A $3 \times 3$ real symmetric matrix has eigenvalues $\lambda_1 = 20$, $\lambda_2 = 21$, and $\lambda_3 = 22$. The trace and determinant of the matrix are:
Options:
(A) Trace = 63, Determinant = 9240
(B) Trace = 9240, Determinant = 63
(C) Trace = 65, Determinant = 9240
(D) Trace = 63, Determinant = 9244
Answer Key:
A
Step-by-Step Solution & Explanation
By the spectral properties of square matrices:
1. The trace is the sum of the eigenvalues:
$\operatorname{tr}(A) = \sum_{i=1}^3 \lambda_i = 20 + 21 + 22 = 63$
2. The determinant is the product of the eigenvalues:
$\det(A) = \prod_{i=1}^3 \lambda_i = 20 \times 21 \times 22 = 9240$
Therefore, Trace = 63 and Determinant = 9240.