If the matrix A=

Solution:

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

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

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

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

L.H.S $=A^{20}+\alpha A^{19}+\beta A=\left[\begin{array}{ccc}4+\alpha+\beta & 0 & 0 \\ 0 & 2^{20}+\alpha 2^{19}+2 \beta & 0 \\ 3 \alpha+3 \beta & 0 & 1-\alpha-\beta\end{array}\right]$

R.H.S $=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1\end{array}\right] \Rightarrow \alpha+\beta=0$ and $2^{20}+\alpha 2^{19}+2 \beta=4$

$\Rightarrow 2^{20}+\alpha\left(2^{19}-2\right)=4$

$\Rightarrow \alpha=\frac{4-2^{30}}{2^{19}-2}=-2$

$\Rightarrow \beta=2$

$\therefore \beta-\alpha=4$