A $3 \times 3$ real symmetric matrix has eigenvalues $\lambda_1 = 35$, $\lambda_2 = 36$, and $\lambda_3 = 37$. The trace and determinant of the matrix are:
Options:
(A) Trace = 108, Determinant = 46620
(B) Trace = 46620, Determinant = 108
(C) Trace = 110, Determinant = 46620
(D) Trace = 108, Determinant = 46624
Answer Key:
A
Step-by-Step Solution & Explanation
By the spectral properties of square matrices:
1. The trace is the sum of the eigenvalues:
$\operatorname{tr}(A) = \sum_{i=1}^3 \lambda_i = 35 + 36 + 37 = 108$
2. The determinant is the product of the eigenvalues:
$\det(A) = \prod_{i=1}^3 \lambda_i = 35 \times 36 \times 37 = 46620$
Therefore, Trace = 108 and Determinant = 46620.