A $3 \times 3$ real symmetric matrix has eigenvalues $\lambda_1 = 14$, $\lambda_2 = 15$, and $\lambda_3 = 16$. The trace and determinant of the matrix are:
Options:
(A) Trace = 45, Determinant = 3360
(B) Trace = 3360, Determinant = 45
(C) Trace = 47, Determinant = 3360
(D) Trace = 45, Determinant = 3364
Answer Key:
A
Step-by-Step Solution & Explanation
By the spectral properties of square matrices:
1. The trace is the sum of the eigenvalues:
$\operatorname{tr}(A) = \sum_{i=1}^3 \lambda_i = 14 + 15 + 16 = 45$
2. The determinant is the product of the eigenvalues:
$\det(A) = \prod_{i=1}^3 \lambda_i = 14 \times 15 \times 16 = 3360$
Therefore, Trace = 45 and Determinant = 3360.