A $3 \times 3$ real symmetric matrix has eigenvalues $\lambda_1 = 26$, $\lambda_2 = 27$, and $\lambda_3 = 28$. The trace and determinant of the matrix are:
Options:
(A) Trace = 81, Determinant = 19656
(B) Trace = 19656, Determinant = 81
(C) Trace = 83, Determinant = 19656
(D) Trace = 81, Determinant = 19660
Answer Key:
A
Step-by-Step Solution & Explanation
By the spectral properties of square matrices:
1. The trace is the sum of the eigenvalues:
$\operatorname{tr}(A) = \sum_{i=1}^3 \lambda_i = 26 + 27 + 28 = 81$
2. The determinant is the product of the eigenvalues:
$\det(A) = \prod_{i=1}^3 \lambda_i = 26 \times 27 \times 28 = 19656$
Therefore, Trace = 81 and Determinant = 19656.