If $A=\begin{bmatrix}1&0&0\\0&1&1\\0&-2&4\end{bmatrix},6A^{-1}=A^2+ cA + dI$, then $(c, d) =$

(-6, 11)

Every square matrix A satisfies its characteristic equation i.e. $|A - λI| = 0$.

Here, $|A - λI| = 0$

$⇒\begin{vmatrix}1&0&0\\0&1- λ&1\\0&-2&4- λ\end{vmatrix}=0$

$⇒(1-λ) \{(1-λ) (4-λ)+2\}=0$

$⇒λ^3-6λ^2+11λ-6=0$

$⇒A^3-6A^2+11A - 6I = 0$

$⇒6I = A^3-6A^2+11A$

$⇒6A^{-1} = A^2 -6A + 11I$ [Multiplying both sides by $A^{-1}$]

$∴6A^{-1} = A^2 + cA + dI⇒c=-6$ and $d=11$