Question

Choose the correct answer in the following question:
If $A=\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha \end{array}\right]$
is such that $A^2=I$. then
$\left(a\right)1+{\alpha }^2+\beta \gamma =0\left(b\right)1-{\alpha }^2+\beta \gamma =0\left(c\right)1-{\alpha }^2-\beta \gamma =0\left(d\right)1+{\alpha }^2-\beta \gamma =0$

Answer 1

Given, $A^2=I$
$\therefore AA=I\Rightarrow \left[\begin{array}{cc}\alpha \beta & \\ \gamma & -\alpha \end{array}\right]\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha \end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]\Rightarrow \left[\begin{array}{cc}{\alpha }^2+\beta \gamma & \alpha \beta -\alpha \beta \\ \alpha \gamma -\gamma \alpha & \gamma \beta +{\alpha }^2\end{array}\right]=\left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]$
On comparing the corresponding elements, we have
${\alpha }^2+\beta \gamma =1\Rightarrow {\alpha }^2+\beta \gamma -1=0\Rightarrow 1-{\alpha }^2-\beta \gamma =1$
$\therefore$ Hence, (c)is the correct option.