A $3 \times 3$ real symmetric matrix has eigenvalues $\lambda_1 = 11$, $\lambda_2 = 12$, and $\lambda_3 = 13$. The trace and determinant of the matrix are:
Options:
(A) Trace = 36, Determinant = 1716
(B) Trace = 1716, Determinant = 36
(C) Trace = 38, Determinant = 1716
(D) Trace = 36, Determinant = 1720
Answer Key:
A
Step-by-Step Solution & Explanation
By the spectral properties of square matrices:
1. The trace is the sum of the eigenvalues:
$\operatorname{tr}(A) = \sum_{i=1}^3 \lambda_i = 11 + 12 + 13 = 36$
2. The determinant is the product of the eigenvalues:
$\det(A) = \prod_{i=1}^3 \lambda_i = 11 \times 12 \times 13 = 1716$
Therefore, Trace = 36 and Determinant = 1716.