`[(1,0,0),(0,1,1),(1,0,0)]=A` then `A^(2025)-A^(2020)=`

Options

A. `A^5`
B. `A^6`
C. `A^6-A`
D. `A^5-A`

Answer

To solve the problem \( A^{2025} - A^{2020} \) where \( A = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{pmatrix} \), we will first find a pattern in the powers of the matrix \( A \).

### Step 1: Calculate \( A^2 \)

To find \( A^2 \), we multiply \( A \) by itself:

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

Calculating each element:

- First row:
- \( (1 \cdot 1 + 0 \cdot 0 + 0 \cdot 1) = 1 \)
- \( (1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0) = 0 \)
- \( (1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0) = 0 \)

- Second row:
- \( (0 \cdot 1 + 1 \cdot 0 + 1 \cdot 1) = 1 \)
- \( (0 \cdot 0 + 1 \cdot 1 + 1 \cdot 0) = 1 \)
- \( (0 \cdot 0 + 1 \cdot 1 + 1 \cdot 0) = 1 \)

- Third row:
- \( (1 \cdot 1 + 0 \cdot 0 + 0 \cdot 1) = 1 \)
- \( (1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0) = 0 \)
- \( (1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0) = 0 \)

Thus, we have:

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

### Step 2: Calculate \( A^3 \)

Now, we calculate \( A^3 = A^2 \cdot A \):

\[ A^3 = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 0 & 0 \end{pmatrix} \cdot \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{pmatrix} \]

Calculating each element:

- First row:
- \( (1 \cdot 1 + 0 \cdot 0 + 0 \cdot 1) = 1 \)
- \( (1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0) = 0 \)
- \( (1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0) = 0 \)

- Second row:
- \( (1 \cdot 1 + 1 \cdot 0 + 1 \cdot 1) = 2 \)
- \( (1 \cdot 0 + 1 \cdot 1 + 1 \cdot 0) = 1 \)
- \( (1 \cdot 0 + 1 \cdot 1 + 1 \cdot 0) = 1 \)

- Third row:
- \( (1 \cdot 1 + 0 \cdot 0 + 0 \cdot 1) = 1 \)
- \( (1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0) = 0 \)
- \( (1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0) = 0 \)

Thus, we have:

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

### Step 3: Identify the Pattern

From the calculations, we can see a pattern emerging:

- \( A^1 = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 0 \end{pmatrix} \)
- \( A^2 = \begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 1 \\ 1 & 0 & 0 \end{pmatrix} \)
- \( A^3 = \begin{pmatrix} 1 & 0 & 0 \\ 2 & 1 & 1 \\ 1 & 0 & 0 \end{pmatrix} \)

The general form for \( A^n \) appears to be:

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

### Step 4: Calculate \( A^{2025} \) and \( A^{2020} \)

Using the identified pattern:

\[ A^{2025} = \begin{pmatrix} 1 & 0 & 0 \\ 2024 & 1 & 1 \\ 1 & 0 & 0 \end{pmatrix} \]

\[ A^{2020} = \begin{pmatrix} 1 & 0 & 0 \\ 2019 & 1 & 1 \\ 1 & 0 & 0 \end{pmatrix} \]

### Step 5: Calculate \( A^{2025} - A^{2020} \)

Now we subtract \( A^{2020} \) from \( A^{2025} \):

\[ A^{2025} - A^{2020} = \begin{pmatrix} 1 & 0 & 0 \\ 2024 & 1 & 1 \\ 1 & 0 & 0 \end{pmatrix} - \begin{pmatrix} 1 & 0 & 0 \\ 2019 & 1 & 1 \\ 1 & 0 & 0 \end{pmatrix} \]

Calculating the subtraction element-wise:

- First row: \( 1 - 1 = 0 \), \( 0 - 0 = 0 \), \( 0 - 0 = 0 \)
- Second row: \( 2024 - 2019 = 5 \), \( 1 - 1 = 0 \), \( 1 - 1 = 0 \)
- Third row: \( 1 - 1 = 0 \), \( 0 - 0 = 0 \), \( 0 - 0 = 0 \)

Thus, we have:

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

### Final Answer

The final result is:

\[ \begin{pmatrix} 0 & 0 & 0 \\ 5 & 0 & 0 \\ 0 & 0 & 0 \end{pmatrix} \]