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 us to analyze two statements regarding a non-zero \( 3 \times 3 \) diagonal matrix \( R \). The diagonal entries of \( R \) are given as trigonometric functions of an angle \( \theta \).

Concept Used:

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

2. Trigonometric Identities: We will use the sum-of-angles formula for sine:

\[ \sin(A + B) = \sin A \cos B + \cos A \sin B \]

3. Adjoint of a Matrix: For an \( n \times n \) matrix \( M \), the second adjoint (adjoint of the adjoint) is given by the formula:

\[ \text{adj}(\text{adj}(M)) = (\det M)^{n-2} M \]

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

Step-by-Step Solution:

Analysis of Statement 1: Trace(R) = 0

Step 1: Write down the expression for the trace of matrix \( R \).

The matrix is \( R = \begin{pmatrix} x & 0 & 0 \\ 0 & y & 0 \\ 0 & 0 & z \end{pmatrix} \). The trace is the sum of its diagonal entries:

\[ \text{trace}(R) = x + y + z \]

Step 2: Substitute the given trigonometric definitions 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.

We know the values:

• \( \cos \left( \frac{2\pi}{3} \right) = -\frac{1}{2}, \quad \sin \left( \frac{2\pi}{3} \right) = \frac{\sqrt{3}}{2} \)
• \( \cos \left( \frac{4\pi}{3} \right) = -\frac{1}{2}, \quad \sin \left( \frac{4\pi}{3} \right) = -\frac{\sqrt{3}}{2} \)

Expanding \( y \) and \( z \):

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

Step 4: Sum the terms to find the trace.

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

This result holds for any value of \( \theta \). Therefore, the statement "Trace(R) = 0" is true.

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

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

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

Since \( R \) is a diagonal matrix, its determinant is the product of its diagonal elements:

\[ \det R = xyz \]

So, the second adjoint is:

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

The trace of this matrix 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 our analysis of Statement 1, we know that \( x + y + z = 0 \). Substituting this into the trace expression:

\[ \text{trace}(\text{adj}(\text{adj}(R))) = xyz(0) = 0 \]

This shows that the hypothesis of Statement 2, \( \text{trace}(\text{adj}(\text{adj}(R))) = 0 \), is always true for any \( \theta \).

Step 3: Check the conclusion of Statement 2: "R has exactly one non-zero entry."

The statement is an implication: If A is true, then B must be true. We have shown A is always true. Now we must check if B is always true. Let's test a value for \( \theta \) that is allowed by the problem conditions, e.g., \( \theta = \frac{\pi}{6} \).

\[ x = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \] \[ y = \sin\left(\frac{\pi}{6} + \frac{2\pi}{3}\right) = \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2} \] \[ z = \sin\left(\frac{\pi}{6} + \frac{4\pi}{3}\right) = \sin\left(\frac{3\pi}{2}\right) = -1 \]

For this value of \( \theta \), the matrix \( R \) has diagonal entries \( \frac{1}{2}, \frac{1}{2}, -1 \). All three entries are non-zero. This provides a counterexample to the conclusion that \( R \) must have exactly one non-zero entry.

Since the hypothesis of the implication is always true but the conclusion is not always true, the implication itself is false.

Final Computation & Result:

Based on the analysis:

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

Therefore, only the first statement 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.