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.

Solution

Maths sir · 4 steps

Write x, y, z as k over a sine. The sines add to zero, but their reciprocals do not.

1. All entries non-zero. Call the common value k ≠ 0, so x = k/(sin θ), y = k/sin(θ + 2π/3), z = k/sin(θ + 4π/3). Since k ≠ 0, none of the three sines is zero, and all three of x, y, z are non-zero.
2. Sine identities. Let a, b, c be the three sines. Then a + b + c = 0 and a² + b² + c² = 3/2, so ab + bc + ca = (0 - 3/2)/2 = -3/4. Also abc = -1/4 sin 3θ, which is non-zero because no sine is zero.

   x + y + z = k(1/a + 1/b + 1/c) = k (ab + bc + ca)/abc = k (-3/4)/(-1/4 sin 3θ) = 3k/(sin 3θ) ≠ 0

3. Check statement II. For a 3 × 3 matrix, adj(adj R) = |R| R, so its trace is xyz(x + y + z). Both factors are non-zero, so this trace can never be 0 for this R. Statement II also needs R to have exactly one non-zero entry, but R always has three. II does not hold for this R, and the answer key marks it false.
4. So. Trace(R) = 3k/(sin 3θ) ≠ 0, so (I) is false, and (II) is false too. The answer is (d).

The idea to keep. When several products like x sin θ are equal, set them to k and write each variable as k over its sine. Also remember: for a 3 × 3 matrix, adj(adj M) = |M| M.

Answer from the NTA final answer key. Solution by PrepWiser.