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(A) Given $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 1 \ 1 & 0 & 0 \end{bmatrix}$.Calculating powers of $A$:$A^2 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1

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$A^{6}-A$

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(A) Given $A = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 1 \ 1 & 0 & 0 \end{bmatrix}$.Calculating powers of $A$:$A^2 = \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 1 \ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 1 \ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 1 & 1 & 1 \ 1 & 0 & 0 \end{bmatrix}$.$A^3 = A^2 \cdot A = \begin{bmatrix} 1 & 0 & 0 \ 1 & 1 & 1 \ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 1 \ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 2 & 1 & 1 \ 1 & 0 & 0 \end{bmatrix}$.$A^4 = A^3 \cdot A = \begin{bmatrix} 1 & 0 & 0 \ 2 & 1 & 1 \ 1 & 0 & 0 \end{bmatrix} \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 1 \ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 & 0 \ 3 & 1 & 1 \ 1 & 0 & 0 \end{bmatrix}$.By induction,$A^n = \begin{bmatrix} 1 & 0 & 0 \ n-1 & 1 & 1 \ 1 & 0 & 0 \end{bmatrix}$.Thus,$A^{2025} - A^{2020} = \begin{bmatrix} 1 & 0 & 0 \ 2024 & 1 & 1 \ 1 & 0 & 0 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \ 2019 & 1 & 1 \ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \ 5 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}$.Now check $A^6 - A = \begin{bmatrix} 1 & 0 & 0 \ 5 & 1 & 1 \ 1 & 0 & 0 \end{bmatrix} - \begin{bmatrix} 1 & 0 & 0 \ 0 & 1 & 1 \ 1 & 0 & 0 \end{bmatrix} = \begin{bmatrix} 0 & 0 & 0 \ 5 & 0 & 0 \ 0 & 0 & 0 \end{bmatrix}$.Therefore,$A^{2025} - A^{2020} = A^6 - A$.

Column $I$	Column $II$
$(A)$ The set $\{\operatorname{Re}(\frac{2 i z}{1-z^2}): \|z\|=1, z \neq \pm 1\}$ is	$(p)$ $(-\infty,-1) \cup(1, \infty)$
$(B)$ The domain of $f(x)=\sin ^{-1}(\frac{8(3)^{x-2}}{1-3^{2(x-1)}})$ is	$(q)$ $(-\infty, 0) \cup(0, \infty)$
$(C)$ If $f(\theta)=\left\|\begin{array}{ccc}1 & \tan \theta & 1 \\ -\tan \theta & 1 & \tan \theta \\ -1 & -\tan \theta & 1\end{array}\right\|$,then the set $\{f(\theta): 0 \leq \theta < \frac{\pi}{2}\}$ is	$(r)$ $[2, \infty)$
$(D)$ If $f(x)=x^{3 / 2}(3 x-10), x \geq 0$,then $f(x)$ is increasing in	$(s)$ $(-\infty,-1] \cup[1, \infty)$
	$(t)$ $(-\infty, 0] \cup[2, \infty)$

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

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