Solution:

$A=\begin{bmatrix} 2 & 4 \\ 0 & 3 \end{bmatrix}$

The characteristic equation is $|A-\lambda I| = 0$

$A=\left |\begin{matrix} 2-\lambda & 4 \\ 0 & 3-\lambda \end{matrix} \right|$=0

$(2-\lambda)(3-\lambda)-0=0$

$\lambda =2,\lambda =3 $

The eigen values of A are 2, 3

Now the eigen vaues of $A^{-1}$ are, $A^{-1}=\frac{1}{2} \ and \ \frac{1}{3} $

For $6A^{-1} = 6(\frac{1}{2}) \ and \ 6 (\frac {1}{3})$

$6A^{-1}=3 \ and \ 2$-------(1)

similarly eigen values of $A^{3} $ are

$A^{3}=(2)^3$ and $(3)^{3}$

$A^{3}$ =9 and 27 --------(2)

and for $2I = 2(1) =2$-----(3)

Hence eigen values for

i) $6A^{-1} + A^3 +2I $

=$3+9+2$

=$14$

ii) $6A^{-1} +A^3+2I$

=$2+27+2$

=$31$

Therefore eigen values for $6A^{-1}+A^{3}+2I$ is $14 \ and \ 31.$