Method $\mathrm{I}$

We evaluate ${\mathrm{A}}^2$ and ${\mathrm{A}}^3$ and write the given equation as ${\mathrm{A}\mathrm{A}}^{–1}=\mathrm{I}=\frac{1}{6}\left[{\mathrm{A}}^3+{\mathrm{c}\mathrm{A}}^2+\mathrm{d}\mathrm{A}\right]$ .

Comparing the corresponding elements on both the sides, we get $\mathrm{c}\mathrm{ }=-6,\mathrm{ }\mathrm{d}=11$.

Method $\mathrm{I}\mathrm{I}$

Characteristic equation for $A$ is $\left|A-xI\right|=0=\left|\begin{array}{ccc}1-x& 0& 0\\ 0& 1-x& 1\\ 0& -2& 4-x\end{array}\right|=0$

$\Rightarrow \left(1-x\right)\left[\left(1-x\right)\left(4-x\right)+2\right]=0$

Hence characteristic equation is $x^3-6x^2+11x-6=0$

Then by Caley Hamilton theorem

$A^3-6A^2+11A-6I=0$

multiply by $A^{-1}$ both the sides,

we get $\frac{1}{6}\left(A^2-6A+11I\right)=A^{-1}$ ...(i)

given $A^{-1}=\frac{1}{6}\left(A^2+cA+d\right)$ ...(ii)

then from equation (i) and (ii)

we get $\begin{array}{cc}c=-6& d=11\end{array}$

then $c+d=5$