Final answer:
To calculate A³ - 2A² + 3A + I for a matrix A with eigenvalues 1, 2, 3, we apply the Cayley-Hamilton theorem to express A³ in terms of lower matrix powers and simplify accordingly. The resulting expression is independent of A's entries and solely determined by its eigenvalues.
Step-by-step explanation:
To compute A³ - 2A² + 3A + I for a 3x3 matrix A with eigenvalues 1, 2, 3, we can make use of the Cayley-Hamilton theorem. This theorem states that a matrix satisfies its own characteristic equation. Since the eigenvalues are 1, 2, and 3, the characteristic polynomial of A is (λ - 1)(λ - 2)(λ - 3) which leads to λ³ - 6λ² + 11λ - 6 = 0 when expanded. Thus, for matrix A, A³ - 6A² + 11A - 6I = 0.
We can rearrange the equation to express A³ as a linear combination of lower powers of A: A³ = 6A² - 11A + 6I. Substituting this expression into the given formula, we get:
A³ - 2A² + 3A + I = (6A² - 11A + 6I) - 2A² + 3A + I.
Simplifying the equation, we combine like terms to get 4A² - 8A + 7I. If we further simplify, we'll notice this result doesn't depend on the entries of matrix A, but only on its eigenvalues.