If A = 0 sin α sin α 0 and det A 2 - 1 2 I = 0 , then a possible value of α is

Step-by-step solution

A 2 = sin 2 α I So, A 2 - I 2 = sin 2 α - 1 2 2 = 0 ⇒ sin α = 1 2 ⇒ α = π 4