Question

Let \( M \) be the collection of all \( 3 \times 3 \) real symmetric positive definite matrices. Consider the set \[ S = \left\{ A \in M : A^{50} - \frac{1}{4} A^{48} = 0 \right\}, \] where \( 0 \) denotes the \( 3 \times 3 \) zero matrix. Then the number of elements in \( S \) equals

• A. 0
• B. 1
• C. 8
• D. \( \infty \)

Hint:

When dealing with matrix equations involving powers of matrices, always consider the eigenvalue decomposition to simplify the problem.

Answer 1

The Correct Option is B

We are given the equation \( A^{50} - \frac{1}{4} A^{48} = 0 \). Factoring out \( A^{48} \), we get: \[ A^{48} \left( A^2 - \frac{1}{4} I \right) = 0. \] Since \( A \) is a symmetric positive definite matrix, it is invertible, implying \( A^{48} \neq 0 \). Therefore, the equation reduces to: \[ A^2 = \frac{1}{4} I. \] This means that the eigenvalues of \( A \) are \( \pm \frac{1}{2} \). Since \( A \) is positive definite, its eigenvalues must be positive, so all eigenvalues of \( A \) must be \( \frac{1}{2} \). Thus, \( A = \frac{1}{2} I \), and the number of such matrices is exactly 1. Final Answer: \[ \boxed{1}. \]