2\beta-\alpha$ is equal to:

If $A = \begin{bmatrix} \cos^2 \alpha & \sin \alpha \cos \alpha \\ \sin \alpha \cos \alpha & \sin^2 \alpha \end{bmatrix}$ and $B = \begin{bmatrix} \cos^2 \beta & \sin \beta \cos \beta \\ \sin \beta \cos \beta & \sin^2 \beta \end{bmatrix}$ are such that $AB$ is a null matrix,then which of the following must be an odd integral multiple of $\frac{\pi}{2}$?

Let $A$ be a $2 \times 2$ matrix with non-zero entries and let $A^2 = I$,where $I$ is the $2 \times 2$ identity matrix. Define $tr(A) = \text{sum of diagonal elements of } A$ and $|A| = \text{determinant of matrix } A$.
Statement $-1: tr(A) = 0$
Statement $-2: \det(A) = 1$

If $A = \begin{bmatrix} 1 & 2 \\ -2 & -5 \end{bmatrix}$ and $\alpha A^2 + \beta A = 2I$ for some $\alpha, \beta \in \mathbb{R}$,then $\alpha + \beta =$