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

$1-\alpha^2-\beta \gamma = 0 $

The correct answer is Option (3) → $1-\alpha^2-\beta \gamma = 0 $

$A^2=\begin{bmatrix} \alpha & \beta \\\gamma & -\alpha \end{bmatrix}\begin{bmatrix} \alpha & \beta \\\gamma & -\alpha \end{bmatrix}=\begin{bmatrix} α^2+βγ & 0 \\0 & α^2+βγ \end{bmatrix}=\begin{bmatrix} 1 & 0 \\0 & 1 \end{bmatrix}$

$1-α^2-βγ=0$