If $A$ is a matrix of order $2$ and $I$ is the identity matrix of order $2$ such that $A^2 - 4A + 3I = 0$,then $(A + 3I)^{-1} =$

The element of the second row and third column in the inverse of $\begin{bmatrix} 1 & 2 & 1 \\ 2 & 1 & 0 \\ -1 & 0 & 1 \end{bmatrix}$ is

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 possible,using elementary row transformations,find the inverse of the following matrix:
$\left[\begin{array}{ccc}2 & -1 & 3 \\ -5 & 3 & 1 \\ -3 & 2 & 3\end{array}\right]$

If matrix $A = \begin{bmatrix} 3 & 2 & 4 \\ 1 & 2 & -1 \\ 0 & 1 & 1 \end{bmatrix}$ and $A^{-1} = \frac{1}{K} \text{adj}(A)$,then $K$ is:

If $A = \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}$ and $A^{-1} = \begin{bmatrix} 3 & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2 \end{bmatrix}$,then the values of $\alpha$ and $\beta$ are respectively: