A $3 \times 3$ real symmetric matrix has eigenvalues $\lambda_1 = 23$, $\lambda_2 = 24$, and $\lambda_3 = 25$. The trace and determinant of the matrix are:
Options:
(A) Trace = 72, Determinant = 13800
(B) Trace = 13800, Determinant = 72
(C) Trace = 74, Determinant = 13800
(D) Trace = 72, Determinant = 13804
Answer Key:
A
Step-by-Step Solution & Explanation
By the spectral properties of square matrices:
1. The trace is the sum of the eigenvalues:
$\operatorname{tr}(A) = \sum_{i=1}^3 \lambda_i = 23 + 24 + 25 = 72$
2. The determinant is the product of the eigenvalues:
$\det(A) = \prod_{i=1}^3 \lambda_i = 23 \times 24 \times 25 = 13800$
Therefore, Trace = 72 and Determinant = 13800.