Let R = x 0 0 0 y 0 0 0 z be a non-zero 3 × 3 matrix, where x sin θ = y sin θ + 2 π 3 = z sin θ + 4 π 3 ≠ 0 , θ ∈ ( 0 , 2 π ) . For a square matrix M , let Trace M denote the sum of all the diagonal entries of M . Then, among the statements: I Trace ( R ) = 0 ( II ) If Trace ( adj ( adj ( R ) ) = 0 , then R has exactly one non-zero entry.

Step-by-step solution

Given, x sin θ = y sin θ + 2 π 3 = z sin θ + 4 π 3 ≠ 0 , ⇒ y = x sin θ sin θ + 2 π 3 , z = x sin θ sin θ + 4 π 3 Now, finding the trace of matrix we get, x + y + z = x + x sin θ sin θ + 2 π 3 + x sin θ sin θ + 4 π 3 ⇒ x + y + z = x sin θ + 2 π 3 sin θ + 4 π 3 + x sin θ sin θ + 4 π 3 + x sin θ sin θ + 2 π 3 sin θ + 2 π 3 sin θ + 4 π 3 ⇒ x + y + z = x sin θ + 2 π 3 sin θ + 4 π 3 + x sin θ sin θ + 4 π 3 + sin θ + 2 π 3 sin θ + 2 π 3 sin θ + 4 π 3 ⇒ x + y + z = x 2 · 2 sin θ + 2 π 3 sin θ + 4 π 3 + x sin θ 2 sin θ + π