If $$A = \left( {\matrix{ 0 & {\sin \alpha } \cr {\sin \alpha } & 0 \cr } } \right)$$ and $\det \left( {{A^2} - {1 \over 2}I} \right) = 0$, then a possible value of $\alpha$ is :

$${A^2} = \left[ {\matrix{ 0 &amp; {\sin \alpha } \cr {\sin \alpha } &amp; 0 \cr } } \right]\left[ {\matrix{ 0 &amp; {\sin \alpha } \cr {\sin \alpha } &amp; 0 \cr } } \right] = \left[ {\matrix{ {{{\sin }^2}\alpha } &amp; 0 \cr 0 &amp; {{{\sin }^2}\alpha } \cr } } \right]$$<br><br>$${A^2} - {1 \over 2}I = \left[ {\matrix{ {{{\sin }^2}\alpha } &amp; 0 \cr 0 &amp; {{{\sin }^2}\alpha } \cr } } \right] - \left[ {\matrix{ {{1 \over 2}} &amp; 0 \cr 0 &amp; {{1 \over 2}} \cr } } \right] = \left[ {\matrix{ {{{\sin }^2}\alpha - {1 \over 2}} &amp; 0 \cr 0 &amp; {{{\sin }^2}\alpha - {1 \over 2}} \cr } } \right]$$<br><br>Given, $\left| {{A^2} - {1 \over 2}I} \right| = 0$<br><br>$$ \Rightarrow \left| {\matrix{ {{{\sin }^2}\alpha - {1 \over 2}} &amp; 0 \cr 0 &amp; {{{\sin }^2}\alpha - {1 \over 2}} \cr } } \right| = 0$$<br><br>$\Rightarrow {\left( {{{\sin }^2}\alpha - {1 \over 2}} \right)^2} = 0$<br><br>$$\Rightarrow {\sin ^2}\alpha = {1 \over 2} \Rightarrow \sin \alpha = {1 \over {\sqrt 2 }}, - {1 \over {\sqrt 2 }}$$<br><br>$\therefore$ $\alpha = {\pi \over 4}$