JEE Mathematics 2024 Question with Solution

Given: $R = \operatorname{diag}(x,y,z)$ where

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

Find: Which of the statements (I) and (II) is true.

For statement (I),

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

Using

$$\sin(A+B) = \sin A \cos B + \cos A \sin B$$

we get

$$\sin\left(\theta + \frac{2\pi}{3}\right) = -\frac{1}{2}\sin\theta + \frac{\sqrt{3}}{2}\cos\theta$$

and

$$\sin\left(\theta + \frac{4\pi}{3}\right) = -\frac{1}{2}\sin\theta - \frac{\sqrt{3}}{2}\cos\theta$$

Therefore,

$$\operatorname{trace}(R) = \sin\theta + \left(-\frac{1}{2}\sin\theta + \frac{\sqrt{3}}{2}\cos\theta\right) + \left(-\frac{1}{2}\sin\theta - \frac{\sqrt{3}}{2}\cos\theta\right) = 0$$

So statement (I) is true.

Now consider statement (II). For a $3 \times 3$ matrix,

$$\operatorname{adj}(\operatorname{adj}(R)) = (\det R)R$$

Since $R$ is diagonal,

$$\det R = xyz$$

Hence,

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

Thus,

$$\operatorname{trace}(\operatorname{adj}(\operatorname{adj}(R))) = x^2yz + xy^2z + xyz^2 = xyz(x+y+z)$$

But from statement (I),

$$x+y+z=0$$

So,

$$\operatorname{trace}(\operatorname{adj}(\operatorname{adj}(R))) = 0$$

for all $\theta$.

To test the conclusion in statement (II), take

$$\theta = \frac{\pi}{6}$$

Then,

$$x = \sin\frac{\pi}{6} = \frac{1}{2}, \quad y = \sin\frac{5\pi}{6} = \frac{1}{2}, \quad z = \sin\frac{3\pi}{2} = -1$$

All three diagonal entries are non-zero, so $R$ does not have exactly one non-zero entry.

Therefore statement (II) is false.

Hence only statement (I) is true. The solution concludes "The Correct Option is A", but this conflicts with the option list, where "Only (I) is true" is option D. Based on the worked solution and the given options, the defensible answer is D.

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

Find: Which statement is correct.

A quick identity gives

$$x+y+z = \sin\theta + \sin\left(\theta + \frac{2\pi}{3}\right) + \sin\left(\theta + \frac{4\pi}{3}\right) = 0$$

So statement (I) is immediately true.

Also,

$$\operatorname{trace}(\operatorname{adj}(\operatorname{adj}(R))) = xyz(x+y+z)$$

Hence this trace is automatically $0$ because $x+y+z=0$. That does not force exactly one of $x,y,z$ to be non-zero.

So statement (II) is false.

Therefore only statement (I) is true, so the correct option is D.