A $3 \times 3$ real symmetric matrix has eigenvalues $\lambda_1 = 2$, $\lambda_2 = 3$, and $\lambda_3 = 4$. The trace and determinant of the matrix are:
Options:
(A) Trace = 9, Determinant = 24
(B) Trace = 24, Determinant = 9
(C) Trace = 11, Determinant = 24
(D) Trace = 9, Determinant = 28
Answer Key:
A
Step-by-Step Solution & Explanation
By the spectral properties of square matrices:
1. The trace is the sum of the eigenvalues:
$\operatorname{tr}(A) = \sum_{i=1}^3 \lambda_i = 2 + 3 + 4 = 9$
2. The determinant is the product of the eigenvalues:
$\det(A) = \prod_{i=1}^3 \lambda_i = 2 \times 3 \times 4 = 24$
Therefore, Trace = 9 and Determinant = 24.