If $A = \begin{bmatrix} -4 & -1 \\ 3 & 1 \end{bmatrix}$,then the determinant of the matrix $(A^{2016} - 2A^{2015} - A^{2014})$ is

Let $A = \begin{bmatrix} 4 & -2 \\ \alpha & \beta \end{bmatrix}$. If $A^2 + \gamma A + 18I = O$,then $\operatorname{det}(A)$ is equal to

The equation $\left| \begin{array}{ccc} (1+x)^2 & (1-x)^2 & -(2+x^2) \\ 2x+1 & 3x & 1-5x \\ x+1 & 2x & 2-3x \end{array} \right| + \left| \begin{array}{ccc} (1+x)^2 & 2x+1 & x+1 \\ (1-x)^2 & 3x & 2x \\ 1-2x & 3x-2 & 2x-3 \end{array} \right| = 0$

For $A = \begin{bmatrix} 0 & 0 & -2 \\ 0 & -2 & 0 \\ -2 & 0 & 0 \end{bmatrix}$,which of the following is true?

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:

If $\alpha , \beta \neq 0$ and $f(n) = \alpha^n + \beta^n$ and $\begin{vmatrix} 3 & 1 + f(1) & 1 + f(2) \\ 1 + f(1) & 1 + f(2) & 1 + f(3) \\ 1 + f(2) & 1 + f(3) & 1 + f(4) \end{vmatrix} = K(1 - \alpha)^2 (1 - \beta)^2 (\alpha - \beta)^2$,then $K = \dots$