$Adj(AB) - (Adj B)(Adj A) = $

The characteristic equation of a matrix $A$ is $\lambda^{3}-5 \lambda^{2}-3 \lambda+2=0$. Then $|\text{adj}(A)|$ is equal to:

If $A = \begin{bmatrix} 1 & -\tan(\theta/2) \\ \tan(\theta/2) & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & \tan(\theta/2) \\ -\tan(\theta/2) & 1 \end{bmatrix}$ then $A^{-1}B$ is equal to

If $A = \begin{bmatrix} \sec \theta & -\tan \theta \\ -\tan \theta & \sec \theta \end{bmatrix}$, $\theta \in (0, \frac{\pi}{2})$ such that $A + \text{adj } A = 4I$, then $\theta =$

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

${\left[ {\begin{array}{*{20}{c}}{ - 6}&5\\{ - 7}&6\end{array}} \right]^{ - 1}}$ =