If A=1000110−24 and I=1 0 00 1 00 0 1 and A−1=16A2+cA+dI, the value of c and d are

We have 6A−1=A2+cA+dI⇒ 6I=A3+cA2+dANow, A2=A⋅A =1000110−241000110−24=1000−150−1014and A3=A2A =1000−150−10141000110−24=1000−11190−3846Now, 6I=A3+cA2+dA gives 6=1+c+d, 0=19+5c+d 6=−11−c+5d, 0=−38−10c−2d 6=46+14c+4dThus, c=-6 and d=11.

If A=1000110−24 and I=1 0 00 1 00 0 1 and A−1=16A2+cA+dI, the value of c and d are

We have 6A−1=A2+cA+dI⇒ 6I=A3+cA2+dANow, A2=A⋅A =1000110−241000110−24=1000−150−1014and A3=A2A =1000−150−10141000110−24=1000−11190−3846Now, 6I=A3+cA2+dA gives 6=1+c+d, 0=19+5c+d 6=−11−c+5d, 0=−38−10c−2d 6=46+14c+4dThus, c=-6 and d=11.

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If $A = (\begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & - 2 & 4 \end{array})$ and $I = (\begin{array}{lll} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array})$ and $A^{- 1} = [\frac{1}{6} (A^{2} + c A + d I)],$ the value of c and d are

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$(- 6, - 1)$

$(6, 11)$

$(- 6, 11)$

$(6, - 11)$

answer is C.

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Detailed Solution

We have $6 A^{- 1} = A^{2} + c A + d I$

$\Rightarrow 6 I = A^{3} + c A^{2} + d A$

Now,

$A^{2} = A \cdot A$

$\begin{array}{l} = (\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & - 2 & 4 \end{array}) (\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & - 2 & 4 \end{array}) \\ = (\begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 1 & 5 \\ 0 & - 10 & 14 \end{array}) \end{array}$

and

$A^{3} = A^{2} A$

$\begin{array}{l} = (\begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 1 & 5 \\ 0 & - 10 & 14 \end{array}) (\begin{array}{rrr} 1 & 0 & 0 \\ 0 & 1 & 1 \\ 0 & - 2 & 4 \end{array}) \\ = (\begin{array}{rrr} 1 & 0 & 0 \\ 0 & - 11 & 19 \\ 0 & - 38 & 46 \end{array}) \end{array}$