Question

If $\alpha ,\beta$ are the roots of the equation $\left[125\right]{\left[\begin{array}{cc}0& \frac{1}{2}\\ -1& \frac{1}{2}\end{array}\right]}^5{\left[\begin{array}{cc}1& -1\\ 2& 0\end{array}\right]}^{10}{\left[\begin{array}{cc}0& \frac{1}{2}\\ -1& \frac{1}{2}\end{array}\right]}^5\left[\begin{array}{c}{x}^2-5x+20\\ x+2\end{array}\right]=\left[40\right]$ , then $(1-\alpha )(1-\beta )-50$ is equal to ___

Answer 1

$\left[\begin{array}{cc}0& \frac{1}{2}\\ -1& \frac{1}{2}\end{array}\right]\left[\begin{array}{cc}1& -1\\ 2& 0\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\left[\begin{array}{cc}1& -1\\ 2& 0\end{array}\right]\left[\begin{array}{cc}0& \frac{1}{2}\\ -1& \frac{1}{2}\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\therefore \left[125\right]I^5{I}^5\left[\begin{array}{c}{x}^2-5x+20\\ x+2\end{array}\right]=\left[40\right]x^2+20x+30=0\Rightarrow \alpha +\beta =-20\alpha \beta =30\therefore (1-\alpha )(1-\beta )=1-(?+?)+??=51$