For a $3 \times 3$ matrix $A$,if $A(\operatorname{adj} A) = \begin{bmatrix} -10 & 0 & 0 \\ 0 & -10 & 2 \\ 0 & 0 & -10 \end{bmatrix}$,then the value of the determinant of $A$ is:

Let $M=\begin{bmatrix} \sin^4 \theta & -1-\sin^2 \theta \\ 1+\cos^2 \theta & \cos^4 \theta \end{bmatrix} = \alpha I + \beta M^{-1}$,where $\alpha = \alpha(\theta)$ and $\beta = \beta(\theta)$ are real numbers,and $I$ is the $2 \times 2$ identity matrix. If $\alpha^*$ is the minimum of the set $\{\alpha(\theta) : \theta \in [0, 2\pi)\}$ and $\beta^*$ is the minimum of the set $\{\beta(\theta) : \theta \in [0, 2\pi)\}$,then the value of $\alpha^* + \beta^*$ is

Let $x, y, z > 1$ and $A = \begin{bmatrix} 1 & \log_x y & \log_x z \\ \log_y x & 2 & \log_y z \\ \log_z x & \log_z y & 3 \end{bmatrix}$. Then $|\operatorname{adj}(\operatorname{adj} A^2)|$ is equal to

$A, P, B$ are $3 \times 3$ matrices. If $|-B|=5, |BA^T|=15, |P^T AP|=-27$, then one of the values of $|P|$ is

If $A$ is a square matrix such that $A^{2} = A$,then $(I + A)^{3} - 7A$ is equal to

Let $P=\begin{bmatrix} -30 & 20 & 56 \\ 90 & 140 & 112 \\ 120 & 60 & 14 \end{bmatrix}$ and $A=\begin{bmatrix} 2 & 7 & \omega^{2} \\ -1 & -\omega & 1 \\ 0 & -\omega & -\omega+1 \end{bmatrix}$,where $\omega=\frac{-1+ i \sqrt{3}}{2}$,and $I_{3}$ is the identity matrix of order $3$. If the determinant of the matrix $(P^{-1}AP - I_{3})^{2}$ is $\alpha \omega^{2}$,then the value of $\alpha$ is equal to: