A $3 \times 3$ real symmetric matrix has eigenvalues $\lambda_1 = 8$, $\lambda_2 = 9$, and $\lambda_3 = 10$. The trace and determinant of the matrix are:
Options:
(A) Trace = 27, Determinant = 720
(B) Trace = 720, Determinant = 27
(C) Trace = 29, Determinant = 720
(D) Trace = 27, Determinant = 724
Answer Key:
A
Step-by-Step Solution & Explanation
By the spectral properties of square matrices:
1. The trace is the sum of the eigenvalues:
$\operatorname{tr}(A) = \sum_{i=1}^3 \lambda_i = 8 + 9 + 10 = 27$
2. The determinant is the product of the eigenvalues:
$\det(A) = \prod_{i=1}^3 \lambda_i = 8 \times 9 \times 10 = 720$
Therefore, Trace = 27 and Determinant = 720.