1. Given matrix

$$A=\begin{pmatrix} 1&0&0\\ 0&1&1\\ 1&0&0 \end{pmatrix}$$

We need to find:

$$A^{2025}-A^{2020}$$

2. Look for a pattern in powers of $A$

Let us compute $A^2$.

$$A^2=A\cdot A =\begin{pmatrix} 1&0&0\\ 0&1&1\\ 1&0&0 \end{pmatrix} \begin{pmatrix} 1&0&0\\ 0&1&1\\ 1&0&0 \end{pmatrix}$$

Now multiply row by column:

• First row: $$(1,0,0)A=(1,0,0)$$
• Second row: $$(0,1,1)A=(1,1,1)$$
• Third row: $$(1,0,0)A=(1,0,0)$$

Hence,

$$A^2=\begin{pmatrix} 1&0&0\\ 1&1&1\\ 1&0&0 \end{pmatrix}$$

Now compute $A^3=A^2A$:

$$A^3= \begin{pmatrix} 1&0&0\\ 1&1&1\\ 1&0&0 \end{pmatrix} \begin{pmatrix} 1&0&0\\ 0&1&1\\ 1&0&0 \end{pmatrix}$$

Again row by column:

• First row: $$(1,0,0)A=(1,0,0)$$
• Second row: $$(1,1,1)A=(2,1,1)$$
• Third row: $$(1,0,0)A=(1,0,0)$$

So,

$$A^3=\begin{pmatrix} 1&0&0\\ 2&1&1\\ 1&0&0 \end{pmatrix}$$

3. Find the general pattern

From the first few powers:

$$A=\begin{pmatrix}1&0&0\\0&1&1\\1&0&0\end{pmatrix}, \quad A^2=\begin{pmatrix}1&0&0\\1&1&1\\1&0&0\end{pmatrix}, \quad A^3=\begin{pmatrix}1&0&0\\2&1&1\\1&0&0\end{pmatrix}$$

This suggests:

$$A^n=\begin{pmatrix} 1&0&0\\ n-1&1&1\\ 1&0&0 \end{pmatrix} \quad (n\ge 1)$$

We can verify by induction.

Assume

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

Then

$$A^{n+1}=A^nA$$

The rows transform as:

• First row remains $(1,0,0)$
• Second row: $$(n-1,1,1)A=(n,1,1)$$
• Third row remains $(1,0,0)$

Thus,

$$A^{n+1}=\begin{pmatrix} 1&0&0\\ n&1&1\\ 1&0&0 \end{pmatrix}$$

So the formula is correct.

4. Compute the required expression

Using the formula,

$$A^{2025}=\begin{pmatrix} 1&0&0\\ 2024&1&1\\ 1&0&0 \end{pmatrix} , \qquad A^{2020}=\begin{pmatrix} 1&0&0\\ 2019&1&1\\ 1&0&0 \end{pmatrix}$$

Therefore,

$$A^{2025}-A^{2020} = \begin{pmatrix} 0&0&0\\ 5&0&0\\ 0&0&0 \end{pmatrix}$$

Now compute $A^6-A$.

From the same formula,

$$A^6=\begin{pmatrix} 1&0&0\\ 5&1&1\\ 1&0&0 \end{pmatrix}$$

Hence,

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

Thus,

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

5. Check options

• A: $A^6-A$ ✅
• B: $A^5$ ❌
• C: $A^5-A$ ❌
• D: $A^6$ ❌

So the correct option is A.