If$A = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}\alpha &\beta \\\gamma &{ - \alpha }\end{array}} \right]$is such that $A^2 = I$, then

.:

Option c is correct

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

Now, ${A^2} = I$
$\Rightarrow$ $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}\alpha &\beta \\\gamma &{ - \alpha }\end{array}} \right]\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}\alpha &\beta \\\gamma &{ - \alpha }\end{array}} \right] = \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0\\0&1\end{array}} \right]$

$\Rightarrow$ $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{{\alpha ^2} + \beta \gamma }&{\alpha \beta - \alpha \beta }\\{\gamma \alpha - \alpha \gamma }&{\gamma \beta + {\alpha ^2}}\end{array}} \right]$ $= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0\\0&1\end{array}} \right]$

$\Rightarrow$ $\left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}{{\alpha ^2} + \beta \gamma }&0\\0&{\gamma \beta + {\alpha ^2}}\end{array}} \right]$ $= \left[ {\begin{array}{rrrrrrrrrrrrrrrrrrrr}1&0\\0&1\end{array}} \right]$

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