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2022 PEA Q20

A,BA, B non-singular of the same order, C=BAB−1C = B A B^{-1}. For scalar λ\lambda, det⁡(C+λI)\det(C + \lambda I) equals

Reveal answer and solution

Answer

C

Solution

  1. 1

    C+λI=BAB−1+λBB−1=B(A+λI)B−1C + \lambda I = B A B^{-1} + \lambda B B^{-1} = B(A + \lambda I) B^{-1}.

  2. 2

    Taking determinants,

  3. 3
    det⁡(C+λI)=det⁡B⋅det⁡(A+λI)⋅det⁡(B−1)=det⁡(A+λI). \det(C + \lambda I) = \det B \cdot \det(A + \lambda I) \cdot \det(B^{-1}) = \det(A + \lambda I).

Answer structure / marking notes

Similar matrices share characteristic polynomials, hence the same det⁡(⋅+λI)\det(\cdot + \lambda I).

Content note

Imported from public/resources/isi/msqe/solutions/pea/2022/ISI_MSQE_PEA_2022_Solutions.tex. Question wording is retained from the available local TeX source; incomplete option blocks or ambiguous source status are flagged for review.