If a $3 \times 3$ real symmetric matrix $A$ has eigenvalues $1, -2,$ and $3$, then the determinant of the matrix $A^2 - 2A + I$ is:
Options:
(A) 0
(B) 16
(C) 36
(D) 72
Answer Key:
A
Step-by-Step Solution & Explanation
For an eigenvalue $\lambda$ of $A$, the eigenvalue of $f(A) = A^2 - 2A + I = (A - I)^2$ is $(\lambda - 1)^2$. For $\lambda_1 = 1$, the eigenvalue is $(1 - 1)^2 = 0$. Since one eigenvalue is $0$, the determinant is the product of eigenvalues: $\det(f(A)) = 0 \times (-3)^2 \times (2)^2 = 0$.