Matrix Determinant Calculation

We are asked to find the determinant of the matrix expression $A^{2025} - 3A^{2024} + A^{2023}$, where $A = \begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix}$.

Key Concepts and Approach

This problem can be efficiently solved using the Cayley-Hamilton theorem. The steps involve finding the characteristic equation of matrix A, applying the theorem to simplify the expression, and then calculating the determinant.

• Matrix Simplification: Factor the expression $A^{2025} - 3A^{2024} + A^{2023}$.
• Cayley-Hamilton Theorem: A matrix satisfies its own characteristic equation.
• Determinant Properties: $\det(kM) = k^n \det(M)$ and $\det(M^p) = (\det(M))^p$, where $n$ is the dimension of the matrix.

Step-by-Step Solution

1. Find the Characteristic Equation

   The characteristic equation is given by $\det(A - \lambda I) = 0$. For matrix A:

   $ \det \begin{bmatrix} 2-\lambda & 3 \\ 3 & 5-\lambda \end{bmatrix} = (2-\lambda)(5-\lambda) - (3)(3) $

   $ = 10 - 7\lambda + \lambda^2 - 9 $

   $ = \lambda^2 - 7\lambda + 1 $

   The characteristic equation is $\lambda^2 - 7\lambda + 1 = 0$.

2. Apply Cayley-Hamilton Theorem

   According to the Cayley-Hamilton theorem, the matrix A satisfies its characteristic equation:

   $ A^2 - 7A + I = 0 $

   Rearranging this gives us $A^2 = 7A - I$.

3. Simplify the Matrix Expression

   Let the expression be $E = A^{2025} - 3A^{2024} + A^{2023}$. Factor out the lowest power of A:

   $ E = A^{2023}(A^2 - 3A + I) $

   Substitute $A^2 = 7A - I$ into the expression:

   $ E = A^{2023}((7A - I) - 3A + I) $

   $ E = A^{2023}(4A) $

   $ E = 4A^{2024} $

4. Calculate the Determinant

   We need to find $\det(E) = \det(4A^{2024})$.

   Using the determinant properties:

   $ \det(4A^{2024}) = 4^n \det(A^{2024}) $

   Since the matrix A is 2x2, $n=2$.

   $ \det(4A^{2024}) = 4^2 \det(A^{2024}) = 16 (\det(A))^{2024} $

5. Calculate $\det(A)$

   $ \det(A) = \det \begin{bmatrix} 2 & 3 \\ 3 & 5 \end{bmatrix} = (2)(5) - (3)(3) = 10 - 9 = 1 $

6. Final Calculation

   Substitute $\det(A) = 1$ back into the determinant expression:

   $ \det(E) = 16 \times (1)^{2024} = 16 \times 1 = 16 $

Final Answer

The determinant of the matrix $(A^{2025} - 3A^{2024} + A^{2023})$ is 16.