If [1000110−24],6A−1=A2+cA+dI , then (c,d) is

6A=A+cA+dI⇒6I=A+cA+dAA2=[1000110−24][1000110−24]=[1000−150−1014]And A3=A2A=[1000−150−1014][1000110−24]=[1000−11190−3846]Now,6I=A3+cA2+dA 6=1+c+d,0=19+5c+d 6=−11−c+d6=46+14c+4d,0=−38−10c−2d ⇒d=11,c=−6

If [1000110−24],6A−1=A2+cA+dI , then (c,d) is

6A=A+cA+dI⇒6I=A+cA+dAA2=[1000110−24][1000110−24]=[1000−150−1014]And A3=A2A=[1000−150−1014][1000110−24]=[1000−11190−3846]Now,6I=A3+cA2+dA 6=1+c+d,0=19+5c+d 6=−11−c+d6=46+14c+4d,0=−38−10c−2d ⇒d=11,c=−6

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

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$(6, 11)$

$(- 6, 11)$

$(11, 6)$

$(- 11, 6)$

answer is A.

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Detailed Solution

$6 A = A + c A + d I$

$\Rightarrow 6 I = A + c A + d A$

$A^{2} = [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & - 2 & 4 \end{matrix}] [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & - 2 & 4 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & - 1 & 5 \\ 0 & - 10 & 14 \end{matrix}]$

And $A^{3} = A^{2} A$
$= [\begin{matrix} 1 & 0 & 0 \\ 0 & - 1 & 5 \\ 0 & - 10 & 14 \end{matrix}] [\begin{matrix} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & - 2 & 4 \end{matrix}] = [\begin{matrix} 1 & 0 & 0 \\ 0 & - 11 & 19 \\ 0 & - 38 & 46 \end{matrix}]$
Now,
$6 I = A^{3} + c A^{2} + d A$
$6 = 1 + c + d, 0 = 19 + 5 c + d$
$6 = - 11 - c + d$
$6 = 46 + 14 c + 4 d, 0 = - 38 - 10 c - 2 d$
$\Rightarrow d = 11, c = - 6$