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

If $A$ and $B$ are square matrices of the same order such that $AB = A$ and $BA = B$,then $(A + I)^5$ is equal to (where $I$ is the identity matrix).

If $A = \begin{bmatrix} \cos^2 \alpha & \cos \alpha \sin \alpha \\ \cos \alpha \sin \alpha & \sin^2 \alpha \end{bmatrix}$ and $B = \begin{bmatrix} \cos^2 \beta & \cos \beta \sin \beta \\ \cos \beta \sin \beta & \sin^2 \beta \end{bmatrix}$ are two matrices such that the product $AB$ is a null matrix,then $\alpha - \beta$ is:

Let $A = \begin{bmatrix} 1 & -1 \\ 2 & \alpha \end{bmatrix}$ and $B = \begin{bmatrix} \beta & 1 \\ 1 & 0 \end{bmatrix}$,where $\alpha, \beta \in \mathbb{R}$. Let $\alpha_{1}$ be the value of $\alpha$ which satisfies $(A + B)^{2} = A^{2} + \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}$ and $\alpha_{2}$ be the value of $\alpha$ which satisfies $(A + B)^{2} = B^{2}$. Then $|\alpha_{1} - \alpha_{2}|$ is equal to:

Let $A$ and $B$ be two non-singular skew-symmetric matrices such that $AB = BA$. Then $A^{2} B^{2} (A^{\top} B)^{-1} (A B^{-1})^{\top}$ is equal to

For a square matrix $B$ of order $3$,if $B^T=B^{-1}$ and $|B|=1$,then $|B-I|=$