If $A=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& −2& 4\end{array}\right],6A^{−1}=A^2+cA+dI$ then (c,d) is

$6A^{−1}=A^2+cA+dI$

$⇒6I=A^3+c{A}^2+dA$

We have 

$A^2=\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& −2& 4\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& −2& 4\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& −1& 5\\ 0& −10& 14\end{array}\right]$

and $\begin{array}{r}{A}^3=A^{2}A\end{array}$

$=\left[\begin{array}{ccc}1& 0& 0\\ 0& −1& 5\\ 0& −10& 14\end{array}\right]\left[\begin{array}{ccc}1& 0& 0\\ 0& 1& 1\\ 0& −2& 4\end{array}\right]=\left[\begin{array}{ccc}1& 0& 0\\ 0& −11& 19\\ 0& −38& 46\end{array}\right]$

Now, 

$\begin{array}{l}6I=A^3+c{A}^2+dA\\ 6=1+c+d,0=19+5c+d\\ 6=−11−c+d\\ 6=46+14c+4d,0=−38−10c−2d\\ ⇒d=11,c=−6\end{array}$