If the matrix A = [ cases 1 & 0 & 0 0 & 2 & 0 3 & 0 & - 1 cases ]...

A = \left[$\begin{cases}1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & - 1\end{cases}$\right]$</p><p>${A}^{2} = \left[$\begin{cases}1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1\end{cases}$\right], {A}^{3} = \left[$\begin{cases}1 & 0 & 0 \\ 0 & 8 & 0 \\ 3 & 0 & - 1\end{cases}$\right]$</p><p>${A}^{4} = \left[$\begin{cases}1 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 1\end{cases}$\right]$</p><p>Hence, </p><p>${A}^{20} = \left[$\begin{cases}1 & 0 & 0 \\ 0 & {2}^{20} & 0 \\ 0 & 0 & 1\end{cases}$\right], {A}^{19} = \left[$\begin{cases}1 & 0 & 0 \\ 0 & {2}^{19} & 0 \\ 3 & 0 & - 1\end{cases}$\right]$</p><p>So${A}^{20} + \alpha{A}^{19} + \betaA = \left[$\begin{cases}1 + \alpha + \beta & 0 & 0 \\ 0 & {2}^{20} + \alpha . {2}^{19} + 2\beta & 0 \\ 3\alpha + 3\beta & 0 & 1 - \alpha - \beta\end{cases}$\right] = \left[$\begin{cases}1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1\end{cases}$\right]$</p><p>Therefore$\alpha + \beta = 0$and${2}^{20} + {2}^{19}\alpha - 2\alpha = 4$</p><p>$⇒ \alpha = \dfrac{4\left(1 - {2}^{18}\right)}{2\left({2}^{18} - 1\right)} = - 2$</p><p>Hence$\beta = 2$</p><p>So, \left(\beta - \alpha\right) = 4$