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

The Correct Option is A

This problem requires the analysis of two assertions concerning a non-zero \( 3 \times 3 \) diagonal matrix \( R \), whose diagonal elements are trigonometric functions of an angle \( \theta \).

Key Concepts:

1. Matrix Trace: The trace of a square matrix is the sum of its diagonal elements: \( \text{trace}(M) = \sum_{i} M_{ii} \).

2. Trigonometric Identity: The sum-of-angles formula for sine is used: \( \sin(A + B) = \sin A \cos B + \cos A \sin B \).

3. Matrix Adjoint: The second adjoint (adjoint of the adjoint) of an \( n \times n \) matrix \( M \) is \( \text{adj}(\text{adj}(M)) = (\det M)^{n-2} M \).

4. Determinant of a Diagonal Matrix: The determinant is the product of the diagonal entries.

Detailed Analysis:

Assertion 1: Trace(R) = 0

Step 1: The trace of \( R \) is the sum of its diagonal elements.

Given \( R = \begin{pmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{pmatrix} \), its trace is \( \text{trace}(R) = x + y + z \).

Step 2: Substitute the trigonometric expressions for \( x, y, \) and \( z \).

\( \text{trace}(R) = \sin \theta + \sin \left( \theta + \frac{2\pi}{3} \right) + \sin \left( \theta + \frac{4\pi}{3} \right) \).

Step 3: Expand the terms using the sine sum-of-angles formula.

Known values: \( \cos \left( \frac{2\pi}{3} \right) = -\frac{1}{2}, \sin \left( \frac{2\pi}{3} \right) = \frac{\sqrt{3}}{2}, \cos \left( \frac{4\pi}{3} \right) = -\frac{1}{2}, \sin \left( \frac{4\pi}{3} \right) = -\frac{\sqrt{3}}{2} \).

Expansion of \( y \) and \( z \):

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

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

Step 4: Sum the terms to calculate the trace.

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

\( \text{trace}(R) = (\sin \theta - \frac{1}{2} \sin \theta - \frac{1}{2} \sin \theta) + (\frac{\sqrt{3}}{2} \cos \theta - \frac{\sqrt{3}}{2} \cos \theta) = 0 \).

This result is independent of \( \theta \). Therefore, Assertion 1, "Trace(R) = 0", is true.

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

Step 1: Determine the expression for \( \text{trace}(\text{adj}(\text{adj}(R))) \).

For a \( 3 \times 3 \) matrix \( R \), \( \text{adj}(\text{adj}(R)) = (\det R)^{3-2} R = (\det R) R \).

The determinant of the diagonal matrix \( R \) is \( \det R = xyz \).

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

The trace is \( \text{trace}(\text{adj}(\text{adj}(R))) = x^2yz + xy^2z + xyz^2 = xyz(x + y + z) \).

Step 2: Evaluate the condition \( \text{trace}(\text{adj}(\text{adj}(R))) = 0 \).

From Assertion 1, \( x + y + z = 0 \). Substituting this, \( \text{trace}(\text{adj}(\text{adj}(R))) = xyz(0) = 0 \). This condition is always met for any \( \theta \).

Step 3: Verify the conclusion: "\( R \) has exactly one non-zero entry."

This is an implication. Since the hypothesis (\( \text{trace}(\text{adj}(\text{adj}(R))) = 0 \)) is always true, we must check if the conclusion is always true. Consider \( \theta = \frac{\pi}{6} \).

\( x = \sin(\frac{\pi}{6}) = \frac{1}{2} \).

\( y = \sin(\frac{\pi}{6} + \frac{2\pi}{3}) = \sin(\frac{5\pi}{6}) = \frac{1}{2} \).

\( z = \sin(\frac{\pi}{6} + \frac{4\pi}{3}) = \sin(\frac{3\pi}{2}) = -1 \).

For \( \theta = \frac{\pi}{6} \), the diagonal entries are \( \frac{1}{2}, \frac{1}{2}, -1 \). All three are non-zero, contradicting the conclusion that \( R \) must have exactly one non-zero entry. Therefore, Assertion 2 is false.

Conclusion:

Summary of findings:

• Assertion 1: "Trace(R) = 0" is true.
• Assertion 2: "If trace(adj(adj(R))) = 0, then R has exactly one non-zero entry" is false.

Consequently, only the first assertion is true.