c
(c) Given $A = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&1\\0&{ - 2}&4\end{array}} \right],\,\,{A^{ - 1}} = \frac{1}{6}\left[ {\begin{array}{*{20}{c}}6&0&0\\0&4&{ - 1}\\0&2&1\end{array}} \right]$
${A^2} = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&1\\0&{ - 2}&4\end{array}} \right]\,\left[ {\begin{array}{*{20}{c}}1&0&0\\0&1&1\\0&{ - 2}&4\end{array}} \right]\, = \left[ {\begin{array}{*{20}{c}}1&0&0\\0&{ - 1}&5\\0&{ - 10}&{14}\end{array}} \right]$
$cA = \left[ {\begin{array}{*{20}{c}}c&0&0\\0&c&c\\0&{ - 2c}&{4c}\end{array}} \right]\,$; $dI = \left[ {\begin{array}{*{20}{c}}d&0&0\\0&d&0\\0&0&d\end{array}} \right]$
$\therefore $ By ${A^{ - 1}} = \frac{1}{6}[{A^2} + cA + dI]$
$ 6 = 1 + c + d$,(By equality of matrices)
$\therefore $ $ (-6,11)$ satisfy the relation.