Let $R = \begin{bmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{bmatrix}$ be a non-zero $3 \times 3$ matrix, <br> where $x \sin \theta = y \sin \left( \theta + \frac{2\pi}{3} \right) = z \sin \left( \theta + \frac{4\pi}{3} \right) \neq 0, \theta \in (0, 2\pi)$. For a square matrix $M$, let trace $(M)$ <br> denote the sum of all the diagonal elements of $M$. Then, among the statements: <br> (I) Trace $(R) = 0$ <br> (II) If trace $(\text{adj}(\text{adj}(R))) = 0$, then $R$ has exactly one non-zero entry.
Step-by-Step Solution
Key Concept: Use the property that x sin(theta) = y sin(theta + 2pi/3) = z sin(theta + 4pi/3) = k implies x = k/sin(theta), y = k/sin(theta + 2pi/3), z = k/sin(theta + 4pi/3). The sum x+y+z = k(1/sin(theta) + 1/sin(theta+2pi/3) + 1/sin(theta+4pi/3)) = 0. For adj(adj(R)), if R is diagonal with entries x, y, z, then adj(R) is diagonal with entries yz, xz, xy, and adj(adj(R)) is diagonal with entries (xz)(xy), (yz)(xy), (yz)(xz) = x^2yz, xy^2z, xyz^2. Trace is xyz(x+y+z) = 0. Since x, y, z are non-zero, trace is 0, but R has three non-zero entries.
Given $x \sin \theta = y \sin \left( \theta + \frac{2\pi}{3} \right) = z \sin \left( \theta + \frac{4\pi}{3} \right) = k \neq 0$. <br> Then $x = \frac{k}{\sin \theta}, y = \frac{k}{\sin(\theta + 2\pi/3)}, z = \frac{k}{\sin(\theta + 4\pi/3)}$. <br> Trace $(R) = x+y+z = k \left( \frac{1}{\sin \theta} + \frac{1}{\sin(\theta + 2\pi/3)} + \frac{1}{\sin(\theta + 4\pi/3)} \right)$. <br> Using the identity $\frac{1}{\sin \theta} + \frac{1}{\sin(\theta + 2\pi/3)} + \frac{1}{\sin(\theta + 4\pi/3)} = 0$, we get Trace $(R) = 0$. So (I) is true. <br> For $R = \text{diag}(x, y, z)$, $\text{adj}(R) = \text{diag}(yz, xz, xy)$. <br> $\text{adj}(\text{adj}(R)) = \text{diag}(x^2yz, xy^2z, xyz^2) = xyz \cdot \text{diag}(x, y, z) = xyz R$. <br> Trace $(\text{adj}(\text{adj}(R))) = xyz(x+y+z) = xyz(0) = 0$. <br> Since $x, y, z \neq 0$, $R$ has three non-zero entries. Thus (II) is false.
Correct Answer: 2