Sum of cubed eigenvalues

For A=(3113)A = \begin{pmatrix} 3 & 1 \\ 1 & 3 \end{pmatrix}, compute λ13+λ23\lambda_1^3 + \lambda_2^3, the sum of the cubes of its eigenvalues - without first solving for the eigenvalues individually.

Show hints (2)+
  1. λ13+λ23=s33ps\lambda_1^3+\lambda_2^3 = s^3 - 3ps where s=trAs=\mathrm{tr}\,A, p=detAp=\det A.
  2. Here s=6s=6, p=8p=8, so 216144=72216 - 144 = 72.

Answer

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72

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Asked at: Multi-Strategy Quant, Data-Driven Research

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