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

6A−1=A2+cA+dI ⇒ 6I=A3+cA2+dAA2=1000110−241000110−24=1000−150−1014and A3=A2A=1000−150−10141000110−24=1000−11190−3846Now, 6I=A3+cA2+dA6=1+c+d, 0=19+5c+d6=−11−c+d6=46+14c+4d, 0=−38−10c−2d⇒ d=11, c=−6

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

6A−1=A2+cA+dI ⇒ 6I=A3+cA2+dAA2=1000110−241000110−24=1000−150−1014and A3=A2A=1000−150−10141000110−24=1000−11190−3846Now, 6I=A3+cA2+dA6=1+c+d, 0=19+5c+d6=−11−c+d6=46+14c+4d, 0=−38−10c−2d⇒ d=11, c=−6

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

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(-11,6)

(11,6)

(6,11)

answer is A.

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Detailed Solution

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