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

Question

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

Accepted Answer

Answer: A. {\pi \over 4} {A^2} = \left[ {\begin{matrix} 0 & {\sin \alpha } \ {\sin \alpha } & 0 \ \end{matrix} } ight]\left[ {\begin{matrix} 0 & {\sin \alpha } \ {\sin \alpha } & 0 \ \end{matrix} } ight] = \left[ {\begin{matrix} {{{\sin }^2}\alpha } & 0 \ 0 & {{{\sin }^2}\alpha } \ \end{matrix} } ight] {A^2} - {1 \over 2}I = \left[ {\begin{matrix} {{{\sin }^2}\alpha } & 0 \ 0 & {{{\sin }^2}\alpha } \ \end{matrix} } ight] - \left[ {\begin{matrix} {{1 \over 2}} & 0 \ 0 & {{1 \over 2}} \ \end{matrix} } ight] = \left[ {\begin{matrix} {{{\sin }^2}\alpha - {1 \over 2}} & 0 \ 0 & {{{\sin }^2}\alpha - {1 \over 2}} \ \end{matrix} } ight] Given, \left| {{A^2} - {1 \over 2}I} ight| = 0 \Rightarrow \left| {\begin{matrix} {{{\sin }^2}\alpha - {1 \over 2}} & 0 \ 0 & {{{\sin }^2}\alpha - {1 \over 2}} \ \end{matrix} } ight| = 0 \Rightarrow {\left( {{{\sin }^2}\alpha - {1 \over 2}} ight)^2} = 0 \Rightarrow {\sin ^2}\alpha = {1 \over 2} \Rightarrow \sin \alpha = {1 \over {\sqrt 2 }}, - {1 \over {\sqrt 2 }} 	herefore \alpha = {\pi \over 4}

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

Answer

Explanation

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