Question

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

• A. \(A^6\)
• B. \(A^5 - A\)
• C. \(A^6 - A\)
• D. \(A^5\)

Hint:

The Cayley-Hamilton theorem is an extremely powerful tool for simplifying high powers of matrices. Always look for it as a potential method when you see expressions like \(A^n\) where \(n\) is large.

Answer 1

The Correct Option is C

Step 1: Understanding the Question
We need to compute the difference of two high powers of a given matrix \( A \). A direct calculation is infeasible, so we should look for a pattern or use the Cayley-Hamilton theorem.

Step 2: Key Formula or Approach
The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation. We will find the characteristic equation of \( A \) and use it to establish a recurrence relation for powers of \( A \).

Step 3: Detailed Explanation
Find the Characteristic Equation:
The characteristic equation is \( \det(A - \lambda I) = 0 \).

\[ \begin{vmatrix} 1-\lambda & 0 & 0 \\ 0 & 1-\lambda & 1 \\ 1 & 0 & -\lambda \end{vmatrix} = 0 \]

\[ (1-\lambda)((1-\lambda)(-\lambda) - 0) - 0 + 0 = 0 \]

\[ (1-\lambda)(-\lambda(1-\lambda)) = 0 \implies -\lambda(1-\lambda)^2 = 0 \implies \lambda( \lambda^2 - 2\lambda + 1) = 0 \implies \lambda^3 - 2\lambda^2 + \lambda = 0 \]

By the Cayley-Hamilton theorem, \( A \) satisfies this equation:

\[ A^3 - 2A^2 + A = O \implies A^3 = 2A^2 - A \]

Find a general pattern for \(A^n\):
\(A^3 = 2A^2 - A\)
\(A^4 = A \cdot A^3 = A(2A^2 - A) = 2A^3 - A^2 = 2(2A^2 - A) - A^2 = 4A^2 - 2A - A^2 = 3A^2 - 2A\)
\(A^5 = A \cdot A^4 = A(3A^2 - 2A) = 3A^3 - 2A^2 = 3(2A^2 - A) - 2A^2 = 6A^2 - 3A - 2A^2 = 4A^2 - 3A\)
By induction, we can see a pattern: \(A^n = (n-1)A^2 - (n-2)A\) for \(n \geq 2\).

Calculate the required expression:
Using the pattern:

\[ A^{2025} = (2025-1)A^2 - (2025-2)A = 2024A^2 - 2023A \]

\[ A^{2020} = (2020-1)A^2 - (2020-2)A = 2019A^2 - 2018A \]

Now, subtract the two:

\[ A^{2025} - A^{2020} = (2024A^2 - 2023A) - (2019A^2 - 2018A) \]

\[ = (2024 - 2019)A^2 - (2023 - 2018)A = 5A^2 - 5A \]

Compare with the options:
We need to check which option equals \(5A^2 - 5A\). Let's evaluate option (C), \(A^6 - A\). Using our pattern for \(A^6\):

\[ A^6 = (6-1)A^2 - (6-2)A = 5A^2 - 4A \]

So, \(A^6 - A = (5A^2 - 4A) - A = 5A^2 - 5A\). Since both \(A^{2025} - A^{2020}\) and \(A^6 - A\) simplify to the same expression \(5A^2 - 5A\), they are equal.

Step 4: Final Answer
\(A^{2025} - A^{2020} = A^6 - A\).