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If the matrix $A=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1\end{array}\right]$ satisfies the equation $A ^{20}+\alpha A ^{19}+\beta A =\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1\end{array}\right]$ for some real numbers $\alpha$ and $\beta$, then $\beta-\alpha$ is equal to ........ .
$6$
$2$
$4$
$0$
Solution
$A =\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1\end{array}\right]$
$A ^{2}=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1\end{array}\right], A ^{3}=\left[\begin{array}{llc}1 & 0 & 0 \\ 0 & 8 & 0 \\ 3 & 0 & -1\end{array}\right]$
$A ^{4}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 1\end{array}\right]$
Hence
$A ^{20}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 2^{20} & 0 \\ 0 & 0 & 1\end{array}\right], A ^{19}=\left[\begin{array}{ccc}1 & 0 & 0 \\ 0 & 2^{19} & 0 \\ 3 & 0 & -1\end{array}\right]$
So $A ^{20}+\alpha A ^{19}+\beta A =\left[\begin{array}{ccc}1+\alpha+\beta & 0 & 0 \\ 0 & 2^{20}+\alpha .2^{19}+2 \beta & 0 \\ 3 \alpha+3 \beta & 0 & 1-\alpha-\beta\end{array}\right]$
$=\left[\begin{array}{lll}1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1\end{array}\right]$
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 $(\beta-\alpha)=4$
