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

We have the matrix

$$A=\begin{bmatrix}1&0&0\\0&1&1\\1&0&0\end{bmatrix}.$$

To understand high powers of $$A$$, we first observe what happens when we multiply $$A$$ by itself. Let us denote $$A^n=\begin{bmatrix}1&0&0\\a_n&b_n&b_n\\1&0&0\end{bmatrix}$$ for some numbers $$a_n$$ and $$b_n$$ (we shall soon find a pattern for them). We start with

$$A^1=A=\begin{bmatrix}1&0&0\\0&1&1\\1&0&0\end{bmatrix},$$

so here $$a_1=0$$ and $$b_1=1$$.

Now we multiply $$A^n$$ by $$A$$ to get $$A^{n+1}$$:

$$A^{n+1}=A^nA =\begin{bmatrix}1&0&0\\a_n&b_n&b_n\\1&0&0\end{bmatrix} \begin{bmatrix}1&0&0\\0&1&1\\1&0&0\end{bmatrix}.$$

We compute each entry row-by-column.

For the first row we get

$$[1,0,0]\times A=[1,0,0],$$

so the first row of every power remains $$[1,0,0]$$. The third row is identical to the first, hence it also stays $$[1,0,0]$$ for every power.

For the second row we multiply $$[a_n,b_n,b_n]$$ with $$A$$:

First column: $$a_n\cdot1+b_n\cdot0+b_n\cdot1=a_n+b_n,$$

Second column: $$a_n\cdot0+b_n\cdot1+b_n\cdot0=b_n,$$

Third column: $$a_n\cdot0+b_n\cdot1+b_n\cdot0=b_n.$$

Hence

$$a_{n+1}=a_n+b_n,\qquad b_{n+1}=b_n.$$

From the initial value $$b_1=1$$ and the recurrence $$b_{n+1}=b_n$$, we see immediately that

$$b_n=1\quad\text{for all }n\ge1.$$

Substituting $$b_n=1$$ into $$a_{n+1}=a_n+1$$ with $$a_1=0$$ gives

$$a_n=n-1.$$ Therefore, for every integer $$n\ge1$$,

$$A^n=\begin{bmatrix} 1&0&0\\ n-1&1&1\\ 1&0&0 \end{bmatrix}.$$

Now we can evaluate the two required powers.

For $$n=2025$$ we get

$$A^{2025}=\begin{bmatrix} 1&0&0\\ 2025-1&1&1\\ 1&0&0 \end{bmatrix} =\begin{bmatrix} 1&0&0\\ 2024&1&1\\ 1&0&0 \end{bmatrix}.$$

For $$n=2020$$ we get

$$A^{2020}=\begin{bmatrix} 1&0&0\\ 2020-1&1&1\\ 1&0&0 \end{bmatrix} =\begin{bmatrix} 1&0&0\\ 2019&1&1\\ 1&0&0 \end{bmatrix}.$$

Their difference is

$$A^{2025}-A^{2020}= \begin{bmatrix} 1-1&0-0&0-0\\ 2024-2019&1-1&1-1\\ 1-1&0-0&0-0 \end{bmatrix} =\begin{bmatrix} 0&0&0\\ 5&0&0\\ 0&0&0 \end{bmatrix}.$$

Next, we compare this result with the matrices that appear in the answer options. Using the same general formula with $$n=6$$ and $$n=1$$, we have

$$A^6=\begin{bmatrix} 1&0&0\\ 6-1&1&1\\ 1&0&0 \end{bmatrix} =\begin{bmatrix} 1&0&0\\ 5&1&1\\ 1&0&0 \end{bmatrix},$$

and obviously

$$A=\begin{bmatrix} 1&0&0\\ 0&1&1\\ 1&0&0 \end{bmatrix}.$$

Subtracting, we find

$$A^6-A= \begin{bmatrix} 1-1&0-0&0-0\\ 5-0&1-1&1-1\\ 1-1&0-0&0-0 \end{bmatrix} =\begin{bmatrix} 0&0&0\\ 5&0&0\\ 0&0&0 \end{bmatrix}.$$

This is exactly the same matrix as $$A^{2025}-A^{2020}$$. Therefore,

$$A^{2025}-A^{2020}=A^6-A.$$

Hence, the correct answer is Option A.