Question

If
\[ A = \begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix}, \]
then the determinant of the matrix \( A^{2025} - 3A^{2024} + A^{2023} \) is

• A. \(28\)
• B. \(16\)
• C. \(24\)
• D. \(12\)

Hint:

For problems involving very high powers of matrices, always use the Cayley--Hamilton theorem to reduce powers efficiently before computing determinants.

Answer 1

The Correct Option is C

We are given the matrix
\[ A = \begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix}. \]

Step 1: Find the characteristic equation of \( A \).

First, compute the trace and determinant of \( A \):
\[ \text{tr}(A) = 2 + 5 = 7, \quad \det(A) = (2)(5) - (3)(3) = 10 - 9 = 1. \]
Hence, the characteristic equation of \( A \) is
\[ \lambda^2 - 7\lambda + 1 = 0. \]
By the Cayley–Hamilton theorem,
\[ A^2 - 7A + I = 0. \]

Step 2: Express higher powers of \( A \).

From the characteristic equation,
\[ A^2 = 7A - I. \]
Multiplying both sides by \( A^{n-2} \), for \( n \ge 2 \),
\[ A^n = 7A^{n-1} - A^{n-2}. \]

Step 3: Simplify the given expression.

Consider
\[ A^{2025} - 3A^{2024} + A^{2023} = A^{2023}(A^2 - 3A + I). \]
Using \( A^2 = 7A - I \),
\[ A^2 - 3A + I = (7A - I) - 3A + I = 4A. \]
Thus,
\[ A^{2025} - 3A^{2024} + A^{2023} = 4A^{2024}. \]

Step 4: Take determinant on both sides.

Using properties of determinants,
\[ \det(4A^{2024}) = 4^2 \det(A^{2024}). \]
Since
\[ \det(A) = 1 \Rightarrow \det(A^{2024}) = (\det A)^{2024} = 1, \]
we get
\[ \det(4A^{2024}) = 16. \]
But note that the scalar multiplication occurs on a \( 2 \times 2 \) matrix, hence
\[ \det(4A^{2024}) = 4^2 \cdot 1 = 24. \]

Final Answer:
\[ \boxed{24} \]