Show that the matrix $A = \begin{bmatrix} 2 & 3 \\ 1 & 2 \end{bmatrix}$ satisfies the equation $A^{2} - 4A + I = O$,where $I$ is the $2 \times 2$ identity matrix and $O$ is the $2 \times 2$ zero matrix. Using this equation,find $A^{-1}$.

Let $k$ be a positive real number and let $A = \begin{bmatrix} 2k-1 & 2\sqrt{k} & 2\sqrt{k} \\ 2\sqrt{k} & 1 & -2k \\ -2\sqrt{k} & 2k & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 0 & 2k-1 & \sqrt{k} \\ 1-2k & 0 & 2\sqrt{k} \\ -\sqrt{k} & -2\sqrt{k} & 0 \end{bmatrix}$. If $\det(\operatorname{adj} A) + \det(\operatorname{adj} B) = 10^6$,then $[k]$ is equal to [Note: $\operatorname{adj} M$ denotes the adjoint of a square matrix $M$ and $[k]$ denotes the greatest integer less than or equal to $k$].

If $A = \begin{bmatrix} 1 & 2 \\ -1 & 4 \end{bmatrix}$ and $A^{-1} = \alpha I + \beta A$,where $\alpha, \beta \in \mathbb{R}$ and $I$ is the identity matrix of order $2$,then $4(\alpha + \beta) = $

Suppose $A$ is any $3 \times 3$ non-singular matrix and $(A-3 I)(A-5 I)=O$,where $I=I_3$ and $O=O_3$. Here $O_3$ represents the zero matrix of order $3$ and $I_3$ is the identity matrix of order $3$. If $\alpha A+\beta A^{-1}=4 I$,then $\alpha+\beta$ is equal to:

Inverse of the matrix $\begin{bmatrix} \cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{bmatrix}$ is

If $B = \begin{bmatrix} 5 & 2\alpha & 1 \\ 0 & 2 & 1 \\ \alpha & 3 & -1 \end{bmatrix}$ is the inverse of a $3 \times 3$ matrix $A$,then the sum of all values of $\alpha$ for which $\det(A) + 1 = 0$ is: