Question

Let $R = \begin{pmatrix} x & 0 & 0 \\\ 0 & y & 0 \\\ 0 & 0 & z \end{pmatrix} \text{ be a non-zero } 3 \times 3 \text{ matrix, where}$

$x = \sin \theta, \quad y = \sin \left( \theta + \frac{2\pi}{3} \right), \quad z = \sin \left( \theta + \frac{4\pi}{3} \right)$

and $\theta \neq 0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi$. For a square matrix $M$, let $\text{trace}(M)$ denote the sum of all the diagonal entries of $M$. Then, among the statements:

1. $\text{Trace}(R) = 0$
2. If $\text{trace}(\text{adj}(\text{adj}(R))) = 0$, then $R$ has exactly one non-zero entry.

Which of the following is true?

• A. Only(I) is true
• B. Only (II) is true
• C. Neither (I) nor (II) is true
• D. Both (I) and (II) are true

Answer 1

Answer: (A)

Only(I) is true

Answer 2

Calculate the trace of $R$: Since $x + y + z = \sin \theta + \sin \left( \theta + \frac{2\pi}{3} \right) + \sin \left( \theta + \frac{4\pi}{3} \right) = 0$, we have:

$\text{trace}(R) = x + y + z = 0.$

Thus, statement (I) is true.

Examine statement (II): $\text{adj}(R) = \begin{pmatrix} yz & 0 & 0 \\\ 0 & xz & 0 \\\ 0 & 0 & xy \end{pmatrix}$. Therefore,

$\text{adj}(\text{adj}(R)) = \begin{pmatrix} x^2yz & 0 & 0 \\\ 0 & xy^2z & 0 \\\ 0 & 0 & xyz^2 \end{pmatrix}.$

The trace of $\text{adj}(\text{adj}(R))$ is $xyz(x + y + z) = 0$, even if $R$ has more than one non-zero entry.

Thus, statement (II) is false.