A $3 \times 3$ real symmetric matrix has eigenvalues $\lambda_1 = 29$, $\lambda_2 = 30$, and $\lambda_3 = 31$. The trace and determinant of the matrix are:
Options:
(A) Trace = 90, Determinant = 26970
(B) Trace = 26970, Determinant = 90
(C) Trace = 92, Determinant = 26970
(D) Trace = 90, Determinant = 26974
Answer Key:
A
Step-by-Step Solution & Explanation
By the spectral properties of square matrices:
1. The trace is the sum of the eigenvalues:
$\operatorname{tr}(A) = \sum_{i=1}^3 \lambda_i = 29 + 30 + 31 = 90$
2. The determinant is the product of the eigenvalues:
$\det(A) = \prod_{i=1}^3 \lambda_i = 29 \times 30 \times 31 = 26970$
Therefore, Trace = 90 and Determinant = 26970.