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Part BCSIR NET December 2025d-is-yours-to-choose-c-is-not-conflating-the-two-breaks-the-construction

D is yours to choose c is not conflating the two breaks the construction

Let I denote the 3×3 identity matrix. Given any three distinct matrices A,B,C∈M3(R)A, B, C \in M_{3}(\mathbb{R}), which of the following statements is necessarily true?

  1. A.There exists D∈M3(R)D \in M_{3}(\mathbb{R}) and a polynomial f∈R[X]f \in \mathbb{R}[X] satisfying f(A) = f(B) = f(C) = 0 and f(D) = I.✓ Correct
  2. B.There exists D∈M3(R)D \in M_{3}(\mathbb{R}) and a polynomial f∈R[X]f \in \mathbb{R}[X] satisfying f(A) = f(B) = 0, f(C) = I and f(D) = I.
  3. C.There exists D∈M3(R)D \in M_{3}(\mathbb{R}) and a polynomial f∈R[X]f \in \mathbb{R}[X] satisfying f(A) = 0, f(B) = f(C) = I and f(D) ≠ 0.
  4. D.There exists D∈M3(R)D \in M_{3}(\mathbb{R}) and a polynomial f∈R[X]f \in \mathbb{R}[X] satisfying f(A) = f(B) = 0, f(C) = I and f(D) = 0.

You have the answer. Trap Analysis is why the other three were written.

Not a worked solution repeated four times — the specific reasoning error each wrong option was built to reward, and why this one tests execution slip.

See pricing

50 are analysed free — try those first.

The trap it tests

Execution slip

The idea was right. The computation was not.

Drill statements like this

Related counterexample: Same characteristic polynomial ⇒ similar

More on this topic

The chapter behind this: Characteristic vs minimal polynomial — what each tells you — free to read

From Eigenvalues and Canonical Forms › Eigenvalues, characteristic & minimal polynomials, Cayley–Hamilton

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