JEE Main 2021 — Mathematics Algebra

$$A=[\begin{matrix}1 & 0 & 0 \\ 0 & 2 & 0 \\ 3 & 0 & -1\end{matrix}]$$A=[103​020​00−1​]

$${A}^{2}=[\begin{matrix}1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1\end{matrix}],{A}^{3}=[\begin{matrix}1 & 0 & 0 \\ 0 & 8 & 0 \\ 3 & 0 & -1\end{matrix}]$$A2=[100​040​001​],A3=[103​080​00−1​]

$${A}^{4}=[\begin{matrix}1 & 0 & 0 \\ 0 & 16 & 0 \\ 0 & 0 & 1\end{matrix}]$$A4=[100​0160​001​]

Hence,

$${A}^{20}=[\begin{matrix}1 & 0 & 0 \\ 0 & {2}^{20} & 0 \\ 0 & 0 & 1\end{matrix}],{A}^{19}=[\begin{matrix}1 & 0 & 0 \\ 0 & {2}^{19} & 0 \\ 3 & 0 & -1\end{matrix}]$$A20=[100​02200​001​],A19=[103​02190​00−1​]

So ${A}^{20}+\alpha {A}^{19}+\beta A=[\begin{matrix}1+\alpha +\beta & 0 & 0 \\ 0 & {2}^{20}+\alpha .{2}^{19}+2\beta & 0 \\ 3\alpha +3\beta & 0 & 1-\alpha -\beta \end{matrix}]=[\begin{matrix}1 & 0 & 0 \\ 0 & 4 & 0 \\ 0 & 0 & 1\end{matrix}]$A20+αA19+βA=[1+α+β03α+3β​0220+α.219+2β0​001−α−β​]=[100​040​001​]

Therefore $\alpha +\beta =0$α+β=0 and ${2}^{20}+{2}^{19}\alpha -2\alpha =4$220+219α−2α=4

$$\Rightarrow \alpha =\frac{4(1-{2}^{18})}{2({2}^{18}-1)}=-2$$⇒α=2(218−1)4(1−218)​=−2

Hence $\beta =2$β=2

So, $(\beta -\alpha )=4$(β−α)=4