Let $A=\begin{bmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{bmatrix}$. Then $A^{2025}-A^{2020}$ is equal to:

$A$ and $B$ are two square matrices such that $A^2B = BA$. If $(AB)^{10} = A^K B^{10}$,then $k$ is:

Let $A = \begin{bmatrix} \cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha \end{bmatrix}$,$\alpha \in R$ such that $A^{32} = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}$. Then a value of $\alpha$ is

Let $ABC = I$. Then $tr(ABC + BCA + CAB)$ is (where the order of matrices $A, B, C$ is $3 \times 3$ and $tr(A)$ is the sum of the diagonal elements in $A$).

If the inverse of $\begin{bmatrix} -x & 14x & 7x \\ 0 & 1 & 0 \\ x & -4x & -2x \end{bmatrix}$ is $\begin{bmatrix} 2 & 0 & 7 \\ 0 & 1 & 0 \\ 1 & -2 & 1 \end{bmatrix}$,then $\left|\begin{array}{ccc} x & x+1 & x+2 \\ x+1 & x+2 & x+3 \\ x+2 & x+3 & x+4 \end{array}\right| = $

For $M=\left[\begin{array}{ll}3 & -4 \\ 1 & -1\end{array}\right]$ and for any $n \in N$,the matrix $M^{n+1}-M^n=$