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If $A = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & - 2 & 4 \end{matrix}] . I = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$
$A^{- 1} = \frac{1}{6} [A^{2} + c A + d I]$
where c, d ∈ R, the pair of values (c, d ) are

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(6, 11)

(6, –11)

(–6, 11)

(–6, –11)

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detailed solution

Correct option is C

GivenA=1000110-24 ,A-1=1660004-1021A2=1000110-241000110-24=1000-150-10-14cA=c000cc0-2c4c , dI=d000d000d∴A-1=16[A2+cA+dI]⇒6=1+c+d, (By equality of matrices)∴ (–6, 11) satisfy the relation

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If A=1000110-24 . I=100010001A-1=16[A2+cA+dI]where c, d ∈ R, the pair of values (c, d ) are

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If $A = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & - 2 & 4 \end{matrix}] . I = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{matrix}]$
$A^{- 1} = \frac{1}{6} [A^{2} + c A + d I]$
where c, d ∈ R, the pair of values (c, d ) are