Mathematics - Matrices Question with Solution | TestHub

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

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

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

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, $\left(\beta -\alpha \right)=4$