1. Key Concepts and Formulas

• Matrix Multiplication: To multiply two matrices $P$ and $Q$ to get $R = P \times Q$, the element $R_{ij}$ (in the $i$-th row and $j$-th column of $R$) is calculated as the sum of the products of corresponding elements from the $i$-th row of $P$ and the $j$-th column of $Q$. This is fundamental for calculating powers of a matrix.
• Pattern Recognition in Matrix Powers: For problems involving high powers of a matrix ($A^n$ where $n$ is large), direct repeated multiplication is impractical. The standard and most efficient approach is to calculate the first few powers ($A^2, A^3, A^4, \dots$) and identify a recurring pattern or a general formula for $A^n$. This method significantly simplifies the problem.
• Matrix Subtraction: To subtract two matrices of the same dimensions, simply subtract their corresponding elements.

2. Step-by-Step Solution

Step 1: Calculate the first few powers of matrix A to identify a pattern. We are given the matrix: $A = \left[ {\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right]$ Our goal is to find a general form for $A^n$. To do this, we compute $A^2$, $A^3$, and $A^4$.

• Calculate $A^2$: We multiply $A$ by itself: $A^2 = A \times A = \left[ {\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right] \left[ {\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right]$ Performing the matrix multiplication (row by column for each element): $A^2 = \left[ {\begin{matrix} (1 \cdot 1 + 0 \cdot 0 + 0 \cdot 1) & (1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0) & (1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0) \\ (0 \cdot 1 + 1 \cdot 0 + 1 \cdot 1) & (0 \cdot 0 + 1 \cdot 1 + 1 \cdot 0) & (0 \cdot 0 + 1 \cdot 1 + 1 \cdot 0) \\ (1 \cdot 1 + 0 \cdot 0 + 0 \cdot 1) & (1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0) & (1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0) \end{matrix} } \right] = \left[ {\begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right]$

• Calculate $A^3$: We multiply $A^2$ by $A$: $A^3 = A^2 \times A = \left[ {\begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right] \left[ {\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right]$ Performing the matrix multiplication: $A^3 = \left[ {\begin{matrix} (1 \cdot 1 + 0 \cdot 0 + 0 \cdot 1) & (1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0) & (1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0) \\ (1 \cdot 1 + 1 \cdot 0 + 1 \cdot 1) & (1 \cdot 0 + 1 \cdot 1 + 1 \cdot 0) & (1 \cdot 0 + 1 \cdot 1 + 1 \cdot 0) \\ (1 \cdot 1 + 0 \cdot 0 + 0 \cdot 1) & (1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0) & (1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0) \end{matrix} } \right] = \left[ {\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right]$

• Calculate $A^4$: We multiply $A^3$ by $A$: $A^4 = A^3 \times A = \left[ {\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right] \left[ {\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right]$ Performing the matrix multiplication: $A^4 = \left[ {\begin{matrix} (1 \cdot 1 + 0 \cdot 0 + 0 \cdot 1) & (1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0) & (1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0) \\ (2 \cdot 1 + 1 \cdot 0 + 1 \cdot 1) & (2 \cdot 0 + 1 \cdot 1 + 1 \cdot 0) & (2 \cdot 0 + 1 \cdot 1 + 1 \cdot 0) \\ (1 \cdot 1 + 0 \cdot 0 + 0 \cdot 1) & (1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0) & (1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0) \end{matrix} } \right] = \left[ {\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right]$

Step 2: Identify the general pattern for $A^n$. Let's list the calculated powers of $A$ along with $A^1$: $A^1 = \left[ {\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right]$ $A^2 = \left[ {\begin{matrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right]$ $A^3 = \left[ {\begin{matrix} 1 & 0 & 0 \\ 2 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right]$ $A^4 = \left[ {\begin{matrix} 1 & 0 & 0 \\ 3 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right]$ Upon careful observation, we can see a clear pattern:

• All elements in the matrix remain constant across different powers of $A$, except for the element in the second row, first column (position (2,1)).
• The element at position (2,1) follows a sequence: $0$ for $A^1$, $1$ for $A^2$, $2$ for $A^3$, $3$ for $A^4$. This indicates that for $A^n$, the element at position (2,1) is $(n-1)$.

Step 3: Formulate the general expression for $A^n$. Based on the pattern identified, the general form of $A^n$ for any positive integer $n \ge 1$ is: $A^n = \left[ {\begin{matrix} 1 & 0 & 0 \\ n-1 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right]$

Step 4: Calculate $A^{2025} - A^{2020}$ using the general formula. Using our generalized formula for $A^n$: $A^{2025} = \left[ {\begin{matrix} 1 & 0 & 0 \\ 2025-1 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right] = \left[ {\begin{matrix} 1 & 0 & 0 \\ 2024 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right]$ $A^{2020} = \left[ {\begin{matrix} 1 & 0 & 0 \\ 2020-1 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right] = \left[ {\begin{matrix} 1 & 0 & 0 \\ 2019 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right]$ Now, we perform the matrix subtraction: $A^{2025} - A^{2020} = \left[ {\begin{matrix} 1 & 0 & 0 \\ 2024 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right] - \left[ {\begin{matrix} 1 & 0 & 0 \\ 2019 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right]$ $A^{2025} - A^{2020} = \left[ {\begin{matrix} 1-1 & 0-0 & 0-0 \\ 2024-2019 & 1-1 & 1-1 \\ 1-1 & 0-0 & 0-0 \end{matrix} } \right] = \left[ {\begin{matrix} 0 & 0 & 0 \\ 5 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} } \right]$

Step 5: Compare the result with the given options. We need to determine which of the options evaluates to the matrix $\left[ {\begin{matrix} 0 & 0 & 0 \\ 5 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} } \right]$.

• (A) $A^6 - A$: Using our general formula for $A^n$: $A^6 = \left[ {\begin{matrix} 1 & 0 & 0 \\ 6-1 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right] = \left[ {\begin{matrix} 1 & 0 & 0 \\ 5 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right]$ Then, $A^6 - A = \left[ {\begin{matrix} 1 & 0 & 0 \\ 5 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right] - \left[ {\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right] = \left[ {\begin{matrix} 1-1 & 0-0 & 0-0 \\ 5-0 & 1-1 & 1-1 \\ 1-1 & 0-0 & 0-0 \end{matrix} } \right] = \left[ {\begin{matrix} 0 & 0 & 0 \\ 5 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} } \right]$ This result matches our calculated value for $A^{2025} - A^{2020}$.

Since option (A) yields the same result, it is the correct answer. (We can quickly verify that other options would not match).

3. Common Mistakes & Tips

• Errors in Matrix Multiplication: This is a frequent source of mistakes. Be meticulous when calculating each element of the product matrix, especially with the sum of products.
• Premature Generalization: Ensure you calculate enough powers ($A^2, A^3, A^4$) to confidently establish a pattern. Sometimes, a pattern might appear after $A^2$ but change for $A^3$.
• Checking Base Case: Always verify that your derived general formula for $A^n$ holds true for $n=1$. In this case, $A^1 = \left[ {\begin{matrix} 1 & 0 & 0 \\ 1-1 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right] = \left[ {\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right]$, which matches the given matrix $A$.
• Alternative Methods: For some matrices, the Cayley-Hamilton theorem or diagonalization can be used to find $A^n$, but pattern recognition is often faster for simpler matrices like this one.

4. Summary

The problem required us to find the value of $A^{2025} - A^{2020}$. The most effective approach involved determining a general formula for $A^n$. By computing $A^2, A^3,$ and $A^4$, we identified a pattern where only the element at position (2,1) changed, specifically taking the value $(n-1)$. This led to the general formula $A^n = \left[ {\begin{matrix} 1 & 0 & 0 \\ n-1 & 1 & 1 \\ 1 & 0 & 0 \end{matrix} } \right]$. Applying this, we found $A^{2025} - A^{2020} = \left[ {\begin{matrix} 0 & 0 & 0 \\ 5 & 0 & 0 \\ 0 & 0 & 0 \end{matrix} } \right]$. Comparing this result with the given options, we found that $A^6 - A$ also yields the same matrix.

5. Final Answer

The final answer is $\boxed{\text{A}}$.