A $3 \times 3$ real symmetric matrix has eigenvalues $\lambda_1 = 32$, $\lambda_2 = 33$, and $\lambda_3 = 34$. The trace and determinant of the matrix are:
Options:
(A) Trace = 99, Determinant = 35904
(B) Trace = 35904, Determinant = 99
(C) Trace = 101, Determinant = 35904
(D) Trace = 99, Determinant = 35908
Answer Key:
A
Step-by-Step Solution & Explanation
By the spectral properties of square matrices:
1. The trace is the sum of the eigenvalues:
$\operatorname{tr}(A) = \sum_{i=1}^3 \lambda_i = 32 + 33 + 34 = 99$
2. The determinant is the product of the eigenvalues:
$\det(A) = \prod_{i=1}^3 \lambda_i = 32 \times 33 \times 34 = 35904$
Therefore, Trace = 99 and Determinant = 35904.