Let $A$ be an invertible square matrix of order $3 \times 3$. Then $|(\text{adj} A) \cdot A|$ is

Let $X=\begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 0 & 0 & 0 \end{bmatrix}$,$Y=\alpha I+\beta X+\gamma X^{2}$ and $Z=\alpha^{2} I-\alpha \beta X+\left(\beta^{2}-\alpha \gamma\right) X^{2}$,where $\alpha, \beta, \gamma \in \mathbb{R}$. If $Y^{-1}=\begin{bmatrix} \frac{1}{5} & \frac{-2}{5} & \frac{1}{5} \\ 0 & \frac{1}{5} & \frac{-2}{5} \\ 0 & 0 & \frac{1}{5} \end{bmatrix}$,then $(\alpha-\beta+\gamma)^{2}$ is equal to

The third element in the second row of the adjoint of a matrix $A = [a_{ij}]_{3 \times 3}$,where $a_{ij} = 2i + j$,is:

Let $P = \begin{bmatrix} 3 & -1 & -2 \\ 2 & 0 & \alpha \\ 3 & -5 & 0 \end{bmatrix}$ where $\alpha \in R$. Suppose $Q = [q_{ij}]$ is a matrix satisfying $PQ = kI_3$ for some non-zero $k \in R$. If $q_{23} = -\frac{k}{8}$ and $|Q| = \frac{k^2}{2}$,then $\alpha^2 + k^2$ is equal to?

Which of the following matrices are invertible?
$A = \begin{bmatrix} 2 & 3 \\ 10 & 15 \end{bmatrix}, B = \begin{bmatrix} 1 & 2 & 3 \\ 2 & -1 & 3 \\ 1 & 2 & 3 \end{bmatrix}, C = \begin{bmatrix} 1 & 2 & 3 \\ 3 & 4 & 5 \\ 4 & 6 & 8 \end{bmatrix}, D = \begin{bmatrix} 2 & 4 & 2 \\ 1 & 1 & 0 \\ 1 & 4 & 5 \end{bmatrix}$

If $A = \begin{bmatrix} 2 & -1 \\ -1 & 2 \end{bmatrix}$,such that $A^{2} - 4A + 3I = 0$,then $A^{-1} =$