If the matrix

$A=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1\end{array}\right]$

$\mathrm{A}^{2}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1\end{array}\right], \mathrm{A}^{3}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 8 & 0 \\ 3 & 0 & -1\end{array}\right]$

$\mathrm{A}^{4}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 1\end{array}\right]$

Hence

$\mathrm{A}^{20}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 2^{20} & 0 \\ 0 & 0 & 1\end{array}\right], \mathrm{A}^{19}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 2^{19} & 0 \\ 3 & 0 & -1\end{array}\right]$

So $\mathrm{A}^{20}+\alpha \mathrm{A}^{19}+\beta \mathrm{A}=\left[\begin{array}{ccc}1+\alpha+\beta & 0 & 0 \\ 0 & 2^{20}+\alpha .2^{19}+2 \beta & 0 \\ 3 \alpha+3 \beta & 0 & 1-\alpha-\beta\end{array}\right]$

$=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1\end{array}\right]$

Therefore $\alpha+\beta=0$ and $2^{20}+2^{19} \alpha-2 \alpha=4$

$\Rightarrow \alpha=\frac{4\left(1-2^{18}\right)}{2\left(2^{18}-1\right)}=-2$

hence $\beta=2$

so $(\beta-\alpha)=4$