A $3 \times 3$ real matrix $A$ has trace equal to $10$ and determinant equal to $24$. If one of the eigenvalues of $A$ is $2$, the sum of the other two eigenvalues is ________ (answer in integer).
Answer Key:
8
Step-by-Step Solution & Explanation
Let the eigenvalues of $A$ be $\lambda_1, \lambda_2, \lambda_3$.
We know that the sum of the eigenvalues equals the trace:
$\lambda_1 + \lambda_2 + \lambda_3 = \operatorname{trace}(A) = 10$
Given that $\lambda_1 = 2$:
$2 + \lambda_2 + \lambda_3 = 10 \implies \lambda_2 + \lambda_3 = 8$
(We can also verify: $\lambda_1 \lambda_2 \lambda_3 = \det(A) = 24 \implies 2(\lambda_2 \lambda_3) = 24 \implies \lambda_2 \lambda_3 = 12$. The roots are $4$ and $6$, and indeed $4 + 6 = 8$).