If $A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$, then $A^{-1} =$

For any $2 \times 2$ matrix $A$,if $A(\text{adj } A) = \begin{bmatrix} 10 & 0 \\ 0 & 10 \end{bmatrix}$,then $|A| = $

The inverse of the matrix $A = \begin{bmatrix} 3 & -2 \\ 1 & 4 \end{bmatrix}$ is:

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

The adjoint matrix of $\begin{bmatrix} 3 & -3 & 4 \\ 2 & -3 & 4 \\ 0 & -1 & 1 \end{bmatrix}$ 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