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if $A = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & - 2 & 4 \end{matrix}] a n d 6 A^{- 1} = A^{2} + c A + d I$ , then $(c, d) =$

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Correct option is A

By Caylay-Hamilton therem, every square Matrix satisfies its characteristic equation.A-λI=0⇒1-λ000λ-λ10-24-λ=0⇒λ3-6λ2+11λ-6=0⇒A3-6A2+11A-6I=0⇒6I=A3-6A2+11AMultiplying with A-1, we get6A−1=A2−6A+11I⇒c=-6 and d=11

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if A=1000110−24 and 6A−1=A2+cA+dI, then c,d=

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if $A = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & - 2 & 4 \end{matrix}] a n d 6 A^{- 1} = A^{2} + c A + d I$ , then $(c, d) =$