Question

If $A=(\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha \end{array})$ is such that $A^2=I$, then

• A. $1+{\alpha }^2+\beta \gamma =0$
• B. $1-{\alpha }^2+\beta \gamma =0$
• C. $1-{\alpha }^2-\beta \gamma =0$
• D. $1+{\alpha }^2-\beta \gamma =0$

Answer 1

The correct option is A $1+{\alpha }^2+\beta \gamma =0$

$A=\left[\begin{array}{cc}\alpha & \beta \\ \gamma & -\alpha \end{array}\right]$

${\Rightarrow A}^2=I$

$\Rightarrow$ Then $1+{\alpha }^2+\beta \gamma =0$

Hence, the answer is $1+{\alpha }^2+\beta \gamma =0.$