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$A = [\begin{array}{rrr} 1 0 0 \\ 0 1 1 \\ 0 - 2 4 \end{array}], I = [\begin{array}{lll} 1 0 0 \\ 0 1 0 \\ 0 0 1 \end{array}] t h e n A^{- 1} = \frac{1}{6} (A^{2} + C A + D I)$ then $C$ and $D$ equal to

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$- 11, 6$

$- 6, - 11$

$- 6, 11$

$6, 11$

answer is B.

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Detailed Solution

$A A^{- 1} = I \Rightarrow \frac{1}{6} [A^{3} + C A^{2} + d A] = I . . . (1)$

Now compute $A^{2}$ and $A^{3}$ and put in (a) and compare the coefficients on both sides $∴ C = - 6, d = 11 .$

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