Let $A$ be a $3 \times 3$ matrix with eigenvalues 1, 2, 3. Then $\det(A)$ equals
A$0$, singular matrix
B$6$, the product
C$5$, the trace value
D$1$, the identity case
Answer & Solution
Correct answer: B. $6$, the product
1. RECALL: $\det(A) = \lambda_1 \lambda_2 \lambda_3$ (product of eigenvalues).
2. $\det(A) = 1 \cdot 2 \cdot 3 = 6$.
3. Sanity: $\text{tr}(A) = 1 + 2 + 3 = 6$ (sum), $\det = 6$ (product). Both checks pass.
4. Option A would mean $A$ is singular (a zero eigenvalue). Option C is the SUM of eigenvalues (option for trace). Option D would mean all eigenvalues are 1.
5. This product formula works regardless of whether eigenvalues are distinct or repeated (counted with multiplicity).
_Source: Sergei Treil, "Linear Algebra Done Wrong", §4.6 (Determinant from eigenvalues)._
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