If $A = \begin{bmatrix} 1 & 2 & 1 \\ 0 & 1 & -1 \\ 3 & -1 & 1 \end{bmatrix}$,then:

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:

Let matrix $A = \begin{bmatrix} 1 & 2 & 3 \\ 0 & 5 & 4 \\ 0 & 3 & 2 \end{bmatrix}$ and $A^3 - 8A^2 + \alpha A + \beta I = O$,then the ordered pair $(\alpha, \beta)$ is

If $A = \begin{bmatrix} 1 & a & 3 \\ b & 2 & c \\ 3 & d & 4 \end{bmatrix}$ is a symmetric matrix and $B = \begin{bmatrix} 0 & 5 & b \\ -5 & 0 & -7 \\ 6 & c & 0 \end{bmatrix}$ is a skew-symmetric matrix, then $AB = $

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:

If $ P=\left|\begin{array}{ll}x & 1 \\ 1 & x\end{array}\right| $ and $ Q=\left|\begin{array}{lll}x & 1 & 1 \\ 1 & x & 1 \\ 1 & 1 & x\end{array}\right| $,then $ \frac{d Q}{d x}= $